Class 9 – Page 3 – Haryana Board Solutions

HBSE 9th Class Social Science Solutions Civics Chapter 1 समकालीन विश्व में लोकतंत्र

October 16, 2023 November 10, 2023 / Class 9 / By Prasanna

Haryana State Board HBSE 9th Class Social Science Solutions Civics Chapter 1 समकालीन विश्व में लोकतंत्र Textbook Exercise Questions and Answers.

Haryana Board 9th Class Social Science Solutions Civics Chapter 1 समकालीन विश्व में लोकतंत्र

HBSE 9th Class Civics समकालीन विश्व में लोकतंत्र Textbook Questions and Answers

प्रश्न 1.
निम्नलिखित विकल्पों में सही विकल्प का चयन कीजिए।
(क) लोगों को संघर्ष
(ख) विदेशी शासन द्वारा आक्रमण .
(ग) उपनिवेशवाद का अंत
(घ) लोगों की स्वतंत्रता की चाह
उत्तर-
(ख) विदेशी शासन द्वारा आक्रमण .

2. आज की दुनिया के बारे में इनमें से कौन-सा कथन सही है?
(क) राजशाही शासन की वह पद्धति है जो अब समाप्त हो गई हैं।
(ख) विभिन्न देशों में बीच संबंध पहले के किसी वक्त से अब नहीं ज्यादा लोकतांत्रिक हैं।
(ग) आज पहले के किसी दौर से ज्यादा देशों में शासकों का चुनाव लोगों के द्वारा हो रहा है।
(घ) आज दुनिया में सैनिक तानाशाह नहीं रह गए हैं।
उत्तर-
(ग) आज पहले के किसी दौर से ज्यादा देशों में शासकों का चुनाव लोगों के द्वारा हो रहा है।

3. निम्नलिखित वाक्यांशाके में से किसी एक का चुनाव करके इस वाक्य को पूरा कीजिए। अंतर्राष्ट्रीय संस्थाओं में लोकतंत्र की जरूरत है ताकि…….
(क) अमीर देशों की बातों का ज्यादा वजन हो।
(ख) विभिनन देशों की बात का वजन उनकी सैन्य शक्ति के अनुपात में हो।
(ग) देशों को उनकी आबादी के अनुपात में समान मिले।
(घ) दुनिया के सभी देशों के साथ समान व्यवहार हो।
उत्तर-
(घ) दुनिया के सभी देशों के साथ समान व्यवहार हो।

4. इन देशों और लोकतंत्र की उनकी राह में मेल बैठाएँ
I. देश — II. लोकतंत्र की ओर
(क) चिले — 1. ब्रिटिश औपनिवेशिक शासन से आजादी
(ख) नेपाल — 2. सैनिक तानाशाही की समाप्ति ।
(ग) पोलैंड — 3. एक दल के शासन का अंत
(घ) घाना — 4. राजा ने अपने अधिकार छोड़ने पर सहमति दी।
उत्तर-
(क-2, ख-4, ग-3, घ-1)

प्रश्न 5.
गैर-लोकतांत्रिक शासन वाले देशों के लोगों को किन-किन मुश्किलों का सामना करना पड़ता है? इस अध्याय में दिए गए उदाहरणों के आधार पर इस कथन मे पक्ष में तर्क दीजिए।
उत्तर-

ऐसे लोगों को मूल अधिकारों से वंचित किया जात है।
उन पर हुए अत्याचारों के विरुद्ध उन्हें आवाज उठाने की आज़ादी नहीं होती।
वह अपना विरोध व्यक्त नहीं कर सकते।
अपनी शिकायतों को ज़ाहिर करने के लिए उन्हें संघ-समुदाय बनाने की अनुमति नहीं होती।
उन्हें स्वतंत्रताएँ प्राप्त नहीं होती।

प्रश्न 6.
जब सेना लोतंत्र को उखाड़ फेंकती हैं, तो सामान्यतः कौन-सी स्वतंत्रताएँ छीन ली जाती हैं?
उत्तर-
जब सेना लोकतंत्र को उखाड़ फेंकती है तो सामान्यतः लोगों की सभी स्वतंत्रताएँ छीन ली जाती हैं। वह अपने विचार नहीं रख सकते, उन विचारों की अभिव्यक्ति नहीं कर सकते। अपने संघ-समुदाय नहीं बना सकते तथा गतिविधियों व आन्दोलन के लिए एकत्रित नहीं हो सकते।

प्रश्न 7.
वैश्विक स्तर पर लोकतंत्र बढ़ाने में किन बातों में मदद मिलेगी? प्रत्येक मामले में अपने जवाब के पक्ष में तर्क दीजिए।
(क) मेरा देश अंतर्राष्ट्रीय संस्थाओं को ज्यादा पैसे देता है इसलिए मैं चाहता हूँ कि मेरे साथ ज्यादा सम्मानजनक व्यवहार हो और मुझे ज्यादा अधिकार मिलें।
(ख) मेरा देश छोटा या गरीब हो सकता है लेकिन मेरी आवाज को समान आदर के साथ सुना जाना चाहिए क्योंकि इन फैसलों का मेरे देश पर भी असर होगा।
(ग) अंतर्राष्ट्रीय मामलों में अमीर देशों की ज्यादा चलनी चाहिए। गरीब देशों की संख्या ज्यादा है, सिफ, इसके चलते अमीर देश अपने हितों का नुकसान नहीं होने दे सकते।
(घ) भारत जैसे बड़े देशों की आवाज का अंतर्राष्ट्रीय संगठनों में ज्यादा वज़न होना ही चाहिए।
उत्तर-
(क) एक ऐश द्वारा अन्तर्राष्ट्रीय संस्थाओं को अधिक धन देने का यह अर्थ नहीं है कि उसे अन्य देशों की अपेक्षा अधिक समान प्राप्त हो तथा दूसरे देशों को उस देश की अपेक्षा कम अधिकार प्राप्त हों। लोकतंत्र धन के बलबूते पर नहीं पनपने चाहिए और न ही ऐसी व्यवस्था में धनियों का शासन हो।
(ख) एक देश छोटा व निर्धन देश हो सकता है। लोकतंत्र के स्वच्छ संचालन के लिए सभी देशों को (छोटे-बड़े, अमीर-अनर्धन आदि) समान व्यवहार मिलना चाहिए। लोकतंत्र में निर्णय सभी देशों द्वारा समान रूप से किए जाने चाहिएँ।
(ग) यदि अमीर देशों को अन्तर्राष्ट्रीय मामलों में अधिक महत्त्व मिलता है तथा उनकी बात अधिक सुनी जाती है तो विश्व मंच पर वह अपवने हितों को प्रोत्साहित करेंगे। यह प्रवृत्ति लोकतंत्र को मज़बूत नहीं करती, अपितु छति पहुँचाती हैं।
(घ) वह देश जो जनंसख्या तथा आकार में बड़े देश – हैं जैसे भारत, ऐसे देशों को आनुपातिक आधार पर प्रतिनिधित्व मिलना चाहिए। मिल ने कहा था कि लोकतंत्र में आनुपातिक आधार पर प्रतिनिधित्व प्राप्त हो, गैर-आनुपातिक आधार पर नहीं।

प्रश्न 8.
नेपाल के संकट पर हुई एक टीवी चर्चा में व्यक्त किए गए तीन विचार कुछ इस प्रकार के थे। इनमें से आप किसे सही मानते हैं और क्यों?
वक्त-1: भारत एक लोकतांत्रिक देश है इसलिए राजशाही के खिलाफ और लोकतंत्र के लिए संघर्ष करने वाले नेपाली लोगों के समर्थन में भारत सरकार को ज्यादा दखल देना चाहिए।
वक्ता 2: यह एक खतरनाक तर्क है। हम उस स्थिति में पहुँच जाएँगे जहाँ इराक के मामले में अमेरिका पहुँचा है। किसी भी बाहरी शक्ति के सहारे लोकतंत्र नहीं आ सकता।
वक्ता-3: लेकिन हमें किसी देश के आंतरिक मामलों की चिंता ही क्यों करनी चाहिए? हमें वहाँ अपने व्यावसायिक हितों की चिंता करनी चाहिए लोकतंत्र की नहीं।
उत्तर-
लोकतंत्र थोपा नहीं जा सकता, थोपा जाना चाहिए भी नहीं। जब लोकतंत्र को थोपा जाता है जैसाकि अमेरिका ने इराक में करने का प्रयास किया है।, यह लोकतंत्र थोपने का प्रयास है भारत सहित अन्य देशों का यह यत्न होना चाहिए कि लोकतंत्र ऊपर से थोपा नहीं जाना चाहिए। कोई किसी को तैरना सिखा सकता है, परन्तु यदि कोई तैरना सीखता ही नहीं चाहता, तो कोई क्या कर सकता है, तीसरे वक्ता के विचार अधिक वज़नी हैं। हमें अन्य देशों में अपने हित सुरक्षित करने चाहिएँ, परंतु अपने हितों के बदले उन्हें लोकतांत्रिक नहीं बनाना चाहिए।

प्रश्न 9.
एक काल्पनिक देश आनंदलोक में लोग विदेशी शासन को समाप्त करके पुराने राजपरिवार को सत्ता सौपते हैं। वे कहते हैं, ‘आखिर ज विदेशियों ने हमारे ऊपर राज करना शुरू किया तब इन्ही के पूर्वज हमारे राजा थे। यह अच्छा है कि हमारा एक मजबूत शासक है जो हमें अमीर और ताकतवर बनने में मदद कर सकता है।’ जब किसी ने लोकतांत्रिक शासन व्यवस्था की बात की तो वहाँ के सयाने लागों ने कह कि यह तो एक विदेशी विचार है। हमारी लड़ाई विदेशियों और उनके विचारों को देश से खदेड़ने की थी। जब किसी ने मीडिया की आज़ादी की माँग की तो बड़े-बुजुर्गों ने कहा कि शासन की ज्यादा आलोचना करने से नुकसान होगा और इससे अपने जीवन स्तर को सुधारने में कोई मदद नहीं मिलेगी। “आखिर महाराज दयावान हैं और अपनी पूरी प्रजा के कल्याण में बहुत दिलचस्पी लेते हैं। उनके लिए मुश्किलें क्यों पैदा की जाएँ? क्या हम सभी खुशहाल नहीं होना चाहते?”
उपरोक्त उद्धरण को पढ़ने के बाद चयन, चंपा और चंदू ने कुछ इस तरह के निष्कर्ष निकाले :
चमन-:-आनंदलोक एक लोकतांत्रिक देश है क्योंकि लोगों ने विदेशी शासकों का उखाड़ फेका औरा राजा का शासन बहाल किया।
चंपा-:-आनंदलोक लोकतांत्रिक देश नही हैं क्योंकि लोग अपने शासन की आलोचना नहीं कर सकते। राजा अच्छा हो सकता है और आर्थिक समृद्धि भी जा सकता है लेकिन राजा लोकतांत्रिक शासन नहीं ला सकता।
चंदू-:-लोगों की खुशहाली चाहिए इसलिए वे अपने शासन को अपनी तरफ से फैसले लेने देना चाहते हैं। अगर लोग खुश हैं तो वहाँ का शासन लोकतांत्रिक ही है। . इन तीनों कथनों के बारे में आपकी क्या राय है? इस देश में सरकार के स्वरूप के बारे में आपकी राय
उत्तर-
लोकतंत्र का अर्थ, लोगों का शासन, लोगों द्वारा तथा लोगों के लिए। एक गुलाम देश कभी स्वतंत्र देश नहीं होता। राष्ट्रीय स्वतंत्रता वहाँ होती है जहाँ लोग विदेशी शासन से मुक्त होते हैं। अंग्रेजों से मुक्ति तथा देश की स्वतंत्रता लोकतंत्र के साथ जुड़े विचार थे। यदि एक देश जब विदेशी ताकत से मुक्त हो जाता है तथा बाद में राजतंत्रीय व्यवस्था को अपना लेता है, तो यह लोकतंत्र नहीं है, क्योंकि राजतंत्र लोकतंत्र नहीं होता।।

वास्तव मे जहाँ शासक लोगों द्वारा आलोचना के दायरे में नहीं आते अर्थात् लोगों को अपने शासकों की आलोचना का अधिकार नहीं होता, वहाँ लोकतंत्र नहीं होता। लोकतंत्र का सार यह है कि वहाँ शासकीय अधिकार अंततः लोगों के पास हों, वह अपने शासकों की आलोचना कर सकते हों, चुनावों में उन्हें बदल सकते हों।

लोकतंत्र तथा सुख एक नहीं होते। ज़रूरी नहीं कि ये सुखी व्यक्ति लोकतांत्रिक व्यक्ति भी हो और कि एक लोकतांत्रिक व्यक्ति सुखी व्यक्ति भी हो। यह अलग बात हैं कि एक स्वदल अर्थव्यवस्था लोकतंत्र को सुदृढ़ करने में विशेष भूमिका निभा सकती है तथा निभाती भी है। लोकतंत्र व अर्थव्यवस्था एक-दूसरे के पूरक है।

एक देश जहाँ राज्य अध्यक्ष कोई सम्राट हो तथा वहाँ राजतंत्र हो, तो यह लोकतंत्र नहीं है। यदि सम्राट मात्र एक संवैधानिक मुखिया है जैसा कि ब्रिटेन में हैं, वहाँ लोकतंत्रीय व्यवस्था हो सकती हैं।

HBSE 9th Class Civics समकालीन विश्व में लोकतंत्र Important Questions and Answers

प्रश्न 1.
आयेंदे का सम्बन्ध किस देश से था?
उत्तर-
चिले से। वे चिले के राष्ट्रपति थे।

प्रश्न 2.
आयेंदे की सरकार का कब तख्ता पलट हुआ था?
उत्तर-
11 सितम्बर, 1973 को।

प्रश्न 3.
आयेंदे को चिले का राष्ट्रपति कब बनाया गया था? .
उत्तर-
1970 में।

प्रश्न 4.
1970 में चिले में किस राजनीतिक दल के पास सत्ता था?
उत्तर-
पापुलर यूनिटी. नातक गठबंधन के पास सत्ता थी।

प्रश्न 5.
चिले में तख्ता पलट के पश्चात् आयेंदे के बाद किसके पास सत्ता आयी थी?
उत्तर-
जनरल ऑगस्तों पिनोशे के हाथों में सत्ता आयी थी।

प्रश्न 6.
कालामा कहाँ स्थित हैं?
उत्तर-
चिले की राजधानी संटियागों से हजार मील दूर।

प्रश्न 7.
कालामा की स्त्रियों ने अपने दःख का इजहार कैसे किया था?
उत्तर-
चुप रह कर, सदैव चुप रह कर।

प्रश्न 8.
आपके देश में कौन-सा राज्य है जो आकार में चिले से मिलता-जुलता हैं?
उत्तर-
केरल।

प्रश्न 9.
जैसा कि चिले में महिलाओं के साथ हुआ, आप संसार के किसी अन्य देश में ऐसे हुए व्यवहार के बारे में जानते हो?
उत्तर-
जारशाही के रूस में महिलाओं के साथ ऐसा कुछ व्यवहार हुआ था।

प्रश्न 10.
कालामा में हुए महिलाओं के बारे में वहाँ के समाचार पत्रों ने क्यों नहीं लिखा/प्रकाशित किया?
उत्तर-
तब समाचार पत्र सरकार के नियंत्रण में थे। इस कारण उनमें महिलाओं के विषय में कुछ प्रकाशित नहीं हो पाया था।

प्रश्न 11.
आज चिले के राष्ट्रपति कौन हैं?
उत्तर-
मिशेल बैशले (जनवरी, 2006)।

प्रश्न 12.
1980 में पोलैंड में कौन-सा राजनीतिक दल शासन करता था?
उत्तर-
पॉलिश यूनाइटिड वर्कर्स पार्टी। वहाँ तक एक-दलीय शासकीय व्यवस्था थी।

प्रश्न 13.
गोलसंक में 1980 में किस फैक्ट्री में हड़ताल हुई थी?
उत्तर-
लेनिन शिपयार्ड।

प्रश्न 14.
1980 में पोलैंड में जिस व्यक्ति ने हड़ताल में प्रवेश किया था, उसका नाम बताइए।
उत्तर-
लेक वालेशा।

प्रश्न 15.
पोलैंड को घेरे कुछ देशों के नाम बताइए।
उत्तर-
जर्मनी, लुथआनिया, बेलारूस, यूक्रेन।

प्रश्न 16.
पोलैण्ड के अतिरिक्त 1980 में उन दो देशों का नाम बताइए। जहाँ साम्यवादी दल का शासन था?
उत्तर-
बुल्गारिया तथा हंगरी।

प्रश्न 17.
पोलैंड में स्वतंत्र सजदूर दल के गठन की आवश्यकता क्यों थी?
उत्तर-
तब तक वहाँ सरकार के अधीन ही मजदूर दलों का गठन होता था जो सरकार की नीतियों का ही प्रचार करती थी।

प्रश्न 18.
इंग्लैंड में शानदारी क्रांति कब घटी थी?
उत्तर-
1688 में।

प्रश्न 19.
किस वर्ष अमेरिका के तेरह उपनिवेशों – ने अपना स्वतंत्रता संग्राम लड़ा था?
उत्तर-
1776 में।

प्रश्न 20.
अमरीकी स्वतंत्रता संग्राम किस देश के विरुद्ध लड़ा गया था?
उत्तर-
इंग्लैंड के विरुद्ध।

प्रश्न 21.
उन प्रयासों का वर्णन कीजिए जो यह बताएँ कि गरीबों की सहायता के लिए आयेंदे सरकार ने अनुकूल कदम उठाए हों?
उत्तर-

शिक्षा प्रणाली में सुधार।
बच्चों को मुफ्त दूध की आपूर्ति।
किसानों में भूमि का पुनः वितरण।

प्रश्न 22.
दो कारण बताएं जिनस यह मालूम हा कि चिले मे आयोंदे की सरकार लोकप्रिय दिखायी देती थी?
उत्तर-

आयोंदे सरकार उन विदेशी ताकतों का विरोध करती थी जो चिले के प्राकृतिक संसाधनों का शोषण कर रही थीं।
उसकी सरकार अमीरों का भी विरोध कर रही थी जो गरीबों के हितों की अनेदखी करते थे।

प्रश्न 23.
1973 में आयेंदे को हराकर पिनोशे शासन ने क्या किया?
उत्तर-

पिनोशे सरकार ने जनसाधारण पर अत्याचार करने शुरू किए, विशेष रूप से उन पर आयोंदे का समर्थन करते थे।
पिनोशे सैनिकों ने लगभग दो हजार व्यक्तियों को मार दिया; हज़ारों लापता हो गए।
लोगों से सभी स्वतंत्रताएँ व अधिकार छीन लिये

प्रश्न 24.
बताइए कि कालामा की महिलाएँ व बच्चे चुप करा दिए गए। इन घटनाओं क विरुद्ध लोगों ने प्रतिक्रिया क्यों नहीं दिखायी थी? ..
उत्तर-
कालामा की महिलाएं व बच्चों को चुप करा दिया गया। लोगों ने इसके विरुद्ध आवाज उठाई कि सैनिक शासन अत्याचारी शासन था तथा लोग उस शासन के अत्याचार से डरते थे।

प्रश्न 25.
1980 में पोलैंड में कौन शासन करता था?
उत्तर-
1980 में पोलैंड में साम्यवादी दल का शासन था। इस दल का नाम था पॉलिश यूनाटिड वर्कर्स पार्टी। यह एक दलीय व्यवस्था की। सभी शासकीय ताकतें इसी दल । के हाथों में थीं। सरकार का पूरी अर्थव्यवस्था पर नियंत्रण था। सरकार ही मज़दूरों के संघों पर नियंत्रण रखती थी। मज़दूर संघ सरकार व दल से अलग नहीं थे।

प्रश्न 26.
गंदान्सक के मजदूरों की क्या माँगें थीं?
उत्तर-
1980 के आसपास गंदान्सक में लेमिन शिपयार्ड में हड़ताल हो गयी। मजदूरों की माँगें निम्नलिखित थीं:

सभी मजदूरों को जिन्हें निकाल दिया गया था, वापस लिया जाए।
मजदूरों ने स्वतंत्र मज़दूर संगठन बनाने के अधिकार की माँग की।
राजनीतिक प्रक्रिया को अंकुश-मुक्त किया जाए।
समाचार पत्रों पर से सैंसरशिप हटायी जाए।

प्रश्न 27.
सरकार व वालेश के बीस हुए समझौते की दो बातें बताइए।
उत्तर-
पॉलिश सरकार तथा वालेशा के नेतृत्व के बीच हुए मजदूरों के बीच हड़ताल के समाप्त होने पर समझौते की दो बातें निम्नलिखित बतायी जा सकती हैं-

मजदूरों को स्वतंत्र मज़दूर संघ बनाने का अधिकार मिल गया;
मजदूरों हड़ताल करने का अधिकार मिल गया।

प्रश्न 28.
पॉलिश यूनारिड वर्कर्स पार्टी की सरकार क्यों कमज़ोर पड़ने लग गयी?
उत्तर-
पॉलिश यूनारिड बर्कर्स पार्टी की पॉलिश सरकार के कमजोर पड़ने के कारणों में निम्नलिखित का उल्लेख किया जा सकता है।

वालेशा की सोलिडेरिटी की सदस्या संख्या एक करोड़ तक पहुँच गयी।
सरकार भ्रष्टाचार की ओर बढ़ने लगी। उसे डर पैदा हो गया कि वालेशा की सोलिडेरिटी उन पर हावी हो जाएगी। घबराहट में सरकार ने सैनिक कानून लागू कर दिया।
सरकार ने अत्याचार आरंभ कर दिए। वालेशा के लोगों को जेल में डाल दिया; उनकी स्वतंत्रताएँ वापस ले लीं।
अर्थव्यवस्था में तंज़ी से गिरावट आने लगी। सरकार वितीय संकट में ग्रस्त होती चली गयी।

प्रश्न 29.
लेक वालेशा ने पोलैण्ड में किस प्रकार सत्ता प्राप्त की?
उत्तर-
1988 में सोलिडेरिटी ने फिर से हड़तालें करवाई और लेक वालेशा ने इनकी अगुवाई की। इस समय पोलैंड की सरकार पहले से कमजोर थी, सोवियत संघ से मदद का भी पहले जैसा भरोसा न था और अर्थव्यवस्था में तेजी से – गिरावट आ रही थी। लेक वालेशा के साथ समझौता-वार्ता का एक और दौर चला और अप्रैल 1989 में जो समझौता हुआ उसमें स्वतंत्र चुनाव कराने की माँग मान ली गई। सोलिडेरिटी ने सीनेट के सभी 100 सीटों के लिए चुनाव लड़ा और उसे 99 सीटों पर सफलता मिली। अक्तूबर 1990 में पोलैंड मे राष्ट्रपति पद के लिए पहली बार चुनाव हुए जिसमें एक से ज्यादा दल हिस्सा ले सकते थे। लेक वालेशा को पोलैंड का राष्ट्रपति चुना गया।

प्रश्न 30.
पोलैण्ड में सोलिडेरिटी के लोकप्रिय होने के कारणों का वर्णन कीजिए।
उत्तर-

सोलिडेरिटी एक ऐसा संगठन था जो मज़दूरों के हितों की प्राप्ति के लिए गठित किया गया था।
यह मजदूरों के अधिकारों के लिए संघर्षरत रही।
लोगों के विचारों की अभिव्यक्ति के लिए भी सोलिडेरिटी लड़ रही थी।
बाद के दिनों में यह पार्टी लोकतंत्र की बहाली के लिए लड़ने लगी थी।

प्रश्न 31.
चिले की पिनोशे सरकार तथा पोलैण्ड के साम्यवादी दल के शासन में अंतर व समानताएँ बताइए।
उत्तर-
जिले में पिनोशे शासन और पोलैंड के साम्यवादी शासन में काफी अंतर है। चिले में सैनिक तानाशाह का राज था। जबकि पोलैंड में एक पार्टी का राज था। पोलैंड की सरकार का दावा था कि वह मजदूर वर्ग की ओर से शासन चला रही है। पिनोशे ने ऐसा कोई दावा नहीं किया और खुलेआम बड़े पूंजीपतियों को लाभ पहुँचाया। इन असमानताओं के बावजूद दोनों में कुछ समानताएँ भी थीं ।

लोग अपने शासकों को चुनाव या उनमें बदलाव नहीं कर सकते थे।
किसी को अपने विचार व्यक्त करने, संगइन बनाने, विरोध करने और राजनैतिक गतिवियिों में हिस्सा लेने की वास्तविक आजादी न थी।

प्रश्न 32.
आयोंदे, वालेश तथा मिशेल (चिले की वर्तमान राष्ट्रपति) में आर्थिक मामलों के समान नजरिया ‘नहीं था? समझाइए।
उत्तर-
आयेंदे, वालेशा तथा मिशेल में आर्थिक व सामाजिक मामलों के संदर्भ में एक जैसा नज़रिया नहीं था। आयेंदे न अर्थव्यवस्था को नियंत्रित ढंग से चलाना पंसद किया जबकि वालेशा चाहते थे कि बाज़ार सरकारी दखल से मुक्त हों। मिशेल इस मामले में कुछ मध्यमार्गी रास्ता अपनाने वाली हैं। फिर भी इन तीनों सरकारों की कुछ विशेषताएँ समान थीं। यहाँ लोगों द्वारा चुनी गई सरकारें ही शासन चला रही थी। बिना चुने हुए नेता या बाहर से संचालित शक्तियाँ या फौज शासन नहीं चला रही थीं। लोगों को विभिन्न प्रकार की बुनियादी राजनैतिक स्वतंत्रता हासिल थीं।

प्रश्न 33.
लोकतंत्र की पहचान के कोई दो तरीके बताइए।
उत्तर-
लोकतंत्र में यह व्यवस्था रहती है कि लोग अपनी मर्जी की सरकार चुनें।
लोकतंत्र में सिर्फ लोगों द्वारा चुने गए नेताओं को ही देश पर शासन करना चाहिए।
लोगों को अभिव्यक्ति की आजादी, संगठन बनाने और विरोध करने की आज़ादी जरूरी हैं।

प्रश्न 34.
जब चिले व पोलैण्ड में फौजी शासन था तब लोगों पर क्या-क्या प्रतिबंध थे?
उत्तर-

लोगों के पास वैयक्तिक स्वतंत्रताएँ नहीं थीं।
उन्हें विचार रखने का अधिकार नहीं था।
वह हड़ताल नहीं कर सकते थे।
वह अपना रोष एवं शिकायतें नहीं बता सकते थे।
वह प्रेस में अपने विचार नहीं रख सकते थे।

प्रश्न 35.
चिले के वर्तमान राष्ट्रपति के विषय में जानकारी दीजिए।
उत्तर-
चिले की वर्तमान राष्ट्रपति मिशेल बैशले आयोंदे सरकार के जनरल बैशेल की बेटी हैं। जनवरी 2006 में उन्होंने सत्ता सम्भाली है। समाजवादी विचारों वाली मिशेल पेशे से डॉक्टर हैं और लातिनी अमेरिका में रक्षा मंत्री के पद पर आने वाली पहली महिला हैं। राष्ट्रपति के चुनाव में उन्होंने जिस व्यक्ति को हराया वह चिले के सर्वाधिक धनी व्यक्तियों में गिना जाता है। चिले में लोकतंत्र के पतन और उत्थान की इस कथा को हम उनके ही भाषण के एक अंश समाप्त करते हैं।
“चूंकि मैं नफरत का शिकर बनी थी इसलिए मैंने अपना यह जीवन नफरत को खत्म करने, सहनशीलता और समझदारी के साथ-साथ प्रेम को बढ़ाने के प्रति समर्पित कर दिया है।”

प्रश्न 36.
लोकतंत्र के विकास में शुरुआती चरणों का वर्णन कीजिए।
उत्तर-
आधुनिक लोकतंत्र की कहानी कम-से-कम दो सदी पहले शुरू हुई। 1789 के जनविद्रोह ने फ्रांस में टिकाऊ और पक्के लोकतंत्र की स्थापना नहीं की थी। पूरी उन्नीसवीं सदी भर फ्रांस में बार-बार लोकतंत्र को उखाड़ फेंका गया और बार-बार इसे बहाल किया गया। लेकिन फ्रांसीसी क्रांति ने पूरे यूरोप में जगह-जगह पर लोकतंत्र के लिए संघर्षों की प्रेरणा दी। ब्रिटेन में लोकतंत्र की तरफ कदम उठने की शुरुआत फ्रांसीसी क्रांति से काफी पहले ही हो गई थी। लेकिन यहाँ प्रगति की रफ्तार काफी कम थी। अठारहवीं और उन्नीसवीं सदी में हुए राजनैतिक घटनाक्रमों ने राजशाही और सामंत – वर्ग की शक्ति में कमी कर दी थी। फ्रांसीसी क्रांति के आसपास ही उत्तर-अमेरिका में स्थित ब्रिटिश उपनिवेशों ने 1776 में खुद को आजाद घोषित कर दिया था। अगले कुछ ही वर्षों में इन उपनिवेशों ने साथ मिलकर संयुक्त राज्यअमेरिका आधुनिक अमेरिका का गठन किया। 1787 में उन्होंने एक लोकतांत्रिक संविधान को मंजूर किया। लेकिन इस व्यवस्था में भी मतदान का अधिकार पुरुषों तक सीमित था।
उन्नीसवीं सदी में लोकतंत्र के लिए होने वाले संघर्ष अकसर राजनैतिक समानता, आजादी और न्याय जैसे मूल्यों को लेकर ही होते थे। एक मुख्य माँग रहा करती थी कि सभी वयस्क नागरिकों को मतदान का अधिकार हो।

प्रश्न 37.
घाना का उदाहरण देते हुए बताइए कि – वहाँ किस प्रकार उपनिवेशवाद का अंत तथा लोकतंत्र की शुरुआत थी?
उत्तर-
पश्चिमी अफ्रीका देश घाना का उदाहरण उपनिवेश रहे देशों के सामान्य अनुभव हो बहुत अच्छी तरह दर्शाता है। यह ब्रिटिश उपनिवेश था और तब इसका नाम .गोल्ड कोस्ट था। यह 1957 मे आज़ाद हुआ। यह अफ्रीका में सबसे पहले आज़ादी पाने वाले देशों में एक था। इससे
अनेक अफ्रीकी देशों को आजादी के लिए संघर्ष करने की प्रेरणा मिली। एक सुनार के पुत्र और शिक्षक रहे वामे एनक्र्मा ने देश की आजादी की लड़ाई से सक्रिय भूमिका निभाई थी।

आज़ादी के बाद एनक्रूमा घाना के पहले प्रधानमंत्री और फिर राष्ट्रपति बने। वे जवाहर लाल नेहरू के मित्र और अफ्रीका में लोकतंत्रादियों के लिए प्रेरणा पुरुष थे। लेकिन वे नेहरू के नवशे-कदम पर टिक नहीं पाए। उन्हेंने अपने आपको आजीवन राष्ट्रपति के रूप में चुनवा लिया। लेकिन थोडे समय बाद ही 1966 में सेना ने उनका तख्तापलट कर दिया।

प्रश्न 38.
भारत में पड़ोसी देश मे लोकतंत्र के अनुभव पर एक टिप्पणी लिखिए।
उत्तर-
भारत के पड़ोस में भी काफी बड़े बदलाव हुए। 1990 के दशक में ही पाकिस्तान और बांग्लादेश में सैनिक शासन की जगह लोकतंत्र का आगमन हुआ। नेपाल में राजा ने अपने अनेक अधिकार, चुने हुए प्रतिनिधियों की – सरकार को सौंपे और खुद संवैधानिक प्रमुख बने रहे।
लेकिन ये बदलाव स्थायी नहीं थे। 1999 में जनरल मुशर्रफ ने पाकिस्तान मे फिर से सैनिक शासन कायम कर लिया। 2005 में नेपाल के नए राजा ने चुनी हुई सरकार को बर्खास्त कर दिया और पिछले दशक में लोगों को दी गई राजनैतिक आज़ादी को समाप्त कर दिया। लेकिन 2006 में फिर लोकतांत्रिक शक्तियों की विजय हुई। राजा का संसद की बैठक बुलानी पड़ी और संसद ने राजा की शक्तियों को घटाकर उसे सिर्फ नाममात्र का शासक बना दिया।
नए देशों में लोकतांत्रिक व्यवस्था की ओर प्रयास तेज हो रहे हैं। सन् 2005 तक करीब 140 देशों में बहुदलीय प्रणाली के दायरे में चुनाव कराए जाते रहे हैं।

प्रश्न 39.
म्यांमार में लोकतंत्र का अनुभव किस प्रकार रहा है? चर्चा कीजिए।
उत्तर-
म्यांमार, जिसे पहले बर्मा कहा जाता था, यह 1948 में ही औपनिवेशिक शासन से आजाद हुआ और इसने लोकतंत्र को अपनाया। लेकिन 1962 में सैनिक तख्तापलट से लोकतंत्र का अंत हो गया। 1990 में लगभग 30 वर्षों बाद पहली बार चुनाव कराए गए। आंग सान सू ची का अगुवाई वाली नेशनल लीग फॉर डेमोक्रेसी कई ने चुनाव जीते। पर म्यांमार के फौजी शासकों ने सत्ता छोड़ने से इंकार कर दिया और चुनाव परिणामों का मान्यता नहीं दी। बल्कि उन्होंने सूची समेत चुने हुए लोकतंत्र समर्थक नेताओं को गिरफ्तार कर जेल में डाल दिया या उनके घर में ही नज़रबंद कर दिया। बहुत छोटी-छोटी राजनैतिक गतिविधियों के लिए भी लोगों को पकड़ कर जेल की सज़ा दी गई। यहाँ सरकार के खिलाफ सार्वजनिक रूप से बोलने या बयान जारी करने वाले किसी व्यक्ति सरकार के खिलाफ सार्वजनिक रूप से बोलने या बयान जारी करने वाने किसी भी व्यक्ति को बीस वर्ष तक की जेल की सज़ा हो सकती है।

म्यांमार की फौजी सरकार की ज्यादतियों से तंग आकर वहाँ के 6 से 10 लाख लोगों ने अपना घर-बार छोड़ दिया है और दूसरी जगहों पर शरणार्थी बनकर रह रहे हैं।
नज़रबंदी की सज़ा झेलने के बावजूद सूची ने लोकतंत्र के लिए अपना अभियान जारी रखा है। उनके अनुसार “बर्मा में लोकतंत्र की मुहिम वहां के लोगों की विश्व समुदाय के स्वतंत्र और बराबर सदस्य के रूप मे पूर्ण और अर्थपूर्ण जीवन जीने का संघर्ष हैं।”

उनके संघर्ष को अंतरराष्ट्रीय स्तर पर मान्यता मिली है। उन्हें नोबल शांति पुरस्कार भी मिला है। फिर भी म्यांमार के लोगों को अपने देश में लोकतांत्रिक सरकार कायम करने का संघर्ष समाप्त नहीं हुआ हैं।

प्रश्न 40.
यहाँ विश्व-लोकतंत्र को मजबूत करने के लिए कुछ सुझाव हैं। क्या आप इन बदलावों का समर्थन करते हैं? क्या ये बदलाव हो सकते हैं? प्रत्येक सुझाव के लिए अपने तर्क दीजिए। सुरक्षा परिषद् के स्थायी सदस्यों की संख्या बढ़नी चाहिए।
संयुक्त राष्ट्र को आम सभा को विश्व संसद के रूप में काम करना चाहिए। जिसमें सदस्य देशों में प्रतिनिधिों की संख्या उस देश की आबादी के आधार पर तय हों। ये प्रतिनिधि एक विश्व सरकार का चुनाव करें।
अलग-अलग देश अपनी सेना नहीं रखें। विभिन्न राष्ट्रों के बीच टकराव की स्थिति में शांति कायम करने के लिए संयुक्त राष्ट्र अपने कार्य दल रखे।
संयुक्त राष्ट्र के प्रमुख का चुनाव दुनिया भर के लोग प्रत्यक्ष मतदान से करें।
उत्तर-
समय बदलता है तथा समय के साथ परिवर्तन भी होने चाहिए। सुरक्षा परिषद् में बेहतर प्रतिनिधित्व के लिए तथा उसके प्रभावकारी कार्य-संचालन के लिए उसके स्थायी सदस्यों की संख्या बढ़ायी जानी चाहिए अथवा सुरक्षा परिषद् की रचना व कार्य प्रणाली मे फेर-बदल होना चाहिए। महासभा को विश्व संसद के रूप में करना चाहिए तथा उसमें आनुपातिक आधार पर प्रतिनिधित्व होना चाहिए। यदि संयुक्त राष्ट्र के पास अपनी सेना हो, तो शांति स्थापना में काफी सहायता हो सकती है। संयुक्त राष्ट्र राष्ट्रपति की व्यवस्था हो सकती हैं तथा चुने हुए सदस्य उसका चुनाव कर सकते हैं।

प्रश्न 41.
क्या हम वैश्विक लोकतंत्र की ओर अग्रसर हैं? उदाहरण दीजिए।
उत्तर-
यह सही है कि लोकतंत्र का विस्तार हुआ है तथा विस्तार भी हो रहा है, यद्यपि कहीं-कहीं अलोकतंत्र के उदाहरण भी मिलते हैं। राष्ट्रों में लोकतंत्र का विस्तार हुआ है। परन्तु अन्तर्राष्ट्रीय संस्थाओं में लोकतंत्र के विस्तार में कुछेक कठिनाइयाँ स्पष्ट दिखायी दे रही हैं।

आज संयुक्त राष्ट्र संघ के 192 सदस्य हैं। यह – एक विश्व संस्था है। यह सभी महासभा के सदस्य भी हैं। क्योंकि यह प्रभुसत्ता सम्पन्न राष्ट्रों की संस्था है, इसमें प्रत्येक देश दूसरे देश के समान है; प्रत्येक सदस्य राज्य को एक मत प्राप्त होता है। दूसरे सदस्य राज्य के समान। वस्तुतः आनुपातिक प्रतिनिधित्व के अभाव में इस तथ्य को पूर्णतयः लोकतांत्रिक नहीं कहा जा सकता।
संयुक्त राष्ट्र सुरक्षा परिषद् एक कार्यपालिका जैसी है। इसके कुल 15 सदस्य हैं 5 : स्थायी तथा 10 अस्थायी। 5 स्थायी सदस्यों अमेरिका, रूस, फ्रांस, ब्रिटेन चीन मे प्रत्येक को वीटो शक्ति प्राप्त है। दूसरे शब्दों में सुरक्षा परिषद् के सभी सदस्य बराबर नहीं हैं। यह बात भी अपने आप मे लोकतांत्रिक नहीं है।
अन्तर्राष्ट्रीय मुद्रा निधि के 173 सदस्य एक-दूसरे के समान नहीं हैं प्रत्येक देश को मिलने वाला धन इस तथ्य पर तोला जाता है कि उस देश ने कितना वित्तीय योगदान दिया है। जी 8 के देशों को अपंखाकृत अधिक-अधिकार प्राप्त हैं।
विश्व व्यापार केंद्र में भले ही सभी देश एक-दूसरे के बराबर हों, परन्तु वास्तविक निर्णय पहले ही अनुपाचिक बैठकों में बड़े-बड़े देशों द्वारा ले लिए जाते हैं। यह तथ्य भी अन्तर्राष्ट्रिय स्तर पर लोकतांत्रिक प्रवृत्ति को धक्का पहुँचा रहा है।

वस्तुनिष्ठ प्रश्न

प्रश्न 1. निम्नलिखित वाक्यों में रिक्त स्थानों को उपयुक्त शब्दों से भरें।

(i) 1973 में पदमुक्त चिले का नेता था…..। (आयेंदे, पिनोशे)
(ii) कालामा ………….से हजारों मील दूर तक स्थान (शिकागो, संटयागो)
(iii) 1980 के दशक मे पोलैण्ड के मजदूर संगठन के नेता का ना………था। (वालेशा, त्युक्मवर्ग)
(iv) फ्रांसीसी क्रांति की घटना…….में हुई थी। (1776, 1789)
(v) सालाजार…………..का एक तानाशाह था। (पुर्तगाल, म्यांमार)
(vi) सूची को…………में नोबेल पुरस्कार मिला था। (अर्थशास्त्र में, शांति)
उत्तर-
(i) आयोंदे,
(ii) संटयागों,
(iii) वालेशा,
(iv) 1789,
(v) पुर्तगाल,
(v) शान्ति।

प्रश्न 2. निम्नलिखित वाक्यों में सही (√) च गलत (x) का चयन कीजिए।

(i) रूस लोकतंत्र-प्रोत्साहन कार्यों में लगा हुआ है।
(ii) म्यांमार बर्मा का बदला हुआ नाम हैं।
(iii) गोल्ड कोस्ट को नामीबिया कहा जाता है।
(iv) सालाजार पुर्तगाल का तानाशाह था।
(v) पोलैण्ड के एक लोकप्रिय/निर्वाचित राष्ट्रपति का नाम था पिनोशे।
(vi) लोकतंत्र सरकार की एक प्रणाली है।
उत्तर-
(i) x,
(ii) √,
(iii) x,
(iv) √,
(v) x,
(vi) √,

प्रश्न 3. निम्नलिखित विकल्पों में सही विकल्प का चयन कीजिए।

(i) आयेंदे के राजनीतिक दल का नाम था
(a) सोलिडेरिटी
(b) पापुलर युनिटी
(c) यूनाइटिड वर्कर्स पार्टी
उत्तर-
(b) पापुलर युनिटी

(ii) म्यांमार को कभी कहा जाता था –
(a) हांग-काग
(b) बर्मा
(c) लाओस
(d) इन्डोनेशिया
उत्तर-
(b) बर्मा

(iii) निम्नलिखित देश लोकतंत्र से अलोकतंत्र में परिवर्तित हुआ था.
(a) अमेरिका
(b) चिले
(c) इंग्लैंड
(d) फ्रांस
उत्तर-
(b) चिले

(iv) वालेश 1990 में निम्नलिखित देश का राष्ट्रपति चुना गया था
(a) चिले
(b) पोलैण्ड
(c) पुर्तगाल
(d) म्यांमार
उत्तर-
(b) पोलैण्ड

(v). वर्ल्ड ट्रेड संस्था का सम्बन्ध निम्नलिखित से हैं
(a) यातायात ,
(b) व्यापार
(c) टेलीविजन
(d) ट्रैफिक
उत्तर-
(b) व्यापार

(vi) निम्नलिखित जी-8 का सदस्य नहीं है
(a) इटली
(b) स्वीडन
(c) जापान
(d) कानाडा।
उत्तर-
(b) स्वीडन

समकालीन विश्व में लोकतंत्र Class 9 HBSE Notes in Hindi

अध्याय का सार

लोकतंत्र में चुनावों का होना अनिवार्य होता है। लोग अपने द्वारा चुनावों के माध्यम से सरकार का गठन करते हैं। लोकतंत्र में लोगों को स्वतंत्रताएँ प्राप्त होता हैं : भाषण की, अभिव्यक्ति की , विचार लिखने की। ऐसी व्यवस्था में लोग अपने आपकों समुदायों व संगठनों में भी एकत्रित कर सकते हैं।

इस अध्याय में लोकतंत्र से जुड़े दो देशों का वर्णन किया गया है। एक देश चिले में लोकतंत्र द्वारा संगठित सरकार जिसके राष्ट्रपति सल्वाडो आयेंदे थे, उन्हें सैनिक तख्ता पलट में पिनोशे ने हटा दिया तथा इस प्रक्रिया में उसकी हत्या कर दी गयी। दूसरा किस्सा पोलैंड का है जहां एक दलीय साम्यवादी दल की सरकार, जो तानाशाही की प्रतीक थी, उसके एक मजदूर नेता लेक वालेशा ने लोकतंत्रीय सरकार की रचना की थी एक किस्सा लोकतंत्र से अलोकतंत्र का तथा दूसरा किस्सा अलोकतंत्र से लोकतंत्र का।

20वीं शताब्दी ऐसे दौर का प्रतीक है जहाँ अनेकों देशों में लोकतंत्र की स्थापना की गयी तथा अनेक अन्य देशों मे लोकतंत्रीय व्यवस्था को चुनौतियों का सामना करना पड़ा। 1945 से 1985 तक जिन अफ्रीकी देशों को विदेशी ताकतों से स्वाधीनता मिली थी, थे: नाइजर (1960) नाइजीरिया (1963), चाड (1960)। इस बीच एशियायी देश जो स्वतंत्र हुए, वे थे: भारत (1947), इन्डोनेशिया, सिंगापुर (1965) 1980 से 2004 के बीच स्वाधीन होने वाले यूरोपीय देशों में चैक गणराज्य (1993), सलोवेनिया (1991-92) स्लोवाकिया (1993) आदि आदि। इस दौरान विश्व में अलोकतंत्रीय सरकारें भी कहीं-कहीं स्थापित थीं : चिले (1973-1989) बोलिविया .(1963-64) पेरू (1969)।

दूसरे विश्व युद्ध के बाद की घटनाएं स्पष्ट करती हैं कि संसार के अनेक देशों में लोकतंत्र का प्रसार हुआ है, यद्यपि यह प्रसार एक जैसा नहीं हुआ है। 1945 से अब तक संसार में लोकतांत्रिक व अलोकतांत्रिक सरकारें दोनों ही देखने को मिलती हैं।

लोकतंत्र का आगमन एकदम नहीं हुआ है। लोकतंत्र के लिए लोगों ने त्याग व कष्ट सहे हैं। इस संदर्भ में क्रांतियों का वर्णन किया जा सकता है: फ्रांसीसी क्रांति (1789) तथा उससे पूर्व अमेरिका का स्वतंत्रता संग्राम आदि का उल्लेख किया जा सकता है। अनेक एशियायी व अफ्रीकी देशों में मुक्ति आन्दोलनों के बाद ही स्वाधीनता प्राप्त हुई है। लोकतंत्र के आगमन के साथ मताधिकार बढ़ा है। उत्तरदायी सरकारों की स्थापना हुई है। लोगों को अधिकारों व स्वतंत्रताओं का आश्वासन मिला है, चुनाव स्वतंत्र, निरपेक्ष तथा सामयिक हुए हैं। यह सही है कि कहीं-कहीं लोकतंत्रीय व्यवस्थाएँ अलोकतंत्रीय व्यवस्थाओं में परिवर्तित हुई है। पुर्तगाल में सालाजार का शासन, म्यांमार में सैनिक व्यवस्था, पाकिस्तान में सैनिक-सत्ता, घाना में सैनिकों का राज आदि ऐसे उदाहरण हैं।

कुछ बड़े देशों ने, जिनमें अमेरिका उल्लेखनीय है, अन्य में लोकतंत्र थोपने के प्रयास किए हैं। ऐसे मामलों में लोकतंत्र को तुरन्त सफलता तो मिल जाती है, परन्त दीर्घकालीन समय में लोकतंत्रीय व्यवस्था में चुनौतियों पनप उठती हैं।

लोकतंत्र के आगमन से एक नयी बात यह है कि जहाँ राष्ट्रों के जीवन में लोकतंत्र का आगमन हुआ है, वहाँ अन्तर्राष्ट्रीय मंच पर लोकतंत्रीय मूल्यों को क्षति पहुँची है। संयुक्त राष्ट्र सुरक्षा परिषद् में कुछ देशों को वीटो अधिकार तथा अन्तर्राष्ट्रीय संस्थाओं जैसे अंतर्राष्ट्रीय मुद्रा निधि में सदस्यों को एक प्रकार की सुविधाओं का न होना अलोकतंत्रीय प्रवृतियाँ हैं।

HBSE 9th Class Social Science Solutions

HBSE 9th Class Social Science Solutions SST Haryana Board

October 15, 2023 November 10, 2023 / Class 9 / By Prasanna

Haryana Board HBSE 9th Class Social Science Solutions

HBSE 9th Class Social Science Solutions in Hindi Medium

HBSE Class 9 Social Science History: India and The Contemporary World – I (इतिहास-भारत और समकालीन विश्व-I)

HBSE Class 9 Social Science Geography: Contemporary India – I (भूगोल-समकालीन भारत-I)

HBSE Class 9 Social Science Civics (Political Science): Democratic Politics – I (राजनीति विज्ञान-लोकतांत्रिक राजनीति-I)

HBSE Class 9 Social Science Economics (अर्थशास्त्र)

HBSE 9th Class Social Science Solutions in English Medium

Haryana Board HBSE 9th Class Social Science Solutions History: India and The Contemporary World – I

Haryana Board HBSE 9th Class Social Science Solutions Geography: Contemporary India – I

Haryana Board HBSE 9th Class Social Science Solutions Civics (Political Science): Democratic Politics – I

HBSE Class 9 Social Science Economics

HBSE 9th Class Maths Notes Chapter 13 Surface Areas and Volumes

December 3, 2022 December 3, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 13 Surface Areas and Volumes Notes.

Haryana Board 9th Class Maths Notes Chapter 13 Surface Areas and Volumes

Introduction
So far, in all our study, we have been dealing with figures that can be easily drawn on page of our note book or blackboards. These are called plane figures. In previous classes, we have learnt about the perimeters and areas of rectangles, squares, rhombus and circles. If we cut out many of these plane figures of same shape and size from cardboard sheet and stack them up in a vertical pile, then by this process, we will obtain some solid figures such as a cuboid, a cylinder etc. In this chapter, we will learn to find the surface areas and volumes of cubes, cuboids, cylinders, cones and spheres.

Units measurement of Area:
1 km² = 1 km × 1 km = 10 hm × 10 hm = 100 hm² or 100 hectares
1 hm² = 1 hm × 1 hm = 10 dam × 10 dam = 100 dam²
1 hm² = 1 hm × 1 hm = 100 m × 100 m = 10000 m²
1 dam² = 1 dam × 1 dam = 10 m × 10 m = 100 m²
1 m² = 1m × 1m = 10 dm × 10 dm = 100 dm²
1 m² = 1m × 1m = 100 cm × 100 cm = 10000 cm²
1 dm² = 1 dm × 1 dm = 10 cm × 10 cm = 100 cm²
1 cm² = 1 cm × 1 cm = 10 mm × 10 mm = 100 mm²
1 km² = 1 km × 1 km = 1000 m × 1000 m = 106 m²
1m² = 1m × 1m = 1000 mm × 1000 mm = 106 mm²

Units measurement of volume:
1 km³ = 1 km × 1 km × 1 km = 1000 m × 1000 m × 1000 m = 109 m³
1 km³ = 1 km × 1 km × 1 km = 10 hm × 10 hm × 10 hm = 1000 hm³
1 hm³ = 1 hm × 1 hm × 1 hm = 10 dam × 10 dam × 10 dam = 1000 dam³
1 hm³ = 1 hm × 1 hm × 1 hm = 100 m × 100 m × 100 m = 106 m³
1 dam³ = 1 dam × 1 dam × 1 dam = 10 m × 10 m × 10 m = 1000 m³
1 m³ = 1 m × 1 m × 1 m = 100 cm × 100 cm × 100 cm = 10⁶ cm³
1 m³ = 1 m × 1 m × 1 m = 1000 mm × 1000 mm × 1000 mm = 10⁹ mm³
1m³ = 1m × 1m × 1m = 10 dm × 10 dm × 10 dm = 1000 dm³
1 cm³ = 1 ml = 1 cm × 1 cm × 1 cm = 10 mm × 10 mm × 10 mm = 1000 mm³
1 litre = 1000 ml = 1000 cm³
1 m³ = 10⁶ cm³ = 1000 litre = 1 kilolitre

Key Words:
→ Solids: Bodies which have three dimensions in space are called solids.

→ Volume: The amount of space occupied by a solid or bounded by a closed surface is known as the volume of solid.

→ Surface area: Surface area is the total sum of all the areas of all the shapes that cover the surface of solid.

→ Lateral surface area: Lateral surface in a solid is the sum of surface areas of all its faces excluding the bases of solid.

→ Cubold: A cuboid is a solid bounded by six rectangular plane regions.

→ Cube: When all the edges of cuboid are equal in length, it is called a cube.

→ Right circular cylinder: If a rectangle is revolved about one of its sides, the solid thus formed is called a right circular cylinder.

→ Right circular cone: If a right angled triangle is revolved about one of the sides containing a right angle, the solid thus generated is called a right circular cone.

→ Sphere: The set of all points in space which are equidistant from a fixed point, is called a sphere.

→ Hemisphere: A plane through the centre of a sphere divides the sphere into two equal parts. Each part is called a hemisphere.

Basic Concepts
Surface Area of a Cuboid and a Cube:
(a) Cuboid: A solid bounded by six rectangular faces is called a cuboid.
e.g., a book, a brick, a matchbox, a tile etc. A cuboid has 6 rectangular faces, 12 edges and 8 vertices.
6 rectangular faces are ABCD, EFGH, BCGF, ADHE, ABFE and DCGH.
ABCD is the bottom face and EFGH is the top face and these are one pair of opposite faces. Similarly, other pairs of opposite faces are BCGF, ADHE and ABFE, DCGH.

Out of these six faces, BCGF, ADHE, ABFE and DCGH are called the lateral faces of the cuboid. Any two faces other than the opposite faces are called adjacent faces.

In the given figure, ABCD and ABFE are adjacent faces.
Similarly, EFGH and ADHE are adjacent faces.
AB, BC, CD, DA, EF, FG, GH, HE, AE, BF, CG and DH are 12 edges of a cuboid. A, B, C, D, E, F, G and H are its 8 vertices.
(i) Total surface area of a cuboid:
Area of the face ABCD = Area of the face EFGH = (1 × b) sq. units
Area of the face BCGF = Area of the face ADHE = (b × h) sq. units
Area of the face ABFE = Area of the face DCGH = (1 × h) sq. units
Total surface area of the cuboid = Sum of the areas of six faces
= 2(l × b) + 2(b × h) + 2(l × h).
= 2(l × b + b × h + h × l)
= 2(lb + bh + hl) sq. units
where l = length, b = breadth and h = height.

(ii) Lateral surface area of the cuboid: Out of the six faces of a cuboid, we only find the area of the four faces, leaving the bottom and top faces. In such a case, area of these four faces is called the lateral surface of the cuboid.
Lateral surface area of the cuboid = Area of face BCGF+ Area of face ADHE + Area of face ABFE + Area of face DCGH
=bxh+bxh+lxh+lxh
= 2(l × h) + 2(b × h)
=2(l + b) × h sq. units
= Perimeter of the base × height.

(iii) Diagonal of a cuboid = \(\sqrt{l^2+b^2+h^2}\) units
(b) Cube: A cuboid whose length, breadth and height are all equal is called a cube eg., ice cubes, dice etc.
Each edge of a cube is called its edge. A cube has six faces, All the six faces of a cube are congruent square faces. Length of each edge of the cube is same. It has also 12 edges and 8 vertices.

(i) Total surface area of the cube : Total surface area of the cube is the sum of the areas of the six congruent square faces. Area of the one face is a³, where a is the edge of a cube.
∴ Total surface of the cube = 6a³ sq. units
(ii) Lateral surface area of the cube = 4a³ sq. units
Diagonal of the cube = \(\sqrt{3}\)a units

(a) Right Circular Cylinder: If we take a number of circular sheets of paper and stack them vertical pile, we will get a solid called a right circular cylinder.

eg., circular pipes, circular pillars, road rollers, gas cylinders, measuring jars etc. In case if circular sheets are not stack up vertically as shown figure 13.8 (I) then the cylinder is not a right circular cylinder. Of course, if we have a cylinder with a non-circular base as shown in figure 13.8 (II), then we also cannot call it a right circular cylinder.

Remarks: 1. From now onwards, the word cylinder would mean a right circular cylinder.

(b) Some Terms Related to the Cylinder:
(i) Base: Two circular ends of cylinder are called its bases.
(ii) Radius: The radius of the circular bases is called the radius of the cylinder. In the adjoining figure AO, OB, A’O’ and O’B’ are equal radii of the cylinder.
(iii) Axis: A line segment joining the centres of two circular bases is the called the axis of the cylinder.
In the figure 13.9, OO’ is the axis of the cylinder.
(iv) Height: The length of the axis of the right circular cylinder is called the height of the cylinder. In the figure, OO’ is the height of the cylinder.

(c) Surface Area of Right Circular Cylinder:
(i) Lateral surface area of right circular cylinder: We take a rectangular sheet of paper, whose length is just enough to go round the cylinder and whose breadth is equal to the height of the cylinder as shown figure below.

If we fold rectangular sheet along its length, we get a right circular cylinder of height h. Lateral surface area of the cylinder is the area of the rectangular sheet. The length of the sheet is equal to the circumference of the circular base which is equal to 2πr. Lateral surface area of a cylinder is also called the curved surface area of the cylinder.
Lateral or curved surface area of the cylinder = Area of the rectangular sheet
= length × breadth
=2πr × h = 2πrh sq. units
Therefore, lateral surface area of a cylinder = 2πrh sq. units.

(ii) Total surface area of the right circular cylinder: If we include areas of top and bottom of the cylinder to its lateral surface area, we get the total surface area of the cylinder.
If r is the radius of the cylinder and h its height.

∴ Total surface area of the cylinder = lateral surface area + 2 bases area
= 2πrh + 2πr²[∵ base area = πr2]
= 2πr(h + r) sq. units
Therefore, total surface area of the cylinder = 2πr(h + r) sq. units

(d) Surface Area of Hollow Cylinder: Hollow cylinder is a solid bounded by two coaxial cylinders of the same height and different radii.

eg., iron pipes, rubber tubes etc.
Let external and internal radii of a hollow cylinder be R and r and h be its height as shown in the given figure.
(i) Area of each base = π(R² – r²) sq, units
(ii) Lateral or curved surface area of the cylinder = External surface area + Internal surface area
= 2πRh + 2πrh = 2πh(R + r) sq units
(iii) Total surface area of the cylinder
= Lateral surface area + 2 area of the base ring
= 2πh(R + r) + 2π(R² – r²)
= 2πh(R + r) + 2π(R + r) (R – r)
= 2π(R + r) [h + R – r] sq. units.

(a) Right Circular Cone: A solid described by the revolution of a right angled triangle about one of its sides (containing the right angle) which remains fixed. In the given figure revolves about side AO. It generates a cone. O is the centre of base BC and A is its vertex.
eg., ice cream cone, clow’s cap, conical vessel etc.

(b) Some Terms Related to the Cone : (i) Radius of the cone: The length segment OB = OC is called radius of the cone. It is usually denoted by r.

(ii) Height of the cone: The length segment AO is called the height of the cone. It is usually denoted by h.
(iii) Slant height of the cone: Slant height of a right circular cone is the distance of its vertex from any point on the circumference of the base. In the given figure, AB and AC represents the slant height of the cone. It is usually denoted by l.

(iv) Vertical angle of the cone: ∠BAC is called the vertical angle of the cone.

(c) Surface Area of a Right Circular Cone: (i) Lateral surface area of a right circular cone: On a sheet of paper, we draw a circle with centre O and radius l. Now cut out a circular region from the sheet of a paper. We obtain a circular disc of paper [see in figure 13.19 (I)].

Now cut cone OAB from the circular disc of paper [see figure 13.19 (II, III) and open it out (see in figure 13.19 (IV)]. AB is the perimeter of base of cone OAB. The cone OAB is cut into small pieces of triangles such as ΔOAb₁, ΔOb₁b₂, ΔOb₂b₃, …… whose height is the slant height of the cone.
Area of each small triangle = \(\frac{1}{2}\)base of each triangle × l
Area of entire piece of paper (cone ΔOAB) = Sum of the areas of all triangles
= \(\frac{1}{2}\)b₁l + \(\frac{1}{2}\)b₂l + \(\frac{1}{2}\)b₃l + ……..
= \(\frac{1}{2}\)l(b₁ + b₂ + b₃ + ………)
\(\frac{1}{2}\) × l × AB
\(\frac{1}{2}\)l × 2πr = πrl
So, curved surface area of a cone = πrl.

(ii) Total surface area of a right circular cone:
Total surface of the cone = Lateral surface area + Area of the base of a cone
= πrl + πr² = πr(l + r)
So, total surface area of the cone = πr(l + r),
where r is its base radius and l its slant height.

(iii) Relation between the height, slant height and radius of cone:
l² = h² + r²
⇒ l = \(\sqrt{h^2+r^2}\)

1. (a) Sphere: A sphere is a solid described by the revolution of a semicircle about its diameter, which remains fixed, e.g., football, volleyball, throwball, etc.
In the given figure 13.27 (i) semicircle ABC by revolving about its diameter AB describes the sphere [(see in figure 13.27 (ii)]

The mid point of AB is the centre. Any line which passes through the centre and is terminated both ways by the surface is a diameter. Any line drawn from the centre to the surface is known as radius. A sphere may also defined as:

A sphere is a three dimensional figure (solid figure) which is made up of all points in the space, which lie at a constant distance from a fixed point is called the radius and the fixed point called the centre of the sphere.

(b) Surface Area of a Sphere: Let us take a rubber ball and drive a nail into it. Let us wind a string around the ball. When you reached the fullest part of the ball, use pins to keep the string in place and continue to wind the string around the remaining part of the ball till it is fully covered [see in figure 13.28 (I) and (II)].

Now, we unwind the string from the surface of the ball.

Start filling the circles one by one, with the string you had wound around the wall (see fig. 13.28). Then, on a sheet of paper, we draw four circles with radius equal to the radius of the ball. Now string is used to completely fill the regions of four circles, all of the same radius as of the sphere. It means, the surface area of a sphere of radius r = 4 times the area of a circle of radius r.
Therefore,surface area of sphere = 4πr².

2. Hemisphere: A plane through the centre of a sphere divides the sphere into two equal parts. Each part is called a hemisphere.

For a hemisphere of radius r, we have
(i) Curved surface area of the hemisphere = 2πr²
(ii) Total surface area of the hemisphere
= 2πr² + πr²
= 3πr².

3. (a) Spherical Shell: The difference of two solid concentric spheres is called a spherical shell.
(b) Surface Area of Spherical Shell: If R and be the external and internal radii of a spherical shell, then
Total surface area of spherical shell = 4π(R² + l²).

1. Volume of a Cuboid: In previous classes, we have learnt about of certain figures. Recall that solids occupy space. The measure of this occupied space is called the volume of the object.
If the object is hollow, then interior is empty and can be filled with air or some liquid that will take the shape of its container. In this case, the volume of the substance that can fill the interior is called the capacity of the container. Thus, we may say that the volume of an object is the measure of the space it occupies and the capacity of an object is the volume of substance its interior can accommodate. Unit of measurement of volume is cubic unit. If we take some rectangular sheets and place it one over the other. If we place these sheets vertically in pile, we get a cuboid. (see figure below)

If area of each rectangle is A, the height upto which the rectangles are stacked is h. Measure of the space occupied by the cuboid
= Area of a rectangular sheet × height
= A × h = l × b × h [∵ A = l × b]
Hence, volume of the cuboid = l × b × h
= length × breadth × height
Also, volume of the cuboid = Area of the base × height

2. Volume of Cube: If length, breadth and height of a cuboid are equal, then it is known as the cube.

Volume of a cube = edge × edge × edge
= a × a × a = a³,
where a is the edge of a cube. Unit of measurement of volume is cubic unit.

(a) Volume of a Cylinder: If we stack the circular sheets of radius r vertically, we will get a solid called a right circular cylinder of radius r and height h.

Volume of the cylinder = The area of each circular sheet × height
= πr² × h = πr²h
or Volume of the cylinder = Area of the base × height
= πr²h.

(b) Volume of the Hollow Cylinder: Let the external and internal radii of the hollow cylinder be R and r respectively and h be its height, then

Volume of the material = External volume – Internal volume.
= πR²h – πr²h
= πh(R² – r²).

Volume of a Right Circular Cone:
Experiment: Take a hollow cylinder and a hollow cone of the same base radius and the same height.

Fill the cone with water to the brim and empty it into the cylinder. Repeat the process two times more. We observe that 3 cone full to brim will fill the cylinder competely. With this, we can safely come to the conclusion that three times the volume of a cone, makes up the volume of a cylinder.

It means volume of a cone of radius ‘r’ and height ‘h’
= \(\frac{1}{3}\) of volume of cylinder of radius ‘r’ and height ‘h’
Volume of a cone = \(\frac{1}{3}\) × πr²h
or Volume of a cone = \(\frac{1}{3}\) × Area of the base × height

1. Volume of a Sphere: We take a cylindrical container and two or three spheres of different radii. Also take a large trough in which we can place the cylindrical container. Then fill the container up to the brim with water.
Now, we place a sphere in the container. Some of the water will overflow into the trough in which it is kept (see in figure given below).

We pour out the water from the trough into a measuring cylinder and find the volume of the over flowed water. If r is the radius of immersed sphere on evaluating \(\frac{4}{3}\)πr³, we find this value atmost equal to the volume of over flowed water.

We repeat this process with two or three spheres of different radii, we find:
Volume of overflowed water = The volume of the sphere immersed in the water
= \(\frac{4}{3}\)πr³
Thus,volume of the sphere = \(\frac{4}{3}\)πr³.

2. Volume of the hollow sphere: If R and r are respectively the outer and inner radii of hollow sphere.
Volume of the material = Volume of outer sphere – Volume of inner sphere
= \(\frac{4}{3}\)πrR³ – \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\)π(R³ – r³)

3. Volume of the hemisphere :
Volume of the hemisphere \(\frac{2}{3}\)πr³
where r is the radius of the hemisphere.

HBSE 9th Class Maths Notes Chapter 12 Heron’s Formula

December 1, 2022 December 1, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 12 Heron’s Formula Notes.

Haryana Board 9th Class Maths Notes Chapter 12 Heron’s Formula

Introduction
In previous classes, we have studied about various plane figures such as squares, rectangles, triangles, quadrilaterals etc. We have also learnt the formula for finding the areas and the perimeters of these figures like rectangle, square, triangle etc. as given below:
1. Triangle :
(i) \(\frac{1}{2}\) × Base × Height
\(\frac{1}{2}\) × BC × AD
(ii) Perimeter = sum of the sides of a triangle
= AB + BC + CA

2. Right angled triangle:
(i) Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × BC × AB
(ii) Perimeter = AB + BC + CA

3. Equilateral triangle: Let each side of an equilateral triangle be a units.
(i) Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × BC × AD
[∵ AD = \(\sqrt{a^2-\frac{a^2}{4}}\)
⇒ AD = \(\sqrt{\frac{3 a^2}{4}}\)
⇒ AD = \(\frac{\sqrt{3} a}{2}\) ]
= \(\frac{1}{2}\) × a × \(\frac{\sqrt{3} a}{2}\)
(ii) Perimeter = a + a + a
3a = 3 × side

4. Parallelogram:
(i) Area = base x height
= AB × DE
(ii) Perimeter = sum of the sides of ||gm
= AB + BC + CD + AD

5. Square: (i)
Area = side × side = side²
or Area = \(\frac{1}{2}\) × d²
(where d = diagonal of a square)
(ii) Perimeter = side + side + side + side
= 4 × side

6. Rectangle :
(i) Area = base × height
= length × breadth
(ii) Perimeter = AB + BC + CD + DA
= L + B + L + B
= 2L + 2B
= 2(L + B)
where L = length of a rectangle
and B = breadth of a rectangle

7. Rhombus :
(i) Area = base × height
or Area = \(\frac{1}{2}\) × d₁ × d₂
where d₁ and d₂ are the diagonals of a rhombus.
(ii) Perimeter = side + side + side + side
= 4 × side

8. Trapezium:
(i) Area = \(\frac{1}{2}\) × (sum of parallel sides) × distance between them
= \(\frac{1}{2}\) × (AB + DC) × DE
(ii) Perimeter = AB + BC + CD + AD
Unit of measurement for length, breadth, height and perimeter is taken as metre (m) or centimetre (cm), milimetre (mm) etc.

Units measurement of Perimeter
1 m = 10 dm, 1 m = 100 cm, 1 m = 1000 mm, 1 km = 10 hm, 1 km = 100 dakm, 1 km = 1000 m Unit of measurement for area of any plane figure is taken as square metre (m²) or square centimetre (cm²) or square millimetre (mm²) etc.
Units measurement of Area
1 km² = 1 km × 1 km = 10 hm × 10 hm = 100 hm²
1 km² = 1 km × 1 km = 1000 m × 1000 m = 1000000 m² = 106 m²
1 hectare = 10000 m²
1 m² = 1m × 1m = 100 cm × 100 cm = 10000 cm²
1m² = 1m × 1m = 10 dm × 10 dm = 100 dm² etc.

Key Words
→ Area: The area of a triangle or a polygon is the measure of the surface enclosed by its sides.

→ Perimeter: The perimeter of a plane figure is the length of its boundary.
OR
A perimeter is the sum of lengths of sides of a polygon.

→ Right angled triangle: A triangle with one angle a right angle (90°) is called a right angled triangle.

→ Scalene triangle: If all three sides of a triangle are unequal or different, it is called scalene triangle.

→ Isosceles triangle: If at least two sides of a triangle are equal, it is called isosceles triangle.

→ Equilateral triangle: If all three sides of a triangle are equal, it is called. equilateral triangle.

→ Quadrilateral: A plane figure bounded by four line segments is called a quadrilateral.

→ Parallelogram: A quadrilateral in which opposite sides are parallel and equal is known as a parallelogram.

→ Square: A quadrilateral whose each angle is 90° and all sides are equal to each other is known as a square.

→ Rectangle: A quadrilateral each of whose angles is 90° is called a rectangle.

→ Rhombus: A quadrilateral having all the four sides equal is called a rhombus.

→ Trapezium: A quadrilateral in which one pair of opposite sides are parallel is called a trapezium.

→ Diagonal: A line joining any two vertices of a polygon that are not connected by an edge and which does not go outside the polygon.

Basic Concepts
Area of a Triangle by Heron’s Formula: Heron was born in about 10 AD possibly in Alexandria in Egypt. He worked in applied mathematics and wrote three books on mensuration. Book I deals with the area of rectangles, squares, triangles, trapezia, various other specialized quadri- laterals, regular polygons, circles, surfaces of cylinders, cones and spheres etc.

He derived the famous formula for finding the area of a triangle in terms of its three sides.
The formula for finding the area of a triangle was given by Heron is stated as below:
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\),
where a, b and c are the sides of a triangle and s = semi-perimeter of the triangle, ie.,
\(\frac{a+b+c}{2}\)
This formula is helpful where it is not possible to find the height of the triangle.

Application of Heron’s Formula in Finding Areas of Quadrilaterals: In this section, we have studied the areas of such figures which are the shape of a quadrilaterals. We need to divide the quadrilateral in two triangles and then use the formula 12.25 area of the triangle.

HBSE 9th Class Maths Notes Chapter 11 Constructions

December 1, 2022 December 1, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 11 Constructions Notes.

Haryana Board 9th Class Maths Notes Chapter 11 Constructions

Introduction
In previous chapters, we have drawn rough diagrams to prove theorems or solving exercise. But sometimes we need to construct accurate figures. For example, to draw a map of building to be constructed, to design the parts of machine and tools etc. To draw such figures the following geometrical instruments are needed:
(i) A graduated scale, (ii) a pair of set squares, (iii) a pair of dividers, (iv) a pair of compasses, (v) a protractor (see in the figure below).

These instruments are used in drawing a geometrical figure, such as a triangle, quadrilateral, trapezium, circle etc. In this chapter we will learn about some simple and basic constructions such as bisector of a line segment, bisector of a given angle, constructions of some standard angle and some constructions of triangles with the help of graduated scale, compass and protractor.

Key Words
→ Geometrical Construction: A geometrical construction is the process of drawing a geometrical figure using only two instruments-an ungraduated ruler and compass.

→ Perpendicular bisector: A line which divides the given line in two equal parts and is perpendicular to the line segment is called the perpendicular bisector.

→ Corresponding: Angles, lines and points in one figure which bear a similar relationship, each to each, to angles, lines and points in another figure.

→ Triangle: A closed figure which bounded by three line segments is called a triangle.

→ Isosceles triangle: If at two sides of a triangle are equal, it is called an isosceles triangle.

→ Equilateral triangle: If all three sides of a triangle are equal, it is called an equilateral triangle.

→ Perimeter of a triangle: Sum of the all sides of a triangle is called the perimeter of a triangle.

→ Median: The median of a triangle, corresponding to any side, is the line joining the midpoint of that side with the opposite vertex.

→ Altitude: An altitude of a triangle, corresponding to any side, is the length of perpendicular drawn from the opposite vertex to that side.

Basic Concepts
Basic Constructions
(a) Construction 11.1: To construct the bisector of a given angle.
We have an ∠ABC, we need to construct its bisector.
Steps of construction:
Step – I: Taking B as centre and any suitable radius draw an arc cutting AB and BC at P and Q respectively.
Step – II: Taking P and Q as centres and radius more than \(\frac{1}{2}\)PQ, draw two arcs intersecting each other at R.
Step-III: Draw the ray BR. The ray BR is the required bisector of the angle ABC.

Justification: Join PR and QR.
In ΔBPR and ΔBQR, we have
PB = QB [Radii of the same arc]
PR = QR [Arcs of equal radii]
BR = BR [Common]
ΔBPR ≅ ΔBQR [By SSS congruence rule]
⇒ ∠PBR = ∠QBR [CPCT]
Hence, BR is the bisector of ∠ABC.

Construction 11.2:
To construct the perpendicular bisector of a given line segment.
We have a line segment AB, we need to construct its perpendicular bisector.
Steps of construction:
Step-I: Draw a line segment AB of given length.
Step-II: Taking A as centre and radius more than \(\frac{1}{2}\)AB, draw arcs one on each side of AB.
Step-III: Taking B as centre and the same radius as in Step (II), draw arcs intersecting previous arcs at P and Q.
Step-IV: Join PQ intersecting AB at M. Then line PQ is the perpendicular bisector of AB.
Justification: Join AP, BP, AQ and BQ.
In ΔPAQ and ΔPBQ, we have
AP = BP Arcs of equal radii]
AQ = BQ [Arcs of equal radii]
PQ = PQ [Common]
∴ ΔPAQ ≅ ΔPBQ [By SSS congruence rule]
⇒ ∠APQ ≅ ∠BPQ [CPCT]
⇒ ∠APM = ∠BPM ……(i)

In ΔPMA and ΔPMB, we have
PA = PB [Arcs of equal radii]
∠APM = ∠BPM [As proved in (i)]
PM = PM [Common]
ΔPMA ≅ ΔPMB (By SAS congruence rule)
⇒ AM = BM
and ∠PMA = ∠PMB (CPCT)
But, ∠PMA + ∠PMB = 180° [By linear pair axiom]
∠PMA + ∠PMA = 180°
2∠PMA = 180°
∠PMA = \(\frac{180^{\circ}}{2}\) = 90°
∠PMA = ∠PMB = 90°.
Hence, PM is the perpendicular bisector of AB.

Construction 11.3:
To construct an angle of 60° at the initial point of a given ray.
Let us take a ray AB with initial point A.
We need to construct a ray AC such that
∠CAB = 60°.

Steps of construction:
Step-I: Draw a ray AB with initial point A.
Step-II: Taking A as centre and suitable radius, draw an arc which intersects AB at P.
Step-III: Taking P as centre and same radius as before, draw an arc intersecting the previous are at point Q.
Step-IV: Draw a ray AC passing through Q. Then ∠CAB is the required angle of 60°.
Justification: Join PQ.
AP = PQ = AQ [By construction]
⇒ ΔPAQ is an equilateral triangle.
∴ ∠QAP = 60°
⇒ ∠CAB = 60°.

Some Constructions of Triangles:
Construction 11.4 To construct a triangle, given its base, a base angle and sum of other two sides.

Given: The base BC, a base angle, say ∠B and the sum AB + AC of the other two sides of a triangle ABC, we need to construct ΔABC.
Steps of construction:
Step-I: Draw the given base BC and the point B construct an angle ∠CBX equal to the given angle with BC.
Step-II: From the ray BX,
cut BD = AB + AC.
Step-III: Join DC and construct
∠DCY = ∠BDC.
Step-IV: Let CY intersects BX at A.
Then, ABC is the required triangle.
Justification: Since,
∠BDC = ∠DCY (By construction)
⇒ ∠ADC = ∠ACD
⇒ AD = AC
[Sides opp. to equal angles are equal]
Now, AB = BD – AD
⇒ AB = BD – AC [∵AD = AC]
⇒ AB + AC = BD
⇒ BD = AB + AC.
Alternative method

Steps of construction:

Step-I: Draw the given base BC and at the point B construct ∠CBX equal to the given angle with BC.
Step-II: From the ray BX, cut BD = AB + AC.
Step-III: Join DC and draw perpendicular bisector PY of CD intersecting BD at point A. Join AC.
Remarks: The construction of the trinagle is not possible if the sum AB + AC ≤ BC.

Construction 11.5: To construct a triangle given its base, a base angle and the difference of the other two sides.
Given: Base BC and a base angle, say ∠B and the difference of other two sides AB – AC or AC – AB. We need to construct AABC. There are following two cases arise:
Case I: If AB > AC that is AB – AC is given.

Steps of construction:
Step-I: Draw the given base BC. At the point B, construct ∠CBX equal to the given angle with BC.
Step-II: From ray BX, cut the line segment BD is equal to AB – AC.
Step-III Join DC and draw perpendicular bisector of CD intersecting BX at A.
Step-IV: Join AC. Then ABC is required triangle.
Justification: Since AP is the perpendicular to CD.
∴ AD = AC
So, BD = AB – AD
⇒ BD = AB – AC.
Case II: If AB < AC that is AC – AB is given.

Steps of construction:
Step-I: Draw the given base BC. At the point B, construct ∠CBX equal to the given angle with BC.
Step-II: Extend XB to D to the opposite side of BC such that
BD = AC – AB.
Step-III: Join CD and draw perpendicular bisector PQ of CD intersecting BX at A.
Step-IV: Join AC, then ABC is required triangle.
Justification: Since PQ is the perpendicular bisector of CD.
∴ AD = AC
Now, BD = AD – AB
⇒ BD = AC – AB.

Construction 11.6:
To construct a triangle, given its perimeter and its two base angles.
Given: Base angles ∠B and ∠C and perimeter of a triangle ABC (ie., BC + CA + AB) are given. We need to construct the triangle ABC.

Steps of construction:
Step – I: Draw a line segment XY equal to BC + CA + AB.
Step – II: Construct ∠LXY equal to ∠B and ∠MYX equal to ∠C with XY.
Step – III: Draw the bisectors of ∠LXY and ∠MYX intersecting at A.
Step – IV: Draw perpendicular bisectors PQ of AX and RS of AY.
Step – V: Let perpendicular bisectors PQ and RS intersect XY at B and C respectively.
Step – VI: Join AB and AC, then ABC is the required triangle.
Justification: Since PQ is the perpendicular bisector of AX.
∴ XB = AB …..(i)
Similarly, RS is the perpendicular bisector of AY.
∴ CY = AC ……(ii)
Now, XY = XB + BC + CY
XY = AB + BC + AC [Using (i) & (ii)]
Since, XB = AB
⇒ ∠AXB = ∠XAB ……(iii)
and CY = AC
⇒ ∠CAY = ∠CYA …..(iv)
∠ABC = ∠AXB + ∠XAB
[∵ Exterior angle is equal to its opp. two interior angles]
⇒ ∠ABC = ∠AXB + ∠AXB [Using (iii)]
⇒ ∠ABC = 2∠AXB
⇒ ∠ABC = ∠LXY
Similarly, ∠ACB = ∠MYX.

HBSE 9th Class Maths Notes Chapter 10 Circles

November 30, 2022 November 30, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 10 Circles Notes.

Haryana Board 9th Class Maths Notes Chapter 10 Circles

Introduction
We used various objects in our daily life which are round in shape, such as coins of dimensions ₹ 1, ₹ 2 and ₹ 5, buttons of shirts, key rings, dials of clocks, wheels of vehicles etc. (see in figure 10.1). In a clock we have observed that the second’s hand goes round the dial of the clock rapidly and its tip moves round path. This path traced by the tip of the second’s hand is called a circle.

If we fix a pencil in compass and put its pointed leg on a point on a sheet of a paper, open the other leg to some distance. Keeping the pointed leg on the same point, rotate the other leg through one revolution we get a closed figure traced by the pencil on the paper which is known as the circle (see in figure 10.2). In this chapter we will study about circles, it’s related terms and some properties of a circle.

Key Words
→ Circle: A circle is a plane figure bounded by one curved line and is such that all straight lines drawn to this line from certain point within it are equal.

→ Radius: A line segment joining the centre and a point on the circle is called its radius.

→ Circular Region: The region consisting of all points which are either on the circle or lies inside the circle is called the circular region or circular disc.

→ Concentric Circles: The circles which have same centre and different radii are called concentric circles.

→ Centroid: The point of concurrency of the three medians of a triangle is called the centroid of the triangle.

→ Circumcentre: The point of concurrency of the perpendicular bisectors of the sides of a triangle is called the circumcentre of the triangle.

→ Circumcircle: The circumcentre O is equidistant from the three vertices of a triangle. A circle with centre O and radius OA is called the circumcircle of the triangle. It passes through all the three vertices.

→ Concyclic: A number of points are concyclic if there is a circle that passes through all of them.

Basic Concepts
Circles and Its Related Terms: A Review
(i) Circle: The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle.
(ii) Centre: The fixed point is called the centre of the circle. In the figure, O is the centre.

(iii) Radius: The constant distance from its centre is called the radius of the circle. The plural of radius is radii.
In the figure, OA, OB and OC are radii of the circle C(O, r).

(iv) Circumference: The length of the complete circle is called its circumference. The perimeter of a circle is also known as the circumference of the circle.

(v) Chord: A line segment joining any two points on the circle is known as the chord of the circle. In the figure, AB and CD are chords of a circle with centre O.

(vi) Diameter: The chord, which passes through the centre of the circle is called a diameter of the circle. A diameter is the longest chord and all diameters are equal in lengths. In the figure, AOB is the diameter.
Diameter = 2 × radius
A diameter divides the circle into two equal arcs. Each of these two arcs is called a semicircle. In the figure, \(\widehat{A P B}\) and \(\widehat{A Q B}\) are two semicircles. The degree measure of a semicircle is 180°.

(vii) Secant: A line which intersects a circle in two distinct points is called a secant of the cricle. In the figure, the line l cuts the circle in two points A and B. So, l is a secant of the circle.

(viii) Arc: A piece of a circle between two points is called an arc.

Let P and Q be two points on the circle. These points P and Q divide the circle into two pieces. Each piece is an arc. These arcs are denoted in anticlockwise direction from P to Q as PQ and from Q to Pas QP. The arc PQ is called the minor are and arc QRP is called the major arc,

where R is any point (figure 10.9) on the arc between P and Q. When arcs PQ and QRP are equal, then each is called a semicircle.

(ix) Interior and Exterior of a circle: A circle divides the plane on which it lies into three parts. They are:

(a) Inside the circle, which is also called the interior of the circle.
(b) Circle.
(c) Outside the circle, which is also called the exterior of the circle (see in figure).
The circle and its interior make up the circular region.
(x) Position of a point with respect to a circle:
(a) Point inside the circle: A point P is said to lie inside the circle C(O, r), if OP < r.

(b) Point on the circle: A point is said to lie on the circle C(O, r), if OP = r.

(c) Point outside the circle: A point is said to lie outside the circle C(O, r), if OP > r.

(xi) Segment of a circle: The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle.

The segment containing the minor area is called the minor segment. Thus APBA is the minor segment of the circle.
The segment containing the major arc is called the major segment. Thus ARBA is the major segment of the circle.
(xii) Sector of a circle: The region between an arc and the two radii, joining the centre to the end points of the arc is called a sector. The minor are corresponds to the minor sector and the major arc corresponds to the major sector. In the figure, the region OPQO is the minor. sector and remaining part of the circular region is the major sector.

(xiii) Quadrant of a circle: One fourth of a circle is called a quadrant of a circle. In the figure, OABO is the quadrant of the circle C(O, r).

(xiv) Central angle: Let C(O, r) be any circle, then any angle whose vertex is O, is called a central angle of the circle. In the figure, ∠AOB is a central angle of the circle C (O, r).

(xv) Degree measure of an arc: Degree measure of a minor arc is the measure of the central angle subtended by the arc. In the given figure 10.17, the measure of the minor are AB is 50° ie., m\(\widehat{A B}\) = 50°. The degree measure of the major are is 360° – 50° = 310°.
⇒ m\(\widehat{B A}\) = 310°
The degree measure of a circle is 360°.
∴ mC(O, r) = m arc AB + m arc BA
= 50° + 310° = 360°.

(xvi) Congruent circles: Two circles are said to be congruent, if and only if one of them can be superposed on the other so as to cover it exactly. This is possible only when the radii of the two circles are equal. Let C(O, r₁) and C(O’, r₂) be two circles such that C(O, r₁) is superposed on C(O’, r₂) so that O’ coincides with O. Then C(O, r₁) will cover C(O’, r₂) completely if and only if r₁ = r₂.

Thus two circles are congruent, if and only if, they have equal radii.

(xvii) Congruent arcs: Two arcs of a circle are said to be congruent if and only if they have the same degree measures.
If arc AB is congruent to arc CD. They are written as
AB = CD

In particular case, arcs of semicircles will be congruent because each has degree. measure = 180°.

Angle subtended by a chord at a point: The angle subtended by chord PQ at a point R (not on the chord PQ) on the circumference of the circle is ∠PRQ. And the angle subtended by chord PQ at the centre O is ∠POQ.

Theorem 10.1:
Equal chords of a circle subtend equal angles at the centre.
Given: Two equal chords AB and CD of a circle with centre O.

To prove: ∠AOB = ∠COD.
Proof: In ΔAOB and ΔCOD, we have
OA = OC (Equal radii of a circle)
OB = OD (Equal radii of a circle)
AB = CD (given)
∴ ΔAOB ≅ ΔCOD (By SSS congruence rule)
⇒ ∠AOB = ∠COD (CPCT)
Hence proved

Theorem 10.2: [Converse of theorem 10.1]
If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Given: Two chords AB and CD of a circle are such that
∠AOB = ∠COD.
To prove: AB = CD.
Proof: In ΔAOB and ΔCOD, we have
OA = OC [Equal radii of a circle]
∠AOB = ∠COD (given)
OB = OD [Equal radii of a circle]
∴ ΔAOB ≅ ΔCOD (By SAS congruence rule)
⇒ AB = CD (CPCT)
Hence proved

Perpendicular from the centre to a chord :
Theorem 10.3:
The perpendicular from the centre of a circle to a chord bisects the chord.

Given: A chord AB of a circle C(O, r) and OM ⊥ to the chord AB.
To prove: AM = MB.
Construction: Join OA and OB.
Proof: In the right ΔOMA and ΔOMB, we have
OA = OB (Equal radii of a circle)
∠OMA = ∠OMB (Each is 90°)
OM = OM (Common)
ΔΟΜΑ ≅ ΔΟΜΒ (By RHS congruence rule)
⇒ AM = MB (CPCT)
Hence proved

Theorem 10.4:
[Converse of theorem 10.3]
The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Given: A chord AB of a circle C(O, r) with mid point M.
To prove: OM ⊥ AB.
Construction: Join OA and OB.
Proof: In ΔOAM and ΔOBM, we have
OA = OB (Equal radii of a circle)
AM = BM
[∵ M is the mid point of AB]
OM = OM (common)
∴ ΔOAM ≅ ΔOBM (By SSS congruence rule)
⇒ ∠OMA = ∠OMB (CPCT) …..(i)
But ∠OMA + ∠OMB = 180° (Linear pair axiom)
⇒ ∠OMA + ∠OMA = 180° [using (i)]
⇒ ∠OMA = 180°
⇒ ∠OMA = \(\frac{180^{\circ}}{2}\) = 90°
∠OMA = ∠OMB = 90°
Hence OM ⊥ AB.
Hence proved

Circle Through Three Points: We take a point P on a paper. There are many circles passing through this point.
Take two points P and Q. There are many infinite number of circle passing through P and Q.
If you take three non-collinear points A, B and C. There is only one circle passing through these points (see in figure below).

Theorem 10.5:
There is one and only one circle passing through three given. noncollinear points.

Given: Three non-collinear points A, B and C.
To prove: There is one and only one circle passing through three points A, B and C.
Construction: Join AB and BC. Draw the 1 bisectors PQ and RS of AB and BC. Since A, B and C are non-collinear, PQ and RS are not parallel and will intersect, say the point O. Join OA, OB and OC.
Proof: Since, O lies on the perpendicular bisector of AB, we have
OA = OB …..(i)
Similarly, O lies on the perpendicular bisector of BC, we have
OC = OB …..(ii)
From (i) and (ii), we get
OA = OB = OC = r (say)
With O as the centre and r the radius, draw circle C(O, r) which will pass through A, B and C.
Hence, there is one and only one circle passing through three non-collinear points.
Remark: If ABC is a triangle, then by Theorem 10.5, there is a unique circle passing through the three vertices A, B and C of the triangle. This circle is called the circumcircle of the ΔABC. Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.

Equal chords and their distances from the centre: Let AB be a line and P be a point (not on the line). There are many infinite number of points L₁, L₂, L₃, L₄, L₅ on the line. If we join these points to P, we will get many line segments PL₁, PL₂, PM, PL₃, PL₄, PL₅, where PM is the perpendicular from P to AB. On measuring these line segments PL₁, PL₂, PM, PL₃, PL₄, PL₅, we find that PM is the shortest line segment.

So, we conclude that the length of the perpendicular from a point to a line is the distance of the line from the point.

Note: If the point lies on the line, the distance of the line from the point is zero.

Theorem 10.6:
Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Given: A circle C(O, r) in which chord AB is equal to chord CD, OL ⊥ AB and OM ⊥ CD.
To prove: OL = OM.
Construction: Join OA and OC.
Proof: We have
AB = CD
\(\frac{1}{2}\)AB = \(\frac{1}{2}\)CD ……(i)
∵ OL ⊥ AB
∴ AL = LB = \(\frac{1}{2}\)AB
(By theorem 10.3)…(ii)
and OM ⊥ CD
∴ CM = MD = \(\frac{1}{2}\)CD
(By theorem 10.3)…(iii)
From (i), (ii) and (iii), we get
AL = CM
Now, right ΔOLA and ΔOMC, we have
Hyp. OA = Hyp. OC [Equal radii of a circle]
AL = CM (as proved above)
∠OLA = ∠OMC (Each is 90°)
ΔOLA ≅ ΔOMC (By RHS congruence rule)
⇒ OL = OM (CPCT)
Hence, equal chords of a circle are equidistant from the centre. Proved.

Theorem 10.7:
Chords equidistant from the centre of a circle are equal in length.

Given: Two chords AB and CD of a circle C(O, r) are such that OL = OM, OL ⊥ AB and OM ⊥ CD.
To prove: AB = CD.
Construction: Join OB and OD.
Proof: ∵ OL ⊥ AB and OM ⊥ CD
∴ LB = \(\frac{1}{2}\)AB and MD = \(\frac{1}{2}\)CD ……(i)
(By theorem 10.3)
Now, right ΔOLB and ΔOMD, we have
Hyp. OB = Hyp. OD [Equal radii of a circle]
∠OLB = ∠OMD (Each is 90°)
OL = OM (given)
∴ ΔΟLΒ ≅ ΔΟΜD (By RHS congruence rule)
⇒ LB = MD (CPCT)
⇒ 2LB = 2MD
⇒ AB = CD [using (i)]
Hence, chords of a circle, which are equidistant from the centre are equal.
Proved

Angle Subtended by an Arc of a Circle: If we cut two arcs corresponding to the equal chords AB and CD respectively. We observe that are AB superimpose the arc CD completely (see in figure). This shows that equal chords make congruent arcs and conversely congruent arcs make equal chords of a circle. You can state it as follows:

If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.
In the given figure, minor are AB subtends ∠AOB at the centre O and major are BA subtends reflex ∠AOB at the centre.

In view of the property above and theorem 10.1, the following result is true:
Congruent ares (or equal arcs) of a circle subtend equal angles at the centre.

Theorem 10.8:
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Given: An arc PQ of a circle subtends ∠POQ at the centre and ∠PAQ at a point A on the remaining part of the circle.
To prove: ∠POQ = 2∠PAQ.
Construction: Join AO and produced it to B.

Proof : Consider three different cases as given in figures. In figure (i) \(\widehat{P Q}\) is minor arc, in (ii) \(\widehat{P Q}\) is a semi-circle and in (iii) \(\widehat{P Q}\) is major arc.
In figure (i) and (ii), In ΔAOP, we have
AO = OP [Equal radii of same circle]
⇒ ∠1 = ∠2 [Angles to the opposite sides are equal]
∠POB = ∠1 + ∠2
[Exterior angle is equal to sum of its two opposite interior angles]
⇒ ∠POB = ∠1 + ∠1
⇒ ∠POB = 2∠1 …..(i)
Similarly, In ΔAOQ, we have
∠3 = ∠4 [∵ AO = OQ]
∠QOB = ∠3 + ∠4
⇒ ∠QOB = ∠3 + ∠3
⇒ ∠QOB = 2∠3 ……(ii)
Adding (i) and (ii), we get
∠POB + ∠QOB = 2∠1 + 2∠3
⇒ ∠POQ = 2(∠1 + ∠3)
⇒ ∠POQ = 2∠PAQ
In figure (iii),
∠POB + ∠QOB = 2 (∠1 + ∠3)
⇒ Reflex ∠POQ = 2∠PAQ. Hence proved

Corollary: The angle in a semicircle is a right angle.
Given: In a circle C(O, r), AB is the diameter with centre O and ∠ACB is an angle in a semicircle.

To prove: ∠ACB = 90°,
Proof: ∠AOB = 2∠ACB (By theorem 10.8)
⇒ 180° = 2∠ACB (∵ AOB is a straight line)
∠ACB = \(\frac{180^{\circ}}{2}\)
∠ACB = 90°
Hence, angle in a semicircle is a right angle. Proved

Theorem 10.9:
Angles in the same segment of a circle are equal.
Given: A circle C(O, r) in which ∠ACB and ∠ADB are two angles made by arc AB in the same segment ACDB of the circle.
To prove: ∠ACB = ∠ADB.
Construction: Join OA and OB.

Proof: In figure (i), by theorem 10.8, we have
∠AOB = 2∠ACB…(i)
and ∠AOB = 2∠ADB…(ii)
From (i) and (ii), we get
2∠ACB = 2∠ADB
∠ACB = ∠ADB
In figure (ii), by theorem 10.8, we have
reflex ∠AOB = 2∠ACB…(iii)
and reflex ∠AOB = 2∠ADB…(iv)
From (iii) and (iv), we get
2∠ACB = 2∠ADB
∠ACB = ∠ADB
In both cases, ∠ACB = ∠ADB
Hence, angles in the same segment of a circle are equal. Hence proved

Theorem 10.10:
If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e., they are concyclic).

Given: AB is a line segment, which subtends equal angles at two points C and D, such that
∠ACB = ∠ADB.
To prove: The points A, B, C and D lie on a circle (i.e., they are concyclic).
Construction: Draw a circle through three non-collinear points A, B, C.
Proof: Suppose circle does not pass through D. Let it will intersect AD (or extended AD) at point, say E (or E’). If points A, C, E and B lie on a circle, then
∠ACB = ∠AEB …(i) (By theorem 10.9)
But, ∠ACB = ∠ADB (given)…(ii)
From (i) and (ii), we get
∠AEB = ∠ADB
This is not possible unless E coincides with D.
Therefore D lies on the circle passing through A, B, C.
Similarly, E’ should also coincide with D.
Hence, the points A, B, C and D are concylic. Hence Proved

Cyclic Quadrilaterals: A quadrilateral ABCD is called cyclic, if all the four vertices of it lie on a circle.
Theorem 10.11:
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
Or
The opposite angles of a cyclic quadrilateral are supplementary.

Given: A cyclic quadrilateral ABCD.
To prove: ∠A + ∠C = 180° and ∠B + ∠D = 180°.
Construction: Join AC and BD.
Proof: ∠ADB = ∠ACB [Angles in a same segment are equal]…(i)
∠ABD = ∠ACD [Angles in a same segment are equal]…(ii)
Adding (i) and (ii), we get
∠ADB + ∠ABD = ∠ACB + ∠ACD
⇒ ∠ADB + ∠ABD = ∠C
Adding ∠BAD on both sides, we get
∠ADB + ∠ABD + ∠BAD = ∠C + ∠BAD
⇒ 180° = ∠C + ∠A
[∵ Sum of angles of a ΔBAD is 180°]
∠A + ∠C = 180° …(iii)
But, we know that the sum of angles of a quadrilateral is 360°.
∴ ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠B + ∠D + 180° = 360° [using (iii)]
⇒ ∠B + ∠D = 360° – 180° = 180°
Hence, ∠A + ∠C = 180° and
∠B + ∠D = 180°. Proved

Theorem 10.12:
If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.
Given: A quadrilateral ABCD in which ∠A + ∠C = 180°.
To prove: ABCD is a cyclic quadrilateral.
Construction: If possible, let ABCD be not a cyclic quadrilateral. Draw a circle, passing through non-collinear points D, A, B. Let this circle meet DE or DE produced at C. Join EB.

Proof: ∠A + ∠C = 180° (given) …(i)
Now ABED is a cyclic quadrilateral.
∠A + ∠E = 180° ….(ii)
[∵ opposite angles of a cyclic quadrilateral is 180°]
From (i) and (ii), we get
∠A + ∠C = ∠A + ∠E
⇒ ∠C = ∠E
This is not possible, since an exterior angle of a triangle can never be equal to its interior opposite angle.
So, ∠C = ∠E is possible only when C coincides with E.
Hence, the circle passing through D, A, B must pass through C also.
Hence, ABCD is a cyclic quadrilateral.
Proved

HBSE 9th Class Maths Notes Chapter 9 Areas of Parallelograms and Triangles

November 29, 2022 November 29, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 9 Areas of Parallelograms and Triangles Notes.

Haryana Board 9th Class Maths Notes Chapter 9 Areas of Parallelograms and Triangles

Introduction
In previous classes, we have studied the areas of plane figure such as, triangle, square, parallelogram, rhombus etc. We may recall that part of plane enclosed by a simple closed figure is called a planar region corresponding to that figure. The magnitude or measure of this planar region is called its area.

Area is always expressed with the help of a number such as 12 cm², 5 cm², 7 hectares etc. In this chapter, we will consolidate the knowledge about these formulae by studying some relationship between the areas of these geometric figures under the condition when they lie on the same base and between the same parallels.

Key Words
→ Base: In a geometry, a particular side or face of geometric figure, such as a triangle, parallelogram, cone etc.

→ Altitude: The perpendicular distance from a vertex to the opposite side is known as altitude.

→ Magnitude: Exter, size, bulk, amount, quantity, greateness. That which is extended in length, breadth and thickness. A thing which can be divided into parts.

→ Monotone: In one tone.

→ Planar region: Of, relating to or situated in a plane.
OR
Having a two dimensional characteristic.

→ Interior of a triangle: The part of the plane enclosed by a triangle is known as the interior of the triangle.

→ Area of a triangle = \(\frac{1}{2}\) × base × height

→ Area of a parallelogram = base × height

→ Area of a square = (side)² OR \(\frac{1}{2}\)d₁²,
[where d₁ is its diagonal]

→ Area of a rectangle = length × breadth

→ Area of a rhombus = base × height or \(\frac{1}{2}\)d₁ × d₂, [where d₁ and d₂ are its diagonals]

→ Area of a trapezium = \(\frac{1}{2}\)(sum of parallelside) × height

Basic Concepts
1. Polygonal Regions:
(a) Triangular region: The union of a triangle and its interior is called a triangular region. By the area of a triangle, we mean the area of its triangular region.

(b) Rectangular region: The part of the plane enclosed by a rectangle is called the interior of a rectangle and the union of a rectangle and its interior is a called a rectangular region. It can be represent as the union of two triangular regions.

(c) Polygonal region: A polygonal region is a plane figure that can be written as a union of a finite number of triangular regions, subject to a constraint. The constraint is that the triangular regions are non-over lapping. That means if two triangular region intersect then their intersetion is either vertex or edge of each of the triangular regions.

2. Area Axioms:
(a) Every polygonal region R has an area, measured in square units.
(b) (i) Congruent Area Axiom If two polygonal regions R₁ and R₂ are such that R₁ ≅ R₂, then
ar (R₁) = ar (R₂).
(ii) Area Monotone Axiom: If two polygonal regions are such that R₁ belongs to R₂ then
ar (R₁) ≤ ar (R₂).
(iii) Area Addition Axiom: If intersection of two polygonal regions R₁ and R₂, is a finite number of points and line segments such that R = R₁∪ R₂, then
ar (R) = ar (R₁) + ar (R₂).

(c) Rectangle Area Axiom: A rectangular region ABCD in which AB = a units and BC = b units then
ar (rectangle ABCD)= ab square units.

3. Figures on the Same Base and Between the Same Parallels:
Look at the following figures:

In figure 9.4 (i) parallelograms ABCD and ABQP have a common side AB. So we say that parallelogram ABCD and ABQP are on the same base AB. Similarly in figure 9.4 (ii) triangles ABC and DBC are on the same base BC, in figure 9.4 (iii) trapezium PQRS and parallelogram PQMN are on the same base PQ. In figure 9.4 (iv) parallelogram ABCD and triangle PDC are on the same base DC.

Now look at the figures (9.5):

We observe that in figure 9.5 (i) triangles CBA and DAB lie on the same base AB and, AB and CD are two parallel lines. In addition to the above, the vertex D of ADAB opposite to the base AB and vertex C of ACBA opposite to the base AB lie on a line CD parallel to the base AB. So we say that ADAB and ACBA are on the same base AB and between the same parallels CD and AB. Similarly in figure 9.5 (ii) trapezium ABCD and parallelogram ABPQ are on the same base AB between the same parallels AB and PD. In figure 9.5 (iii) trapezium ABCD and parallelogram ABFE are on the same base AB and between the same parallels AB and DF. In figure 9.5 (iv) parallelograms PQRS and PQMN are on the same base PQ and between the same parallels PQ and SM.

So, two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.

Now consider the following figures:

We observe that in figure 9.6 (i) triangles ABC and DBC are the same base BC but do not lie between the same parallels. In figure, 9.6 (ii) parallelogram ABCD and triangle PQR are lie between the same parallels AB and CD but they are not on the same base. In figure 9.6 (iii) the parallelograms ABCD and ABFE are on the same base AB but do not lie between the same parallels.

Parallelograms on the Same Base and Between the Same Parallels:
Theorem 9.1:
Parallelograms on the same base and between the same parallels are equal in area.

Given: Two parallelogram ABCD and ABFE are on the same base AB and are between the same parallels AB and DF.
To prove:
ar (||gm ABCD) = ar (||gm ABFE).
Proof: In ΔAED and ΔBFC, we have
AD = BC
[∵ AD and BC are the opposite sides of parallelogram ABCD]
∠DAE = ∠CBF
[∵ AD || BC and AE || BF,
∴ angle between AD and AE = angle between BC and BF]
and AE = BF
(∵ AE and BF are the opposite sides of the parallelogram ABFE)
ΔAED ≅ ΔBFC
(By SAS congruence rule)
⇒ ar (ΔAED) = ar (ΔBFC)
(By congruent area axiom) …(i)
Now
ar (||gm ABCD) = ar (☐ ABCE) + ar (ΔAED)
⇒ ar (||gm ABCD) = ar (☐ ABCE) + ar (ΔBFC) [using (i)]
⇒ ar (||gm ABCD) = ar (||gm ABFE)
Hence proved

Corollary 1: A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
Corollary 2: If a triangle and a parallelogram are on the same base and between the same parallels, then area of the triangle is equal to half the area of the parallelogram.

Corollary 3: Area of parallelogram = base × height.
Given: A parallelogram ABCD in which AB is the base and AP is the corresponding height.
To prove : Area of parallelogram ABCD = AB × AP.
Construction: Draw BQ ⊥ CD so that rectangle ABQP is formed.

Proof: Since parallelogram ABCD and rectangle ABQP are on the same base AB and between the same parallels AB and PC.
∴ ar (||gm ABCD) = ar (rectangle ABQP)
⇒ ar (||gm ABCD) = AB × AP
[∵ area of rectangle = length × breadth]
⇒ ar (||gm ABCD) = base × height.

Triangles on the Same Base and Between the Same Parallels:
Theorem 9.2:
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Given: Two triangles ABC and PBC on the same base BC and between the same parallels BC and AP.
To prove : ar (ΔABC) = ar (ΔPBC).
Construction: Through B draw BD || CP, intersecting PA produced in D and through C, draw CQ || BA, intersecting AP produced in Q.
Proof: Since, BC || AP and AB || CQ.
∴ ABCQ is a parallelogram.
Again, BC || AP and BD || CP
∴ BCPD is a parallelogram.
Now, parallelograms ABCQ and DBCP are on the same base BC and between. the same parallels BC and DQ.
∴ ar (||gm ABCQ) = ar (||gm DBCP) …(i)
But, a diagonal of a parallelogram divides it into two triangles are of equal areas.
∴ ar (ΔABC) = ar (||gm ABCQ) …(ii)
Similarly, ar (ΔPBC) = \(\frac{1}{2}\)ar (||gm DBCP) …(iii)
From (i), (ii) and (iii), we get
ar (ΔABC) = ar (ΔPBC)
Hence proved

Corollory: Area of a triangle = \(\frac{1}{2}\)base × height.
Given: A ΔABC in which AP is the height to the side BC.
To prove: ar (ΔABC) = \(\frac{1}{2}\)BC × AP.

Construction: Through A and C, draw AD || BC and CD || BA, intersecting each other at D.
Proof : Since, AD || BC and CD || BA
∴ ABCD is a parallelogram.
AC is the diagonal of parallelogram ABCD. It divides the parallelogram ABCD into two triangles of equal areas.
∴ ar (ΔABC) = \(\frac{1}{2}\)ar (||gm ABCD)…(i)
Since BC is a side of ||gm ABCD and AP is its corresponding height.
ar (||gm ABCD) = BC × AP …..(ii)
From (i) and (ii), we get
ar(ΔABC) = \(\frac{1}{2}\)BC × AP.
Hence proved.

Corollary: Area of a trapezium = \(\frac{1}{2}\) × (sum of the parallel sides) × (distance between them)
Given: A trapezium ABCD in which AB || CD and CP ⊥ AB. Let CP = h.
To prove :
ar (trapezium) = \(\frac{1}{2}\)(AB + CD) × h.
Construction: Draw AQ ⊥ CQ, intersecting CD produced at Q. Join AC.

Proof: ar (trapezium ABCD) = ar (ΔABC) + ar (ΔACD)
= \(\frac{1}{2}\) × AB × CP + \(\frac{1}{2}\) × CD × AQ
= \(\frac{1}{2}\) × (AB + CD) × h, [∵ CP = AQ = h]
∴ Area of a trapezium = \(\frac{1}{2}\)(sum of the parallel sides) × (distance between them)

Theorem 9.3:
Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
OR
If two triangles have equal areas and one side of the triangle is equal to one side of the other, then their corresponding altitudes are equal.
Given: Two triangles ABC and DEF are such that
ar (ΔABC) = ar (ΔDEF) and AB = DE and CM and FN are the altitudes corresponding to AB and DE respectively of the two triangles.
To prove: CM = FN.
Proof: In AABC, CM is the altitude to the corresponding side AB.

ar (ΔABC)= \(\frac{1}{2}\) × AB × CM …..(i)
[∵ ar (Δ) = \(\frac{1}{2}\) base × height]
Similarly,
ar (ΔDEF) = \(\frac{1}{2}\) × DE × FN …(ii)
But, ar (ΔABC) = ar (ΔDEF) (given)…(iii)
From (i), (ii) and (iii), we get
\(\frac{1}{2}\) × AB × CM = \(\frac{1}{2}\) × DE × FN
⇒ \(\frac{1}{2}\) × AB × CM = \(\frac{1}{2}\) × AB × FN
[given DE = AB]
⇒ CM = FN. Hence proved
As CM = FN (Proved above)
and CM ⊥ AB and FN ⊥ AB
∴ CF || AB

Hence, two triangles having the same base and equal area lie between the same parallels.

HBSE 9th Class Maths Notes Chapter 8 Quadrilaterals

November 29, 2022 November 29, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 8 Quadrilaterals Notes.

Haryana Board 9th Class Maths Notes Chapter 8 Quadrilaterals

Introduction
In previous chapters 6 and 7, we have studied about triangles and their properties. We know that figure obtained by joining three non-collinear points in pairs, is a triangle. If we have four points in a plane such that no three of them are collinear.

If we joined these points, we obtain a figure with four sides (see figure 8.1).

Key Words
→ Intercept: To cut off.

→ Internal bisector: Name sometimes given to the bisector of the interior angle of a triangle (or polygon).

→ External bisector: The bisector of the exterior angle of a triangle (or polygon) is sometimes called the external bisector of the angle of the triangle (or polygon).

Basic Concepts
1. Quadrilateral: A closed figure which bounded by four line segments is called quadrilateral. Figure 8.2 represents a quadrilateral with vertices A, B, C and D. It is denoted by the symbol quad. ABCD or ☐ABCD.

Four straight lines AB, BC, CD and DA are called the sides of the quadrilateral ABCD and ∠A, ∠B, ∠C and ∠D are the four angles of it.
The straight lines joining the opposite vertices of a quadrilateral are called its diagonals. Thus, a quadrilateral ABCD has two diagonals; namely AC and BD.

In a quadrilateral, we have the following definitions:
(a) Consecutive or adjacent sides: Two sides with common vertex are called the adjacent sides. In fig. 8.2 AB and BC, BC and CD, CD and DA and DA and AB are four adjacent sides of the quadrilateral ABCD.
(b) Opposite sides: Two sides of a quadrilateral having no common end point are called its opposite sides.
In fig. 8.2, AB and CD, AD and BC are two pairs of opposite sides of quad. ABCD.
(c) Consecutive angles: Two angles of a quadrilateral having a common side are called its consecutive angles.
In fig. 8.2, ∠A and ∠B, ∠B and ∠C, ∠C and ∠D and ∠D and ∠A are four pairs of consecutive angles of ☐ABCD.
(d) Opposite angles: Two angles of a quadrilateral having no common side are called its opposite angles.
In figure 8.2, ∠A and ∠C, ∠B and ∠D are two pairs of opposite angles of ☐ABCD.

2. Angle Sum Property of a Quadrilateral:
Let us now recall the angle sum property of a quadrilateral. The sum of the angles of a quadrilateral is 360°. This can be verified by drawing a diagonal and dividing the quadrilateral into two triangles.

Let ABCD be a quadrilateral and AC be a diagonal (see figure 8.3).
In ΔABC, we have
∠1 + ∠B + ∠2 = 180°…(1)
I. Sum of interior angles of a triangle 180°]
In ΔADC, we have
∠3 + ∠D + ∠4 = 180°…(ii)
Adding (i) and (ii), we get
∠1 + ∠3 + ∠B + ∠2 + ∠4 + ∠D = 180° + 180°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
[∵ ∠1 + ∠3 = ∠A and ∠2 + ∠4 = ∠C]
Hence, sum of angles of a quadrilateral is 360°.

3. Types of Quadrilaterals:
(i) Trapezium: A quadrilateral in which one pair of opposite sides is parallel but the other pair of opposite sides is non-parallel is called a trapezium.
The figure 8.4, represents a trapezium ABCD.

In which AB || CD and AD and BC are non-parallel sides. If the non-parallel sides AD and BC of the trapezium are equal, it is called an isosceles trapezium. In this case, ∠D = ∠C and ∠A = ∠B.

(ii) Parallelogram: A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.

The figure 8.5, represents a parallelogram ABCD, since AD || BC and AB || CD.

(iii) Rectangle: A parallelogram in which each angle is a right angle is called a rectangle.

The figure 8.6, represents a rectangle ABCD in which ∠A = ∠B = ∠C = ∠D = 90°.

(iv) Square: A parallelogram in which all sides are equal and each angle is equal to 90° is called a square.

The figure 8.7, represents a square ABCD in which AB = BC = CD = DA and ∠A = ∠B = ∠C = ∠D = 90°.

(v) Rhombus: A parallelogram in which all sides are equal is called a rhombus.

The figure 8.8, represents a rhombus. ABCD in which AB = BC = CD = DA.

(vi) Kite: A quadrilateral in which both pairs of adjacent sides are equal is called a kite.

The figure 8.9, represents a kite ABCD in which AD = DC and AB = BC.

Note:
(i) Every parallelogram is a trapezium but every trapezium is not a parallelogram.
(ii) Every square is a rectangle and also a rhombus but every rectangle or rhombus is not a square.
(iii) A kite is not a parallelogram.
(iv) Every square, rectangle and rhombus are parallelograms.

4. Properties of a Parallelogram:
Theorem 8.1:
A diagonal of a parallelogram divides it into two congruent triangles.
Given: ABCD is a parallelogram and AC its diagonal.
To prove :
ΔABC ≅ ΔCDA.

Proof: ABCD and AC cuts them
∠BAC = ∠DCA …..(i)
(Alternate interior angles)
and BC || AD and AC cuts them.
∠BCA = ∠DAC (Alternate interior angles)
Now, in ΔABC and ΔCDA, we have
∠BAC = ∠DCA, [From (i)]
AC = AC, (Common)
and ∠BCA = ∠DAC, [From (ii)]
ΔABC ≅ ΔCDA [By ASA congruence rule]
Hence proved

Theorem 8.2:
In a parallelogram, opposite sides are equal.
Given: ABCD is a parallelogram.
To prove: AB = CD and AD = BC.
Construction: Join AC.

Proof: ABDC and AC cuts them.
∠BAC = ∠DCA …(i) (Alternate interior angles)
Again AD || BC and AC cuts them.
∠BCA = ∠DAC …(ii)
(Alternate interior angles).
Now, in AABC and ACDA, we have
∠BAC = ∠DCA, [From (i)]
AC = AC, [Common]
and ∠BCA = ∠DAC, [From (ii)]
ΔABC ≅ ΔCDA (By ASA congruence rule)
AB = CD and AD = BC, (CPCT)
Hence proved

Theorem 8.3:
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Given: A quadrilateral ABCD in which AB = CD and BC = DA.
To prove: Quadrilateral ABCD is a parallelogram.

Construction: Join AC.
Proof: In ΔABC and ΔCDA, we have
AB = CD, (Given)
BC = AD, (Given)
and AC = AC, (Common)
ΔABC ≅ ΔCDA (By SSS congruence rule)
⇒ ∠BAC = ∠DCA, (CPCT)
⇒ AB || CD …..(i)
and ∠BCA = ∠DAC, (CPCT)
AD || BC …..(ii)
From (i) and (ii), we get
AB || CD and AD || BC
Hence, ABCD is a parallelogram. Hence Proved

Theorem 8.4:
In a parallelogram, opposite angles are equal.
Given: ABCD is a parallelogram.
To prove: ∠A = ∠C and ∠B = ∠D.
Proof : BC || AD and AB cuts them.

⇒ ∠A + ∠B = 180° …(1)
[Sum of co-interior angles is 180°]
Again, AB || CD and AD intersects them.
⇒ ∠A + ∠D = 180° …..(ii)
[Sum of co-interior angles is 180°]
From (i) and (ii), we get
∠A + ∠B = ∠A + ∠D
⇒ ∠B = ∠D
Similarly, ∠A = ∠C.
Hence proved

Theorem 8.5:
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Given: A quadrilateral ABCD in which ∠A = ∠C and ∠B = ∠D.
To prove: Quad. ABCD is a parallelogram.

Proof: We have
∠A = ∠C…(1)
and ∠B = ∠D…(ii)
Adding (i) and (ii), we have
∠A + ∠B = ∠C + ∠D…(iii)
In a quadrilateral ABCD, we have
∠A + ∠B + ∠C + ∠D = 360°,
[∵ Sum of angles of a quadrilateral is 360°]
⇒ ∠A + ∠B + ∠A + ∠B = 360°, [Using (iii)]
⇒ 2(∠A + ∠B) = 360°
⇒ ∠A + ∠B = \(\frac{360^{\circ}}{2}\) = 180°…(iv)
Now, line segments BC and AD cut by AB,
such that
∠A + ∠B = 180°
⇒ BC || AD
Again, ∠A + ∠B = 180°.
⇒ ∠C + ∠B = 180° [From (i) ∠A = ∠C]
Now, line segments AB and CD cut by BC,
∠B + ∠C = 180°
⇒ AB || CD
Thus, AB || CD and BC || AD.
Hence, ABCD is a parallelogram. Proved

Theorem 8.6:
The diagonals of a parallelogram bisect each other.
Given: A parallelogram ABCD such that its diagonals AC and BD intersect at O.
To prove: OA = OC and OB = OD.
Proof: Since ABCD is a parallelogram.

Therefore AB || CD and BC || AD.
Now, AB || CD and AC intersects.
∠BAC = ∠DCA, (Alternate interior angles)
∠BAO = ∠DCO …(1)
Again, AB || CD and BD intersects them.
∠ABD = ∠CDB, (Alternate interior angles)
⇒ ∠ABO = ∠CDO…(ii)
Now, in ΔAOB and ΔCOD, we have
∠BAO = ∠DCO, [From (i)]
AB = CD, (Opposite sides of parallelogram)
and ∠ABO = ∠CDO, [From (ii)]
∴ ΔAOB ≅ ΔCOD, (By ASA congruence rule)
⇒ OA = OC and OB = OD. Hence proved

Theorem 8.7:
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Given: ABCD is a quadrilateral in which the diagonals AC and BD bisect each other at O i.e., AO = OC and BO = OD.
To prove : Quadrilateral ABCD is a parallelogram.

Proof: In ΔAOB and ΔCOD, we have
AO = OC, (Given)
∠AOB = ∠COD, (vertically opposite angles)
and BO = OD, (Given)
∴ ΔAOB ≅ ΔCOD, (by SAS congruence rule)
⇒ ∠BAO = ∠DCO, (CPCT)
⇒ ∠BAC = ∠DCA
But these are alternate interior angles.
∴ AB || CD
and similarly, AD || BC
Hence, quad. ABCD is a parallelogram. Hence Proved

5. Another Conditions for a Quadrilateral to be a parallelogram:
In this chapter, we have studied many properties and verified that if in a quadrilateral any one of those properties is satisfied, then it will be a parallelogram. Now, we study a required condition for a quadrilateral to be a parallelogram. It is stated in the form of a theorem as given below.

Theorem 8.8:
A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
Given: A quadrilateral ABCD in which AB = CD and AB || CD
To prove: Quadrilateral ABCD is a parallelogram.
Construction : Join AC.

Proof: We have
AB || CD and AC intersects them.
⇒ ∠BAC = ∠DCA …(i)
(Alternate interior angles)
In ΔABC and ∠CDA, we have
AB = CD, (Given)
∠BAC = ∠DCA, [From (i)]
and AC = AC, (Common)
∴ ΔABC ≅ ΔCDA (By SAS congruence rule)
⇒ ∠BCA = ∠DAC, (CPCT)
But these are alternate interios angles.
∴ AD || BC
and AB || CD, (Given)
Hence, quad. ABCD is a parallelogram.
Hence Proved

The Mid Point Theorem: In this section we shall discuss some more properties of a triangle using properties of a parallelogram which is related to the mid points of sides of a triangle.

Theorem 8.9:
The line segment joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.

Given: D and E are the mid points of the sides AB and AC of the triangle ABC.
To prove : (i) DE || BC, (ii) DE = \(\frac{1}{2}\)BC.
Construction: Produce DE to F, such that EF = DE and join CF.
Proof: In ΔAED and ΔCEF, we have
AE = CE, (∵ E is the mid point of AC)
∠AED = ∠CEF, (Vertically opposite angles)
and DE = EF, (By construction)
∴ ΔAED ≅ ΔCEF, (By SAS congruence rule)
⇒ ∠1 = ∠2, (CPCT)
But these are alternate interior angles.
AB || FC ⇒ BD || FC …(i)
and AD = FC (CPCT)
But AD = BD
(∵ D is the mid point of AB)
∴ BD = FC ……(ii)
From (i) and (ii), we get
BD || FC and BD = FC
∴ BCFD is a parallelogram. [By theorem 8.8]
⇒ DF || BC,
(∵ Opposite sides of a parallelogram)
⇒ DE || BC
and DF = BC,
(∵ Oppsite sides of a parallelogram)
⇒ 2DE = BC,
[∵ DE = EF ⇒ DF = DE + EF = 2DE]
⇒ DE = \(\frac{1}{2}\)BC
Hence, DE || BC and DE = BC. Proved

Theorem 8.10:
The line drawn through the mid point of one side of a triangle, parallel to another side bisects the third side.

Given: A ΔABC, D is the mid point of AB and line through D parallel to BC intersects AC in E.
To prove: E is the mid point of AC.
Construction: Draw CM || BA intersects line l in F.
Proof: Since, AB || CM (By construction)
∴ BD || CF
and DF || BC (Given)
∴ BDFC is a parallelogram.
⇒ BD = CF,
(Opposite sides of a parallelogram)
But, BD = AD,
(∵ D is the mid point of AB)
∴ AD = CF …..(i)
Now, AB || CM and AC intersects them.
∠1 = ∠2,
(Alternate interior angles) …(ii)
In ΔADE and ΔCFE, we have
∠1 = ∠2 [From (i)]
∠AED = ∠CEF
(Vertically opposite angles)
and AD = CF [From (i)]
∴ ΔADE ≅ ΔCFE
(By AAS congruence rule)
⇒ AE = EC
Hence, E is the mid point of AC. Proved

HBSE 9th Class Maths Notes Chapter 7 Triangles

November 28, 2022 November 28, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 7 Triangles Notes.

Haryana Board 9th Class Maths Notes Chapter 7 Triangles

Introduction
In previous classes, we have learnt about triangles and their various properties. Recall that, a closed figure formed by three line segments, is called a triangle. It is usually denoted by the greek letter Δ (delta).

The figure 7.1 shows a triangle ABC denoted as a ΔABC. A triangle has three sides namely AB, BC and CA, three vertices namely A, B and C and has three angles namely ∠A, ∠B and ∠C.

In this chapter we shall learn about the congruence of triangles, criteria for congruence of triangles, some more properties of triangles and inequalities in a triangle.

Key Words
→ Equalities: The logical relation expressing identity or sameness and denoted by the sign “=”.

→ Axiom: A statement whose truth is either to be taken as self evident or to be assumed.

→ Inequalities: Refers to a statement that a quantity is either greater than or less than but not equal to them.

→ Perimeter: The perimeter of a two dimensional region is the length of its boundary.

→ Concentrate: On one point, to pay attention upon.

→ Slides: A gliding seat or surface for sliding.

→ Swing: To move with to and fro motion.

→ Hexagon: It is a polygon bounded by six straight line. (six-sided figure).

Basic Concepts
1. (a) Definition of Congruence: If two geometrical figures coincide exactly, by placing one over the other, the figures are said to be congruent to each other.
(i) Two line segments are congruent if and only if their lengths are equal. (See fig. 7.2).

Two line segments AB and CD are congruent, if AB = CD.

(ii) Two angles are congruent if and only if their measures are equal. (See fig. 7-3)

Two angles ABC and PQR are congruent if m∠ABC = m∠PQR.

(iii) Two circles are congruent if and only if lengths of their radii are equal. (See fig. 7.4)

Two circles C and C2 are congruent, if r1 = r2.

(iv) Two squares are congruent if and only if they have equal sides in length. (See fig. 7.5)

Two squares ABCD and EFGH are congruent, if AB = BC = CD = DA = EF = FG = GH = HE.

(b) Congruence of Triangles: The three sides and three angles of a triangle are called its six parts or six elements. Two triangles are said to be congruent to each other, if on placing one over the other, they exactly coincide.

Let ΔABC is placed over ΔDEF, such that vertex A falls on vertex D and side AB falls on side DE, then if the two triangles coincide with each other in a such a way that B falls on E, C falls on F; side BC coincides with side EF and side AC coincides with side DF, then the two triangles are congruent to each other. (See fig. 7.6)

The symbol used for congruence is “≅”.
ΔABC is congruent to ΔDEF is written as:
ΔABC ≅ ΔDEF.
In case of congruent ΔABC and ΔDEF as given above, the sides of the two triangles, which coincide with each other are called corresponding sides. Thus the sides AB, BC and CA of AABC are corresponding to the side DE, EF and FD of ADEF respectively. In the same way, the angles of the two traingles which coincide with each other, are called corresponding angles. Thus three pairs of corresponding angles are ∠A and ∠D, ∠B and ∠E and ∠C and ∠F.

It follows from the above discussion that if ΔABC coincide ΔDEF exactly such that the vertices of ΔABC fall on the vertices of ΔDEF, in the order
A ↔ D, B ↔ E, C ↔ F.

Then we have the following six equalities:
(i) AB = DE, BC = EF and CA = FD i.e. corresponding sides are equal (congruent).
(ii) ∠A = ∠D, ∠B = ∠E and ∠C = ∠F i.e. corresponding angles are equal (congruent). Hence, following is the general condition for congruence of two triangles:
Two triangles are congruent if and only if there exists a correspondence between their sides and vertices such that the corresponding sides and the corresponding angles of two triangles are equal.

If ΔABC is congruent to ΔDEF and the correspondence ABC ↔ DEF makes the six pairs of corresponding parts of the two triangles congruent, then we write ΔABC ≅ ΔDEF.
But it will not be correct to write ΔCBA ≅ ΔDEF because corresponding parts of ΔCBA and ΔDEF will not be equal i.e., congruent.
Note: The corresponding parts of congruent triangles (abbreviated as CPCT).

(c) Congruence relation in the set of all triangles: From the definition of congruence triangles and notation above, we obtain the following results:
(i) ΔABC ≅ ΔABC
[Congruence relation is reflexive]
(ii) If ΔABC ≅ ΔDEF,
then ΔDEF ≅ ΔABC,
[Congruence relation is symmetric]
(iii) If ΔABC ≅ ΔDEF
and ΔDEF ≅ ΔKLM,
then ΔABC ≅ ΔKLM,
[Congruence relation is transitive]

2. Criteria for Congruence of Triangles:
Axiom 7.1 (SAS congruence rule): Two triangles are congruent if two sides and included angle of one triangle are equal to the two sides and the included angle of the other triangle.
Given: ΔABC and ΔDEF in which AB = DE, AC = DF and ∠A = ∠D. (See fig. 7.7).

To prove : ΔABC ≅ ΔDEF.
Proof: Place ΔABC over ΔDEF such that vertex A falls on vertex D and side AB falls on side DE. Since side AB side DE. Therefore vertex B falls on vertex E. Since ∠A = ∠D, therefore AC will fall on DF.
But AC = DF and A falls on D.
∴ C will fall on F.
Thus AC will coincide with DF.
Now B falls on E and C falls on F.
∴ BC will coincide with EF.
∴ ΔABC coincide with ΔDEF.
Hence, by definition of congruence ΔABC ≅ ΔDEF.
Note: (1) This axiom refers two sides and the angle between them so it is known as (side-angle-side) or the “SAS congruence axiom” or the “SAS” criteria

(2) SAS axiom is not applicable when two sides and one angle ‘not included between these sides’ of one triangle are equal to two sides and one angle of the other triangle.

Theorem 7.1 (ASA congruence rule):
Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.
Given: Two triangles ABC and DEF, such that
∠B = ∠E, ∠C = ∠F and BC = EF.
To prove : ΔABC ≅ ΔDEF.
Proof: For proving the congruence of the two triangles there are three cases arise:
Case I: Let AB = DE. (See figure 7.8)

In this case, we have
AB = DE, (Assumed)
∠B = ∠E, (Given)
and BC = EF, (Given)
So, ΔABC ≅ ΔDEF, (By SAS axiom)

Case II: Let if possible AB > DE, so we can take a point P on AB such that PB = DE. Now consider ΔPBC and ΔDEF.

In ΔPBC and ΔDEF, we have
PB = DE, (By construction)
∠B = ∠E, (Given)
and BC = EF, (Given)
So, ΔPBC ≅ ΔDEF,(By SAS axiom)
⇒ ∠PCB = ∠DFE, (CPCT)
But ∠ACB = ∠DFE, (Given)
∴ ∠ACB = ∠PCB
This is possible only when P coincide with A, therefore AB must be equal to DE.
Thus, in ΔABC and ΔDEF, we have
AB = DE, (As proved above)
∠B = ∠E, (Given)
and BC = EF, (Given)
So, ΔABC ≅ ΔDEF, (By SAS axiom)

Case III: If AB < DE, we can choose a point M on DE such that ME = AB and repeating the arguments as given in Case (II), we can conclude that AB = DE and so we can prove that

ΔABC ≅ ΔDEF
Hence, in all three cases, we have
ΔABC ≅ ΔDEF. Proved

Corollary (AAS congruence rule):
If any two angles and non-included side of one triangle are equal to the corresponding angles and side of another triangle, then two triangles are congruent.
Given: Two triangles ABC and DEF, such that
∠A = ∠D, ∠B = ∠E
and BC = EF.
To prove : ΔABC ≅ ΔDEF
Proof: Since, we know that sum of the interior angles of a triangle is 180°.

∴ ∠A + ∠B + ∠C = ∠D + ∠E + ∠F = 180°…(i)
But ∠A = ∠D and ∠B = ∠E, (given) …….(ii)
∠A + ∠B + ∠C = ∠A + ∠B + ∠F,
[From (i) and (ii)]
⇒ ∠C = ∠F
In ΔABC and ΔDEF, we have
∠B = ∠E, (Given)
∠C = ∠F, (As proved above)
and BC = EF, (Given)
So, ΔABC ≅ ΔDEF
(By ASA congruence rule)

Some Properties of a Triangle: In chapter 6 we have studied the triangle and their kinds. Recall that, if a triangle has two sides equal, then it is called an isosceles triangle. Now we shall study some results related to isosceles triangles.

Theorem 7.2:
Angles opposite to equal
Given: In ΔABC, AB = AC.
To prove: ∠B = ∠C. sides of an isosceles triangle are equal.

Construction: Draw AD, the bisector of ∠A which meets BC on D.
Proof: In ΔABD and ΔACD, we have
AB = AC, (Given)
∠BAD = ∠CAD, (By construction)
and AD = AD, (Common)
ΔABD ≅ ΔACD,
(By SAS congruence rule)
⇒ ∠B = ∠C. Hence proved.

Theorem 7.3:
[Converse of Theorem 7.2]: The sides opposite to equal angles of a triangle are equal.
Given: In ΔABC,
∠B = ∠C.
To prove: AB = AC.

Construction: Draw AD perpendicular to BC. It meets BC on D.
Proof: In ΔABD and ΔACD, we have
∠ABD = ∠ACD, (Given)
∠ADB = ∠ADC,
(By construction each = 90°)
and AD = AD, (Common)
∴ ΔABD ≅ ΔACD, (By AAS congruence rule)
⇒ AB = AC (CPCT)
Hence proved

Some More Criteria for Congruence of Triangles :
Theorem 7.4 (SSS congruence rule): If three sides of one triangle are equal to the three sides of another triangle, then two triangles are congruent.
Given: ΔABC and ΔPQR such that AB = PQ, BC = QR and AC = PR.
To prove : ΔABC = ΔPQR.

Construction: Suppose BC is the longest side of ΔABC. Draw QS and RS such that ∠SQR = ∠ABC and ∠SRQ = ∠ACB. Join P to S.
Proof: In ΔABC and ΔSQR, we have
BC = QR, (Given)
∠ABC = ∠SQR, (By construction)
and ∠ACB = ∠SRQ, (By construction)
∴ ΔABC ≅ ΔRQS,
(By ASA congruence rule)
⇒ ∠BAC = ∠QSR, (CPCT) …(i)
AB = QS, (CPCT)
and AC = SR, (CPCT)
Now AB = PQ (given)
and AC = PR (given) …(ii)
AB = QS
and AC = SR (as proved above)…(iii)
From (ii) and (iii), we get
PQ = QS and PR = SR
⇒ ∠QPS = ∠QSP…(iv)
and ∠RPS ∠RSP…(v)
Adding (iv) and (v), we get
∠QPS + ∠RPS = ∠QSP + ∠RSP
⇒ ∠QPR = ∠QSR
and ∠BAC = ∠QSR, [From (i)]
∴ ∠BAC = ∠QPR …(vi)
Now in ΔABC and ΔPQR, we have
AB = PQ, (Given)
∠BAC = ∠QPR, [From (vi)]
and AC = PR, (Given)
∴ ΔABC = ΔPQR (By SAS congruence rule).
Hence proved

Theorem 7.5.
[RHS congruence rule]: If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.
Given: In ΔABC and ΔDEF,
∠B = ∠E = 90°, AC = DF and BC = EF.
To prove: ΔABC ≅ ΔDEF.

Construction: Produce DE to G so that GE = AB. Join G to F.
Proof: In ΔABC and ΔGEF, we have
AB = GE, (by construction)
∠ABC = ∠GEF, [Each = 90°]
and BC = EF, (Given)
∴ ΔABC ≅ ΔGEF, (By SAS congruence rule)
⇒ ∠A = ∠G, (CPCT) …(i)
AC = GF, (CPCT) …(ii)
Now, AC = DF, (Given)
AC = GF, [from (ii)]
∴ DF = GF
⇒ ∠D = ∠G …(iii)
(Angles opposite to equal sides are equal)
From (i) and (iii), we get
∠A = ∠D…(iv)
In ΔABC and ΔDEF, we have
∠B = ∠E, (Each = 90°)
∠A = ∠D, [from (iv)]
∴ ∠C = ∠F …(v)
[third angles of the triangles]
Now in ΔABC and ΔDEF, we have
BC = EF, (Given)
∠C = ∠F, [From (v)]
and AC = DF, (Given)
∴ ΔABC = ΔDEF, (by SAS congruence rule)
Hence proved
Note: RHS stands for Right angle-Hypotenuse-Side.

Inequalities in a Triangle: In this section, we shall learn about some inequality relations among sides and angles of a triangle. Consider scalene triangle ΔABC such that AB = 7 cm, BC = 6 cm and AC = 5 cm. Now measure the three angles A, B and C with the help of protector, we observe that:

(i) AB > AC and ∠C > ∠B
i.e., the longer side has greater angle opposite to it.
(ii) AC < BC and ∠B < ∠A.
i.e., shorter side has smaller angle opposite to it.
(iii) AB > BC and ∠C > ∠A
i.e., angle opposite the larger side is greater.
Below we give some important results of inequalities in a triangle.
It is stated in the form of a theorem.

Theorem 7.6:
If two sides of a triangle are unequal, the angle opposite to the longer side is larger or (greater). (See fig. 7.62)

Given: A triangle ABC in which AB > AC.
To prove: ∠ACB > ∠ABC.
Construction: From AB, cut AD = AC.
Join C and D.
Proof: In ΔACD, we have
AC = AD, (By construction)
⇒ ∠ACD = ∠ADC …..(1)
(Angles opposite to equal sides are equal)
In ΔBDC, we have
Exterior ∠ADC > ∠B
(exterior angle of a triangle is always greater than each its opposite interior angle)
⇒ ∠ACD > ∠B, [From (i)]
⇒ ∠ACB > ∠B, [∵ ∠ACB > ∠ABC]
⇒ ∠ACB > ∠ABC

Theorem 7.7:
In any triangle, the side opposite to the larger (greater) angle is longer.
Given: A triangle ABC in which ∠ABC > ∠ACB.
To prove: AC > AB.

Proof: There are following possibilities.
(i) AC = AB
(ii) AC < AB
(iii) AC > AB
There is only one possibilities must be true.
(i) If AC = AB
⇒ ∠ABC = ∠ACB
[Angles opposite to equal sides are equal]
It is impossible, for it is given that
⇒ ∠ABC > ∠ACB
(ii) If AC < AB
⇒ AB>AC
⇒ ∠ACB > ∠ABC
This is also impossible, being contrary to hypothesis.
(iii) Since AC is neither less than nor equal to AB.
∴ AC must > AB.
Hence AC > AB. Proved.

Theorem 7.8:
The sum of any two sides of a triangle is greater than the third side. (See fig. 7.64).

Given: A triangle ΔABC.
To prove : (i) AB + AC > BC
(ii) AB + BC > AC
(iii) AC+ BC > AB.
Construction: Produce BA to D such that AD = AC. Join C to D

Proof: In ΔACD, we have
AD = AC (By construction)
⇒ ∠ADC = ∠ACD …..(i)
∠BCD > ∠ACD,
[∵ ∠BCD = ∠ACB + ∠ACD]
⇒ ∠BCD > ∠ADC, [from (i), ∠ACD = ∠ADC]
⇒ ∠BCD > ∠BDC
⇒ BD > BC (By theorem 7·7)
⇒ AB + AD > BC
⇒ AB + AC > BC (∵ AD = AC)
Similarly,
AB + BC > AC and AC + BC > AB.
Hence proved

Corollary 1.
The difference between the lengths of any two sides of a triangle is always less than the third side.

HBSE 9th Class Maths Notes Chapter 6 Lines and Angles

November 26, 2022 November 26, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 6 Lines and Angles Notes.

Haryana Board 9th Class Maths Notes Chapter 6 Lines and Angles

Introduction
In previous chapter, we have already studied axioms, postulate and some terms allied to geometry. In this chapter, we shall study the angles and their properties and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points.

An architect cannot draw the plan of the multistoried building without the knowledge of the properties of parallel lines, intersecting lines and angles.

In science, to study the refraction property of light when a ray enters from one medium to other medium, we use the properties of parallel lines and intersecting lines. We have to represent forces by direct line segments to study the net effect of the forces on the body.

Key Words
→ Architect: One who designs building and supervises their construction.

→ Congruent: Having the same size and shape.

→ Adjacent: Lying near to.

→ Intersect: Two geometrical figures or curves are said to intersect if they have at least one point in common.

→ Converse: The converse of an implication p ⇒ q is the implication q ⇒ p. If an implication is true, then its converse may or may not be true.

→ Concurrent: A number of lines are said to be concurrent, if there is a point through which they all pass.

→ Incident Ray: The incident ray is the ray of light that strikes the surface before reflection, transmission or absorption.

→ Reflected Ray: A ray extending outward from a point.

→ Angle of Incidence: The angle between the direction of an approaching emission and the normal to the surface upon which it is incident.

→ Angle of Reflection: The angle between the direction of propagation of a reflected emission and the normal to the reflecting surface.

1. Basic Terms and Definitions : (a) Angle: An angle is formed when two rays or two line segments have a common end point. The two rays which form an angle are called the arms and the point at which they meet is called the vertex of the angle.

The adjacent figure represent an angle ABC or ∠ABC or ∠CBA or simply B. AB and BC are the arms of the angle and their common point B is the vertex.

∠ABC divides the plane containing it into two parts:
(i) Interior of an angle
(ii) Exterior of an angle.
(i) Interior of an angle: The interior of an angle ABC is the set of all points in its plane, which lie on the same side of BC as A and also on the same side of AB as C. In the adjoining figure, P is a point in the interior of an angle ABC.

(ii) Exterior of an angle : The exterior of an ∠ABC is the set of all points in its plane, which do not lie on the angle or in its interior.
In the adjoining figure, the point Q is exterior of the ∠ABC.
Note: Point R lies on the angle ∠ABC.

(b) Measure of an angle: The amount of turning of a ray from the initial position of BC to BA is called the measure of ∠ABC which is written as m∠ABC. The unit of angle measure is a standard angle, called a degree.

If the measure of an angle ABC is x degree, then we write
m∠ABC = x°.
Note: 1° = 60′ (i.e., 60 minutes),
1′ = 60″ (60 seconds)

(c) Congruent or equal angles: If the measures of the angles are same then they said to be equal or congruent.
For examples:
If m∠ABC = m∠PQR,
then ∠ABC ≅ ∠PQR
or if ∠ABC ≅ ∠DEF,
then m∠ABC = m∠DEF.

(d) Types of angles: (i) Right angle: The angle which measures equal to 90°, is called a right angle. In the figure 6.4, ∠PQR is a right angle.

(ii) Acute angle: The angle which measures less than one right angle (i.e., less than 90°) is called an acute angle.
In the figure 6.5, ∠PQR is an acute angle.

(iii) Obtuse angle: The angle which measures more than 90° and less than 180° is called an obtuse angle. In the figure 6.6, ∠PQR is an obtuse angle.

(iv) Straight angle: The angle which measures equal to two right angles (i.e., equal to 180°) is called a straight angle.
In the figure 6.7, ∠PQR is a straight angle.

(v) Reflex angle: The angle which measures more than 180° and less than 360° is called a reflex angle.
In the figure 6.8, ∠PQR is a reflex angle.

(vi) Complete angle: The angle which measures is 4 right angles (i.e., 360°) is called a complete angle.
In the figure 6.9, ∠QPR is a complete angle.

(vii) Complementary angles: If the sum of two angles is 90°, they are called complementary angles eg., if ∠a + ∠b= 90°, then ∠PQS and ∠RQS are complementary angles.

(viii)Supplementary angles: If the sum of two angles is 180°, they are called. supplimentary angles eg., if ∠x + ∠y = 180°, then ∠PQS and ∠RQS are supplementary angles.

(e) Bisetor of an angle: A ray is called the bisector of an angle if it divides the angle into two equal parts.

In the figure 6.12, ray QS is the bisector of ∠PQR, then
∠PQS = ∠RQS.

2. Pairs of Angles: (i) Adjacent angles: Two angles are called adjacent angles, if:
(i) they have the same vertex.
(ii) they have a common arm.
(iii) other arms of these angles are on opposite sides of the common arm.

In the figure 6.13, ∠AOC and ∠BOC have the common vertex O and have a common arm OC and other arms OA and OB lie on the opposite sides of the common arm OC. Therefore, ∠AOC and ∠BOC are adjacent angles.

(ii) Linear pair angles: Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays.

In the figure 6.14, ∠AOC and ∠BOC are two adjacent angles and OA and OB are two opposite rays. Therefore, ∠AOC and ∠BOC form a linear pair of angles.

Axiom 6.1: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.
Recall that when the sum of two adjacent angle is 180°, then they are called a linear pair of angles.

In the figure 6.15, ray OB stands on the line AC. It forms adjacent angles ∠AOB and ∠COB.
∵ ∠AOB + ∠COB = 180° and OA and OC are opposite rays. Thus, two angles form a linear pair of angles. The above axiom can be stated in the reverse ways as below:
Axiom 6.2: If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.

In the figure 6.16, ∠AOB and ∠COB are adjacent angles, such that
∠AOB + ∠COB = 180°
⇒ ∠AOC = 180°,
[∵ ∠AOC = ∠AOB + ∠COB]
⇒ ∠AOC is a straight line.
Hence, AOC is a straight line.
The two axioms above together is called the linear pair axiom.

(iii) Vertically opposite angles: Two angles are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays.

In the figure 6.17, two lines AB and CD intersect at O, then two pairs of vertically opposite angles are formed.
(i) ∠AOC and ∠BOD
(ii) ∠AOD and ∠BOC.

Theorem 6.18.
If two lines intersect each other, then the vertically opposite angles are equal.
Given: Two lines AB and CD intersect at a point O.
To prove : (i) ∠AOC = ∠BOD
(ii) ∠AOD = ∠BOC.
Proof. Since ray OA stands on line CD.
∴∠AOC + ∠AOD = 180°,
(Linear pair axiom) …….(i)
Ray OD stands on line AB.
∴ ∠AOD + ∠BOD = 180°,
(Linear pair axiom) …….(ii)
From (i) and (ii), we have
∠AOC + ∠AOD = ∠AOD + ∠BOD
⇒ ∠AOC = ∠BOD
Similarly, we can prove that
∠AOD = ∠BOC. Hence proved

Parallel Lines and a Transversal: A line which intersects two or more given lines at distinct points is called a transversal of the given lines.

In the figure 6.32, AB and CD are two lines and a transversal EF intersects them at P and Q respectively. The three lines determine eight angles, four angles at P namely ∠1, ∠2, ∠3, ∠4 and remaining four angles at Q namely ∠5, ∠6, ∠7 and ∠8.

In the figure, ∠2, ∠3, ∠5 and ∠8 are called interior angles and ∠1, ∠4, ∠6 and ∠7 are called exterior angles. We classify these eight angles in the following groups:
(a) Corresponding angles: (i) ∠1 and ∠5, (ii) ∠4 and ∠8, (iii) ∠2 and ∠6, (iv) ∠3 and ∠7.
(b) Alternate interior angles: (i) ∠2 and ∠8, (ii) ∠3 and ∠5.
(c) Alternate Exterior angles: (i) ∠1 and ∠7, (ii) ∠4 and ∠6.
(d) Interior angles on the same side of the transversal: (i) ∠2 and ∠5, (ii) ∠3 and ∠8.
Interior angles on the same side of the transversal are also referred to as consecutive interior angles or allied angles or co-interior angles.

Some Important Relations
If AB || CD and transversal XY cuts them (see in the figure 6.33) then

(i) Corresponding angles are equal i.e., ∠1 = ∠5, ∠4 = ∠8, ∠2 = ∠6 and ∠3 = ∠7.
(ii) Alternate interior angles are equal i.e., ∠2 = ∠8 and ∠3 = ∠5.
(iii) Alternate exterior angles are equal i.e., ∠1 = ∠7 and ∠4 = ∠6.
(iv) Co-interior angles are supplementary i.e., ∠2 + ∠5 = 180° and ∠3 + ∠8 = 180°.
From this we conclude the following axioms:
Axiom 6.3: If a transversal intersects two parallel lines, then each pair of corresponding angles are equal. It is called corresponding angles axiom also.
Axiom 6.4: If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other. It is called converse of corresponding angles axiom.

Theorem 6.2:
If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Given: A transversal line XY cuts the parallel lines AB and CD at P and Q respectively.

To prove : ∠2 = ∠8 and ∠3 = ∠5.
Proof :
∠4 = ∠2 ……(i)
(vertically opposite angles)
∠4 = ∠8 ……(ii)
(corresponding angles axiom)
From (i) and (ii), we get
∠2 = ∠8
Again, ∠1 = ∠3 ……(iii)
(vertically opposite angles)
∠1 = ∠5 ……(iv)
(corresponding angles axiom)
From (iii) and (iv), we get
∠3 = ∠5
Hence, ∠2 = ∠8 and ∠3 = ∠5.
Proved

Theorem 6.3:
(Converse of theorem 6.2) : If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

Given: A transversal line XY intersects two lines AB and CD at P and Q respectively such that ∠2 and ∠8 are a pair of alternate interior angles and ∠2 = ∠8.
To prove : AB || CD.
Proof :
∠2 = ∠8, (given) ……(i)
∠2 = ∠4 …….(ii)
(vertically opposite angles)
From (i) and (ii), we get
∠4 = ∠8
Thus, a pair of corresponding angles 24 and 28 are equal. Therefore, by converse of corresponding angle axiom, we have
AB || CD.
Hence proved

Theorem 6.4:
If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Given: A transversal line XY intersects two parallel lines AB at P and CD at Q making two pairs of consecutive interior angles ∠2, ∠5, ∠3 and ∠8.
To prove :
∠2 + ∠5 = 180° and ∠3 + ∠8 = 180°.
Proof: Ray PB stands on line XY.
∴ ∠3 + ∠4 = 180
(Linear pair axiom)
But ∠8 = ∠4
(corresponding angles axiom)
∴ ∠3 + ∠8 = 180° ……(i)
Now, ray QC stands on line XY.
∴ ∠5 + ∠6 = 180°,
(Linear pair axiom)
But ∠6 = ∠2,
(corresponding angles axiom)
∴ ∠5 + ∠2 = 180°
⇒ ∠2 + ∠5 = 180° ……(ii)
Hence, ∠2 + ∠5 = 180°
and ∠3 + ∠8 = 180°.
Proved.

Theorem 6.5.
(Converse of theorem 6.4): If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

Given: A transversal line XY intersects two lines AB at P and CD at Q such that ∠2 and ∠5 are pair of consecutive interior angles and ∠2 + ∠5 = 180°.
To prove : AB || CD.
Proof: Since, Ray PA stands on line XY.
∴ ∠1 + ∠2 = 180°
(Linear pair axiom) …..(i)
∠2 + ∠5 = 180° (Given) …..(ii)
From (i) and (ii), we get
∠1 + ∠2 = ∠2 + ∠5
⇒ ∠1 = ∠5
Thus, a pair of corresponding angles ∠1 and ∠5 are equal.
Therefore, by converse of corresponding angles axiom, we have
AB || CD. Hence proved.

Lines Parallel to Same Line:
Theorem 6.6:
Lines which are parallel to the same line are parallel to each other.

Given: l || m and n || m.
To prove : l || m
Construction: Draw a transversal lines t for the lines l, m and n.
Proof: Since, l || m (Given)
∠1 = ∠2 …….(i)
(Corresponding angles axiom)
Since, n || m (Given)
∠3 = ∠2 …….(ii)
(Corresponding angles axiom)
From (i) and (ii), we get
∠1 = ∠3
But these are corresponding angles.
⇒ l || n,
[By converse of corresponding angles]
Hence proved

1. Triangle: A plane figure bounded by three straight lines, is called a triangle. It is usually denoted by the Greek letter Δ (delta).
The figure, given along side, shows a triangle ABC (ΔABC) bounded by three sides AB, BC and CA. Triangle ABC has three vertices, namely A, B and C and it has three angles, namely ∠A, ∠B and ∠C.
Note: If side BC of ΔABC is produced upto any point D, the angle ∠ACD so formed, is called exterior angle at C, where as the angles A and B are called its interior opposite angles.

2. Kinds of Triangles: (a) With regard to their sides:
(i) Equilateral triangle: If all the sides of a triangle are equal, it is called an equilateral triangle.
In the given figure, ΔABC is an equilateral triangle because AB = BC = CA. Also, all the angles of an equilateral triangle are equal to each other and so each angle = 60°.

(ii) Isosceles triangle: If at least two sides of a triangle are equal, it is called an isosceles triangle. In the given figure, ΔABC is an isosceles triangle, because AB = AC.

In an isosceles triangle, two angles of opposite equal side are also equal.

(iii) Scalene triangle: If all the sides of a triangle are unequal, it is called an scalene triangle.
The figure below shows a scalene triangle because AB ≠ BC ≠ AC.

In a scalene triangle, all its angles are also unequal.

(b) With regard to their angles: (i) Acute angled triangle: When each angle of a triangle is acute (i.e., less than 90°), it is called an acute angled triangle.
The given figure 6.60, shows an acute angled triangle, because each angle of the triangle is less than 90°.

(ii) Right angled triangle: When one angle of a triangle is right angle (ie., equal to 90°), it is called right angled triangle, the side opposite to the right angle is called its hypotenuse and the remaining two sides are called its legs.

The given figure 6.61, shows a right angled triangle because ∠B = 90°. In triangle ABC, AC is hypotenuse and AB and BC are its legs. Each of the other two angles of the right triangle is acute.

(iii) Obtuse angled triangle: When one angle of a triangle is obtuse (i.e., greater than 90° but less than 180°), it is called an obtuse angled triangle. The given figure 6-62, shows an obtuse angled triangle, because B is obtuse.

In an obtuse angled triangle, each of the other two angles is acute.

3. Some Important Terms Related to the Traingle: (i) Median : The median of a triangle, corresponding to any side, is the line joining the mid point of that side with the opposite vertex. e.g.,

AL is the median corresponding to side BC

CN is the median corresponding to side AB

A triangle has three medians and all three medians are always concurrent i.e., they intersect each other at one point only.

(ii) Centroid: The point of intersection of all the three medians of a triangle is called its centroid. In the given tigure 6-66, three medians AD, BE and CF intersect at the point G. Therefore, G is the centroid of the triangle ABC.

(iii) Altitude: An altitude of a triangle, corresponding to any side, is the length of perpendicular drawn from the opposite vertex to that side. e.g.,

AL is the altitude corresponding to side BC

BM is the altitude corresponding to side AC

CN is the altitude co-rresponding to side AB A triangle has three altitudes and all three altitudes are always concurrent i.e., they intersect each other at one point only.

(iv) Ortho centre: The point of intersection of all the three altitudes of a triangle is called ortho centre.
In the given figure 6.69, O is the ortho centre of triangle ΔABC.

4. Angle sum property of a triangle: In previous classes, we have studied that sum of the angles of a triangle is 180°. We can prove this statement using the theorem related to parallel lines.

Theorem 6.7:
The sum of the angles of a triangle is 180°.

Given: A triangle ΔABC and ∠1 + ∠2 + ∠3 are its angles
To prove : ∠1 + ∠2 + ∠3 = 180°.
Construction: Through A, draw a line PQ parallel to BC.
Proof: PQ || BC and AB is the transversal.
∠2 = ∠4 ……(i)
(Alternate interior angles)
Again, PQ || BC and AC is the transversal.
∴ ∠1 = ∠5, ……(ii)
(Alternate interior angles)
Adding (i) and (ii), we get
∠2 + ∠1 = ∠4 + ∠5
⇒ ∠1 + ∠2 + ∠3 = ∠4 + ∠3 + ∠5,.
(Adding ∠3 on both sides)
But, ∠4 + ∠3 + ∠5 = 180°,
(Linear pair axiom)
∴ ∠1 + ∠2 + ∠3 = 180°
Hence, the sum of three angles of a triangle is 180°.
Hence Proved

Theorem 6.8:
If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Given: A ΔABC whose side BC has been produced to D such that ∠4 is its exterior angle.
To prove : ∠4 = ∠2 + ∠3.
Proof: We know that,
Sum of the angles of a triangle is 180°,
(By theorem 6.7)
⇒ ∠1 + ∠2 + ∠3 = 180°…(i)
∠1 + ∠4 = 180°…(ii)
(Linear pair axiom)
From (i) and (ii), we get
∠1 + ∠4 = ∠1 + ∠2 + ∠3
⇒ ∠4 = ∠2 + ∠3. Hence proved

Corollary: An exterior angle of a triangle is greater than either of its interior opposite angles.

Proof: We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles (By theorem 6.8)
∴ ∠4 = ∠2 + ∠3
∴ ∠4 > ∠2 and ∠4 > ∠3
Hence, an exterior angle of a triangle is greater than each interior opposite angle.

HBSE 9th Class Maths Notes Chapter 5 Introduction to Euclid’s Geometry

November 25, 2022 November 25, 2022 / Class 9 / By Bhagya

Haryana State Board HBSE 9th Class Maths Notes Chapter 5 Introduction to Euclid’s Geometry Notes.

Haryana Board 9th Class Maths Notes Chapter 5 Introduction to Euclid’s Geometry

Introduction
The word ‘geometry’ is derived from the Greek words ‘geo’, means the ‘earth’, and ‘metrein’, means ‘to measure’. So, it appears to have originated from the need for measuring land. Many ancient civilization of Egypt, Babylonia, India, Greece, China have studied the ‘geometry’ in various forms. They short out the several practical problems which required the development of geometry in various ways.

In Indus Valley Civilization, the ratio length : breadth : thickness of the bricks was taken as 4 : 2 : 1.

Construction of altars (or vedis) and fireplaces for performing Vedic rites were originated in Vedic period. Square and circular altars were used for household rituals, while shape of altars were the combination of rectangles, triangles and trapeziums. They are used for public worship. The ‘sriyantra’ consists of nine interwoven isosceles triangles which are arranged in such a way that they produce 43 subsidiary triangles. A Greek mathematician, Thales, (640 BC) gave the first known proof of the statement that a circle is bisected (cut into two equal parts) by its diameter.

Pythagoras (572 BC) was the most famous pupil of Thales discovered many geometric properties and developed the theory of geometry to a great extent. This process continued till 300 BC. His important theorem is; In right triangle, the square of hypotenuse is equal the sum of the square of base and square of perpendicular.

Euclid (300 BC) was a mathematics teacher who was born in Alexandria in Greece introduced the method of proving mathematical results by using deductive logical reasoning and the previously proved results. The geometry of plane figure is known as “Euclidean Geomery.”

The Indian Mathematician, Aryabhatta (born 476 AD) worked out the area of an isosceles triangle, the volume of a pyramid and approximate value of π.

Brahmagupta (born 598 AD) discovered the formula for finding the area of cyclic quadrilateral.

Bhaskara II (born 1114 AD) gave a dissection proof of Pythagoras’ theorem.

In this chapter, we shall study the Euclid’s approach to geometry.

Key Words
→ Version – Translation.

→ Altars – Raised plateform used for sacrifice or religious offering.

→ Rituals – Way of performing religious services.

→ Rite – Religious ceremony.

→ Subsidiary – Accessory.

→ Allied – Related.

→ Extent – Large space or tract size.

→ Volume – Space occupied, mass.

→ Cyclic Quadrilateral – A quadrilateral whose four vertices lie on a circle.

→ Proof – The course of reasoning which establishes the truth of a statement.

→ Logic – The science or art of reasoning.

→ Physical – Belonging to physics.

→ Universal – Belonging to all.

→ Fact – A thing known to be true.

→ Coincide – To occupy the same space.

→ Proposition – A statement of something to be done.

→ Interpreted – To exposed the meaning of.

→ Demonstrate – To prove with certainly.

→ Consequence – result.

→ Superposed – To put or place over or above.

→ Assumption – A supposition, thing assumed.

Basic Concepts
Euclid’s Definitions, Axioms and Postulates:
(a) Euclid summarised the work of ‘geometry’ as definitions. He listed 23 definitions in Book 1 of ‘Elements’. A few of them are given below:

A point is that which has no part.
A line is breadthless length.
The ends of a line are points.
A straight line is a line which lies evenly with the points on itself.
A surface is that which has length and breadth only.
The edges of surface are lines.
A plane surface is a surface which lies evenly with the straight lines on itself.

In these definitions, we observe that, point, line, breadth, length, plane etc. are undefined terms. The only thing is that we can represent them intuitively or explain them with the help of ‘physical models.’

Starting with these definitions, Euclid assumed certain properties, which were not to be proved. These assumptions are actually “obvious universal truths”. He divided them into two types: axioms and postulates. He used term ‘postulate’ for the assumptions that were specific to geometry. Common notions (often called axioms), on the other hand, were. assumption used throughout mathematics and not specifically linked to geometry.

(b) Axioms: The basic facts which are taken for granted, without proof, are called axioms.
Some of the Euclid’s axioms (not in order) are given below:
(i) Things which are equal to the same thing are equal to one another.
i.e., If p = q and r = q, then p = r.
(ii) If equals are added to equals, the wholes are equal.
i.e., If p = q, then p + r = q + r.
(iii) If equals are subtracted from equals, the remainders are equal.
i.e., If p = q, then p – r = q – r.
(iv) Things which coincide with one. another are equal to one another.
(v) The whole is greater than the part ie., If p>q, then there exists r(r ≠ 0) such that
P = q + r.
(vi) Things which are double of the same things are equal to one another. i.e., If p = q, then 2p = 2q.
(vii) Things which are halves of the same things are equal to one another.
Le. If p = q, then \(\frac{p}{2}=\frac{q}{2}\)

(c) Postulate: A statement whose validity is accepted without proof, is called a postulate.
Euclid gave five postulates stated as below:
Postulate 1: A straight line may be drawn from any one point to any other point.
This postulate tells us that at least one straight line passes through two distinct points which is shown in the figure below:

Axiom 5.1. Given two distinct points, there is a unique line that passes through them.

In the given figure, P and Q are two distinct points. Out of all lines passing through the point P, there is exactly one line ‘l’ which also passes through Q. Also out of all lines passing through the point Q, there is exactly one line l which also passes through the point P. Hence, we find there is a unique line l which passes through the points P and Q.

Postulate 2: A terminated line can be produced indefinitely.

Postulate 3: A circle can be drawn with any centre and any radius.
Postulate 4: All right angles are equal to one another.
Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

For example, in the figure line PQ falls on the lines AB and CD such that ∠1 + ∠2 < 180° on the left side of PQ. Therefore, the lines AB and CD will eventually intersect on the left side of PQ.

(d) Statement: A sentence which can be judged either true or false is called a statement.
For example:
(i) The sum of the angles to a quadrilateral is 360°, is a true statement.
(ii) The sum of the angles of a triangle is 90°, is a false statement.
(iii) y + 5 > 9 is a sentence but not a statement.

Theorem: In mathematics, a theorem is a statement that has been proven on the basis of previously established statements, such as other theorems and previously accepted statements, such as axioms. The derivation of a theorem is often interpreted as a proof of the truth of the resulting expression, but different deductive systems can yield other interpretations, depending on the meanings of the derivation rules. The proof of a mathematical theorem is logical argument demonstrating that the conclusion.

For example:
(i) The area of a triangle is equal to half its base, multiplied by its attitude.
(ii) An exterior angle of a triangle is equal to sum of its opposite interior angles.
Corollary: A proposition, whose truth can easily be deducted from a preceeding theorem, is called its corollary.

(e) Some terms allied to Geometry:
(i) Point: A point is a mark of position, which has no length, no breadth and no thickness. We represent a point by a capital letters A, B, C, P, Q etc., as shown in the figure.

(ii) Plane: A plane is a every surface such that point of the line joining any two points on it, lies on it.
The surface of the top of the table, surface of the smooth blackboard and surface of a sheet of paper are the some close examples of a plane surfaces are limited in extent but geometrical plane extends endlessly in all directions.

(iii) Line: A line has length but no breadth or thickness. A line has no ends points. A line has no definite length. It can be extended infinitely in both the directions. The given figure shows line \(\bar{AB}\).

Some times we shall denote lines by small letters such as l, m, n, p etc.

(iv) Ray: It is a straight line which starts from a fixed point and moves in the same direction. A ray has one endpoint and it has no definite length.
The given figure shows a ray \(\overrightarrow{A B}\).

(v) Line segment: It is a straight line with its both ends fixed. The given figure shows a line segment \(\bar{AB}\), with fixed ends A and B. A line has a definite length.

So, a line segment is the shortest distance between two fixed points.

(vi) Collinear points: Three or more than three points are said to be collinear, if they lie on the straight line.
The given figure shows the collinear points P, Q, R while A, B, C are non-collinear.

(vii) Parallel lines: The straight lines which lie in the same plane and do not meet at any point on producing on either side, are called parallel lines.
If l and m are parallel lines in a plane, we write l || m.

The distance between the two parallel lines always remains same.

(viii) Intersecting lines: Two lines having a common point are known as intersecting lines.
The common point (O, in figure 5.18) is known as the point of intersection.

(ix) Concurrent lines: In the figure 5.19, the lines pass through the same point. In this case, lines are called concurrent lines.

(x) Congruence of line segment: If the given two line segments are of the same size, they are said to be ‘congruent. Two lines AB and CD are congruent, if the trace copy of one can be superposed on the other so as to cover it completely and exactly.
Symbolically, \(\bar{AB}\) ≅ \(\bar{CD}\) or line segment AB ≅ line segment CD.

Congruence relation in the set of all line segments:
(i) AB ≅ AB
(ii) AB ≅ CD ⇒ CD ≅ AB
(iii) AB ≅ CD and CD ≅ EF, then AB ≅ EF.

Theorem 5.1.
Two distinct lines cannot have more than one point in common.
Proof. Here, we are given two lines I and m. We need to prove that they have only one point in common. If possible, let P and Q two points common to the given lines l and m. Then the line I passes through the points P and Q and line m passes through the points P and Q. But this assumption clashes with the axiom that one and only one line can pass through two distinct points. So, our supposition that two lines pass through two distinct points is wrong.
Hence, two distinct lines cannot have more than one point in common.

Equivalent Versions of Euclid’s Fifth Postulate:
There are several equivalent versions of fifth postulate of Euclid. One of them is ‘Playfair’s Axiom’ (given by Scottish mathematician John Playfair in 1729), as stated below:

Playfair’s Axiom: “For every line l and for every point P not lying on l, there exists a unique line m passing through P and parallel to l.”
From figure 5.52, we observe that all the lines passing through the point P, only line m is parallel to line l.

This result can also be stated as: Two distinct intersecting lines cannot be parallel to the same line.