Author name: Bhagya

Haryana State Board HBSE 7th Class Hindi Solutions Vasant Chapter 8 शाम एक किशान Textbook Exercise Questions and Answers.

Haryana Board 7th Class Hindi Solutions Vasant Chapter 8 शाम एक किशान

HBSE 7th Class Hindi शाम एक किशान Textbook Questions and Answers

कविता से

शाम एक किसान शब्दार्थ HBSE Class 7 प्रश्न 1.
इस कविता में शाम के दृश्य को किसान के रूप में दिखाया गया है-यह एक रूपक है। इसे बनाने के लिए पाँच एकरूपताओं की जोड़ी बनाई गई है। उन्हें उपमा कहते हैं। पहली एकरूपता आकाश और सोफे में दिखाते हुए कविता में ‘आकाश का साफा’ वाक्यांश आया है। इसी तरह तीसरी एकरूपता नदी और चादर में दिखाई गई है, मानों नदी चादर-सी हो। अब आप दूसरी, चौथी और पाँचवी एकरूपताओं को खोजकर लिखिए।

• इस प्रकार की दूसरी एकरूपता दिखाने वाली उपमाओं को कविता से खोजकर सूची बनाइए।

उत्तर :
एकरूपता दिखाने वाली उपमाएँ
1. पहाड़ – किसान
2. नदी – चादर
3. पलाश का जंगल – दहकती अँगीठी
4. अंधकार – भेड़ों का गल्ला।

शाम एक किसान कविता का सारांश HBSE Class 7 प्रश्न 2.
शाम का दृश्य अपने घर की छत या खिड़की से देखकर बताइए
(क) शाम कब से शुरू हुई ?
(ख) तब से लेकर सूरज डूबने में कितना समय लगा?
(ग) इस बीच आसमान में क्या-क्या परिवर्तन आए?
उत्तर :
(क) सायं छह बजे शुरू हुई।
(ख) आधे घंटे का समय लगा।
(ग) आसमान में लालिमा छा गई, आकाश पीला-पीला हो गया और अंत में सूरज आँखों से ओझल हो गया।

शाम एक किसान के प्रश्न उत्तर HBSE Class 7 प्रश्न 3.
मोर के बोलने पर कवि को लगा जैसे किसी ने कहा हो-‘सुनते हो’। आगे दिए गए पक्षियों की बोली सुनकर उन्हें भी एक या दो शब्दों में बाँधिए –
कबूतर, कौआ, मैना, तोता, चील, हंस.
उत्तर:

1. कबूतर-गुटरगूं
2. कौआ-काँव-काँव
3. चील-चें-चें

कविता से आगे

Class 7 Hindi Chapter 8 HBSE प्रश्न 1.
इस कविता को चित्रित करने के लिए किन-किन रंगों का प्रयोग करना होगा ?
उत्तर :
पीला, सुनहरा, सफेद, लाल रंगों का प्रयोग करना होगा।

शाम एक किसान कविता की व्याख्या HBSE Class 7 प्रश्न 2.
शाम के समय ये क्या करते हैं, पता लगाइए और लिखिए
पक्षी, खिलाड़ी, फलवाले, माँ, पेड़-पौधे, पिताजी, किसान, बच्चे.
उत्तर :

1. पक्षी : घोंसलों की ओर लौट जाते हैं।
2. खिलाड़ी : खेलना बंद कर देते हैं।
3. फल वाले : फल बेचते हैं।
4. माँ : बच्चों के लिए खाना पकाने की तैयारी करती है।
5. पेड़-पौधे : सोने लगते हैं।
6. पिताजी : घर लौट आते हैं।
7. किसान : खेतों से लौट आते हैं। गाय-भैंसों का दूध निकालते हैं।
8. बच्चे : गृह कार्य करते हैं।

शाम एक किसान के शब्दार्थ HBSE Class 7 प्रश्न 3.
हिंदी के एक प्रसिद्ध कवि सुमित्रानंदन पंत ने संध्या का वर्णन इस प्रकार किया है
संध्या का झुटपुट, कौओं का झुटपुट, चहक रही है चिड़ियाँ, टी वी टी टुर-टुर,
ऊपर दी गई कविता और सर्वेश्वरदयाल की कविता में आपको क्या मुख्य अंतर लगा ? लिखिए।
उत्तर :
सर्वेश्वर दयाल सक्सेना ने संध्याकालीन दृश्य का किसान के रूप में रूपक बाँधकर प्रस्तुत किया है जबकि कवि पंत ने संध्याकाल में विभिन्न पक्षियों की आवाजों को सुनवाया है।

HBSE 7th Class Hindi शाम एक किशान Important Questions and Answers

अति लघुत्तरात्मक प्रश्न

शाम एक किसान प्रश्न उत्तर HBSE Class 7 प्रश्न 1.
कविता में पहाड़ को किस रूप में दर्शाया गया है?
उत्तर :
कविता में पहाड़ को साफा बाँधे किसान के रूप में दर्शाया गया है।

शाम एक किसान कविता के प्रश्न उत्तर HBSE Class 7 प्रश्न 2.
इस किसान के घुटनों पर क्या है?
उत्तर :
इस किसान के घुटनों पर चादर है।

शाम एक किसान Question Answer HBSE Class 7 प्रश्न 3.
किसे अंगीठी बताया गया है और क्यों?
उत्तर :
पलाश के जंगल को अँगीठी बताया गया है क्योंकि पलाश के लाल-लाल फूल आग की तरह दहकते प्रतीत होते हैं।

शाम एक किसान व्याख्या HBSE Class 7 प्रश्न 4.
‘भेड़ों के गल्ले-सा’ किसे कहा गया है?
उत्तर :
अंधकार को भेड़ों के गल्ले-सा कहा गया है।

Chapter 8 Shaam Ek Kisan HBSE Class 7 प्रश्न 5.
मोर की आवाज़ कैसी लगी?
उत्तर :
जैसे किसी के कहा हो-‘सुनते हो?’

लघुत्तरात्मक प्रश्न

Shaam Ek Kisan Class 7th HBSE प्रश्न 1.
किसको किस रूप में चित्रित किया गया है ?
पहाड़, नदी, पलाश के जंगल को, डूबते सूरज को, आकाश को
उत्तर :

1. पहाड़ को : किसान के रूप में
2. नदी को : चादर के रूप में
3. पलाश के जंगल को : दहकती अँगीठी के रूप में।
4. डूबते सूरज को : चिलम के रूप में।
5. आकाश को : किसान के साफे के रूप में।

शाम एक किसान HBSE Class 7 प्रश्न 2.
चिलम औंधी क्यों हो गई?
उत्तर :
सूरज चिलम के रूप में था। अंधकार होते ही सूरज डूब गया और इसके साथ चिलम का आकार भी गायब हो गया। इससे ऐसा लगा कि किसान की चिलम औंधी हो गई है।

शाम एक किशान काव्यांशों की सप्रसंग व्याख्या

1. आकाश का …………… गल्ले-सा।

शब्दार्थ: आकाश = आसमान (Sky)सूरज = सूर्य (Sun), बहकना = जलना (To burn)

सप्रसंग व्याख्या:
प्रसंग :
प्रस्तुत पक्तियाँ हमारी पाठ्यपुस्तक ‘वसंत भाग-2’ में संकलित कविता ‘शाम : एक किसान’ से ली गई हैं। इस कविता के रचयिता सर्वेश्वरदयाल सक्सेना हैं। इसमें कवि ने जाई की शाम में पहाड़ को एक किसान के रूप में चित्रित किया है।

व्याख्या :
कवि जाड़े की शाम के प्राकृतिक दृश्य का चित्रण करते हुए कहता है कि इस दृश्य में पहाड़ एक किसान की तरह दिखाई देता है। आकाश उसके सिर के साफे के समान है। सूरज चिलम के समान प्रतीत होता है जिसे वह दम लगाकर खींचता – है। पहाड़ के नीचे नदी बहती रहती है और यह नदी पहाड़ रूपी किसान के घुटनों पर पड़ी चादर के समान प्रतीत होती है। सर्दी भगाने के लिए इस किसान के पास अंगीठी दहक रही है। पलाश के लाल-लाल फूलों को अंगीठी का रूप दिया गया है। पूरब दिशा में क्षितिज पर होता अंधकार झुंड में बैठी भेड़ों जैसा प्रतीत होता है। पश्चिम दिशा में डूबता सूरज इस पहाड़ रूपी किसान की चिलम है। यह पूरा दृश्य शांत है।

विशेष :
1. प्रकृति का मनोहारी चित्रण किया गया है।
2. ‘भेड़ों का गल्ला-सा’ में उपमा अलंकार है।

अर्थग्रहण संबंधी प्रश्न :
1. कॉन, किस रूप में बैठा है?
2. नदी किसके समान प्रतीत होती है?
3. किसे दहकती अंगीठी बताया गया है?
4. अंधकार कैसा लग रहा है?
उत्तर:
1. पहाड़ एक किसान के रूप में बैठा है। उसने सिर पर साफा बाँध रखा है तथा चिलम पी रहा है।
2 पहाड़ के नीचे की नदी किसान के घुटनों पर पड़ी चादर के समान प्रतीत होती है।
3. पलाश के लाल-लाल फूलों को दहकती अंगीठी बताया गया है।
4. अंधकार दूर सिमटा भेड़ों के गल्ले के समान लग रहा है।

बहुविकल्पी प्रश्न सही उत्तर चुनकर लिखिए

1. इस कविता के रचयिता कौन है?
(क) सर्वेश्वरदयाल सक्सेना
(ख) सुमित्रानंदन पंत
(ग) रघुवरदयाल सक्सेना
(घ) मैथिलीशरण गुप्त
उत्तर :
(ख) सुमित्रानंदन पंत

2. पहाड़ किस रूप में है?
(क) किसान के रूप में
(ख) साफा बाँधे हुए
(ग) चिलम खींचता
(घ) ये सभी रूप
उत्तर :
(घ) ये सभी रूप

3. कौन-सी अंगीठी दहक रही है?
(क) पलाश के जंगल की
(ख) कोयले की
(ग) लकड़ी की
(घ) प्रकृति की
उत्तर :
(क) पलाश के जंगल की

4. ‘भेड़ों के गल्ले-सा’ में कौन-सा अलंकार है?
(क) अनुप्रास
(ख) उपमा
(ग) रूपक
(घ) यमक
उत्तर :
(ख) उपमा

2. अचानक – बोला …………………….. छा गया।

सप्रसंग व्याख्या
प्रसंग :
प्रस्तुत पंक्तियाँ हमारी पाठ्यपुस्तक ‘वसंत भाग-2’ में संकलित कविता ‘शाम : एक किसान’ से ली गई हैं। इस कविता के रचयिता सर्वेश्वरदयाल सक्सेना है। इसमें कवि ने जाडे की शाम में पहाड़ को एक किसान के रूप में चित्रित किया है।

व्याख्या :
कवि जाड़े की संध्या का चित्रण करते हुए बताता है कि अभी तक सारा वातावरण शांत था। अचानक मोर बोल उठता है, मानो किसी ने आवाज़ लगाई हो-‘सुनते हो?’ इसके बाद प्रकृति में सारा दृश्य घटना में बदल जाता है–चिलम उलट जाती है, आग बुझ जाती है, धुआँ उठने लगता है.सूरज डूब जाता है, शाम ढल जाती है और रात का अँधेरा छा जाता है।

विशेष : संध्याकालीन वातावरण का सजीव चित्रण किया गया है।

अर्थग्रहण संबंधी प्रश्न :
1. इस कविता के रचयिता कौन हैं?
2 इस कविता में किस वातावरण का चित्रण है?
3. दृश्य में क्या परिवर्तन आ जाता है?
उत्तर:
1. इस कविता के रचयिता सर्वेश्वरदयाल सक्सेना हैं।
2. इस कविता में जाड़े की एक संध्या के वातावरण का चित्रण है।
3. मोर के बोलते ही अर्थात् संध्या के होते ही सारा प्राकृतिक वातावरण बदल जाता है-सूरज डूब जाता है, रात का अंधेरा छाने लगता है। लगता है, चिलम औंधी हो गई है, आग बुझ गई है।

बहुविकल्पी प्रश्न सही उत्तर चुनकर लिखिए

1. अचानक कौन बोल उठा?
(क) मोर
(ख) किसान
(ग) तांता
(घ) किसान की पत्नी
उत्तर :
(क) मोर

2. “चिलम औंधी होना’ किसका प्रतीक है?
(क) सूरज डूबने का
(ख) सूरज चमकने का
(ग) दिन छपने का
(घ) रात होने का
उत्तर :
(क) सूरज डूबने का

3. कौन-सा शब्द ‘सूरज’ का पर्यायवाची नहीं है
(क) रवि
(ख) दिनकर
(ग) भास्कर
(घ) शशि
उत्तर :
(घ) शशि

शाम एक किशान Summary in Hindi

शाम एक किशान कवि-परिचय

प्रश्न : कवि सर्वेश्वरदयाल सक्सेना के जीवन एवं साहित्य का संक्षिप्त परिचय दीजिए।
उत्तर :
जीवन एवं साहित्यिक परिचय :
आधुनिक हिंदी साहित्य की नई कविता के कवियों में श्री सर्वेश्वर दयाल सक्सेना का नाम अग्रगण्य है। आपका जन्म 15 सितंबर, 1927 को उत्तर ‘प्रदेश स्थित बस्ती में हुआ। आपने एंग्लों संस्कृत विद्यालय, बस्ती से हाईस्कूल की परीक्षा उत्तीर्ण की और क्वीस कॉलेज वाराणसी में प्रवेश लिया।

तत्पश्चात् आपने इलाहाबाद विश्वविद्यालय से एम. ए. की परीक्षा उत्तीर्ण की। शिक्षा क्षेत्र में अध्यापन करने के उपरांत आपने आकाशवाणी में सहायक प्रोड्यूसर के रूप में कार्य किया। सर्वेश्वरदयाल सक्सेना ने कुछ दिन दिनमान के प्रमुख उपसंपादक पद पर भी कार्य किया। बाद में बच्चों की पत्रिका ‘पराग’ का संपादन कार्य भी किया। आपका देहावसान 25 सितंबर, 1984 को हुआ।

सर्वेश्वरदयाल सक्सेना ‘तीसरा तारसप्तक’ के कवि हैं। छायावाद के उपरांत नई कविता के कवियों में महत्त्वपूर्ण स्थान रखते हैं। सुमित्रानंदन पंत ने सक्सेना जी की साहित्य कला दृष्टि की सराहना की है तथा उन्हें सहज-प्रयत्न कवि और नए कवियों में कलाबोध का पारखी बताया है। उनके काव्य में रोमानियत और सम-सामयिकता पाई जाती है। वे समष्टि चेतना और व्यष्टि चेतना के प्रति सजग हैं। आपने कविता के नए-नए विषयों की ओर ध्यान दिया है। सक्सेना जी ने जीवन के विविध पक्षों को अपनी कविता में नए रंग-ढंग से व्यक्त किया है।

भाषा-शैली :
सर्वेश्वर दयाल सक्सेना खड़ी बोली के कवि हैं। उन्हें छंदमुक्त कविता अभीष्ट है। आपकी भाषा सरल और स्पष्ट है। कविता में कहीं-कहीं तीखे व्यंग्य पाए जाते हैं। कविता में बिंब योजना और प्रतीकात्मकता विशेष रूप से परिलक्षित होती है। भाषा प्रसाद गुण युक्त है। भाषा में तत्सम शब्दों के साथ स्थानीय शब्दों का अदभुत समन्वय है तथा कल्पना की प्रधानता है। चित्रात्मकता के साथ मानवीकरण उनकी अनूठी विशेषता है।

रचनाएँ : सर्वेश्वरदयाल सक्सेना बहुमुखी प्रतिभा के धनी साहित्यकार हैं। आपकी साहित्यिक कृतियाँ निम्न प्रकार हैं

काव्य कृतियाँ : काठ की घंटियाँ, बाँस का पुल, एक सूनी नाव, गर्म हवाएँ, जंगल का दर्द, खूटियों पर टंगे लोग आदि।

नाटक : बकरी, कल फिर भात आएगा, लड़ाई, अब गरीबी हटाओ, राज-बाज बहादुर और रानी रूपमती आदि।

उपन्यास : पागल कुत्तों का मसीहा, सोया हुआ जल। लेख संग्रह : चरचे और चरखे आदि।

शाम एक किशान कविता का सार

‘शाम : एक किसान’ शीर्षक कविता सर्वेश्वरदयाल सक्सेना द्वारा रचित है। इस कविता में कवि ने जाड़े की शाम के दृश्य को एक किसान के रूप में दर्शाया है। यह कविता एक रूपक है। इस कविता में शाम के दृश्य और किसान के बीच एकरूपता दिखाई गई है।

जिस प्रकार किसान सिर पर साफा बाँधकर चिलम खींचता या पीता है, उसी प्रकार पहाड़ (जो एक किसान के रूप में है) आकाश रूपी साफा बाँधकर और सूरज की चिलम को खींचता दिखाई देता है। जिस प्रकार किसान के घुटनों पर चादर पड़ी रहती है उसी प्रकार शाम के घुटनों पर नदी पड़ी रहती है। सर्दी भगाने के लिए किसान अँगीठी जलाता है।

शाम की अँगीठी है-पलाश के जंगल। इनमें लाल-लाल फूल आग की तरह दहकते हैं। अंधकार भेड़ों के गल्ले के समान दूर छिपकर बैठा रहता है। तभी अचानक बोल उठा, मानो कोई आवाज़ दे रहा हो। उस समय वह चिलम उल्टी हो गई, धुआँ उठने लगा और सूरज डूब गया तथा चारों ओर अंधेरा छा गया।

Haryana State Board HBSE 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
Solution:
Let required ratio be k : 1 and line 2x + y – 4 divides the line segment.
Join A(2, – 2) and B(3, 7) at the point P(x, y).
Then co-ordinates of P are.

⇒ x = \(\frac{3 k+2}{k+1}\)
y = \(\frac{7 k-2}{k+1}\)
Thus, the co-ordinates of P are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\).
Since, point P lies on the given line 2x + y – 4 = 0.
Therefore, 2 × \(\left(\frac{3 k+2}{k+1}\right)+\frac{7 k-2}{k+1}\) – 4 = 0
\(\frac{6 k+4+7 k-2-4 k-4}{k+1}\) = 0
\(\frac{9 k-2}{k+1}\) = 0
9k – 2 = 0
k = \(\frac{2}{9}\)
k : 1 = 2 : 9
Hence, required ratio is 2 : 9.

Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
Let A(x, y), B(1, 2)and C(7, 0) are given points
Since, the given points are coimear.
Therefore,Area of ∆ABC = 0
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2)] = 0
⇒ \(\frac{1}{2}\) [2x – y + 7y – 14] = 0
⇒ \(\frac{1}{2}\) [2x + 6y – 14] = 0
⇒ 2x + 6y – 14 = 0
⇒ x + 3y – 7 = 0
Hence, required relation is x + 3y – 7 = 0.

Question 3.
Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
Solution :
Let P(x, y) be the centre of a circle which passes through A(6, – 6), B(3, – 7) and C(3, 3)
∴ PA = PB = PC (equal radii of circle)
Now PA = PB
⇒ PA² = PB²
⇒ (6 – x)² + (- 6 – y)² = (3 – x)² + (- 7 – y)²
⇒ 36 + x² – 12x + 36 + y² + 12y = 9 + x² – 6x + 49 + y² + 14y

⇒ x² + y² – 12x + 12y + 72 = x² + y² – 6x + 14y + 58
⇒ x² + y² – 12x + 12y – x² – y² + 6x – 14y = 58 – 72
⇒ – 6x – 2y = – 14
⇒ 3x + y = 7 ……………..(1)
and PB = PC
⇒ PB² = PC²
⇒ (3 – x)² + (- 7 – y)² = (3 – x)² + (3 – y)²
⇒ (3 – x)² + (- 7 – y)² – (3 – x)² = (3 – y)²
⇒ (- 7 – y)² = (3 – y)²
⇒ 49 + y² + 14y = 9 + y² – 6y
⇒ y² + 14y – y² + 6y = 9 – 49
20y = – 40
y = – \(\frac{40}{20}\)
y = – 2
Substituting the value ofy in equation (1), we get
3x – 2 = 7
3x = 7 + 2 = 9
x = \(\frac{9}{3}\)
x = 3
Hence, centre of the circle is (3, – 2).

Question 4.
The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the co-ordinates of the other two vertices.
Solution :
Let the opposite vertices of a square be A(- 1, 2) and C(3, 2) and unknown vertex be B(x, y) then
AB = BC (sides of square)
⇒ AB² = BC²
⇒ (x + 1)² + (y – 2)² = (3 – x)² + (2 – y)²
⇒ x² + 1 + 2x + y² + 4 – 4y = 9 + x² – 6x + 4 + y² – 4y
⇒ x² + 2x – 4y + y² + 5 = x² + y² – 6x – 4y + 13

x² + 2x – 4y + y² – y² + 6x + 4y = 13 – 5
8x = 8
x = \(\frac{8}{8}\)
x = 1 ……………….(1)
In right ∆ABC, we get
AC² = AB² + BC² [By Pythagoras theorem]
⇒ (3 + 1)² + (2 – 2)² = (x + 1)² + (y – 2)² + (3 – x)² + (2 – y)²
⇒ 16 = x² + 1 + 2x + y² + 4 – 4y + 9 + x² – 6x + 4 + y² – 4y
⇒ 16 = 2x² + 2y² – 4x – 8y + 18
⇒ 2x² + 2y² – 4x – 8y = 16 – 18 = – 2
⇒ x² + y² – 2x – 4y= – 1
⇒ (1)² + y² – 2 × 1 – 4y = – 1 [∵ From (1) x = 1]
⇒ 1 + y² – 4y – 2 = – 1
y² – 4y = – 1 – 1 + 2
⇒ y(y – 4) = 0
y= 0 or y -4 = 0
y = 0 or y = 4
Hence, required vertices of the square are (1, 0) and (1, 4).

Question 5.
The class X students of a secondary school in Krishinagar have been alloted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaininq area of the plot.
(i) Taking A as origin, find the co-ordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of ∆ PQR if C is the origin ? Also, calculate the area-s of the triangles in these cases. What do you observe?
Solution :
(i) Taking A as origin the coordinates of the vertices of P, Q and R are (4, 6), (3, 2) and (6, 5) respectively.
(ii) Taking C as the origin, co-ordinates of P, Q and R are (12, 2), (13, 6) and (10, 3).
In 1st condition area of ∆PQR = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]
= \(\frac{1}{2}\) [4 × (- 3) + 3 × (- 1) + 6 × 4]
= \(\frac{1}{2}\) (- 12 – 3 + 24)
= \(\frac{1}{2}\) × 9
= \(\frac{9}{2}\) square units.

In 2nd condition area of ∆PQR = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
= \(\frac{1}{2}\) [12 × 3 + 13 × 1 + 10 × (- 4)]
= \(\frac{1}{2}\) (36 + 13 – 40)
= \(\frac{1}{2}\) × 9
= \(\frac{9}{2}\) square units
Hence, coordinates of ∆PQR.
(i) Taking A as origin and taking AD as x-axis, AB as y-axis are (4, 6), (3, 2) and (6, 5).
(ii) Taking C as origin and taking CB x-axis, CD as y-axes are (12, 2), (13, 6) and (10, 3)
Areas in two cases are \(\frac{9}{2}\) square units.

Question 6.
The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\). Calculate the area of the ∆ADE and compare it with the area of ∆ABC. (Recall theorem 6.2 and theorem 6.6).
Solution:
We have,
\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\)
⇒ \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}=4\)
⇒ \(\frac{\mathrm{AB}}{\mathrm{AD}}-1=\frac{\mathrm{AC}}{\mathrm{AE}}-1\) = 4- 1
⇒ \(\frac{\mathrm{AB}-\mathrm{AD}}{\mathrm{AD}}=\frac{\mathrm{AC}-\mathrm{AE}}{\mathrm{AE}}\) = 3
⇒ \(\frac{\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{EC}}{\mathrm{AE}}\) = 3
⇒ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{1}{3}\)

⇒ AD : DB = AE : EC = 1 : 3.

Point D divides the AB in the ratio 1 : 3.
Co-ordinates of D

Point E divides the AC in the ratio 1 : 3.
Co-ordinates of E

Area of ∆ABC
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [4(5 – 2) + 1(2 – 6) + 7(6 – 5)]
= \(\frac{1}{2}\) (4 × 3 + 1 × – 4 + 7 × 1)
= \(\frac{1}{2}\) (12 – 4 + 7)
= \(\frac{1}{2}\) × 15
= \(\frac{15}{2}\) square units
\(\frac{\text { Area of } \triangle \mathrm{ADE}}{\text { Area of } \triangle \mathrm{ABC}}=\frac{\frac{15}{32}}{\frac{15}{2}}\)
\(\frac{\text { Area of } \triangle \mathrm{ADE}}{\text { Area of } \triangle \mathrm{ABC}}=\frac{15}{32} \times \frac{2}{15}=\frac{1}{16}\)
Hence, Area of ∆ADE = \(\frac{15}{32} \) square units
and Area of ADE : Area of ∆ABC = 1 : 16.

Question 7.
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC.
(i) The median from A meets BC at D. Find the co-ordinates of the point D.
(ii) Find the co-ordinates of the points P on AD such that AP : PD = 2 : 1.
(iii) Find the co-ordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe ?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC, find the co-ordinates of the centroid of the triangle.
Solution :
(i) D is the mid-point of BC.

Co-ordinates of point D are \(\left(\frac{6+1}{2}, \frac{5+4}{2}\right)\) i.e., \(\left(\frac{7}{2}, \frac{9}{2}\right)\).

(ii) P divides the AD in the ratio 2 : 1.
Co-ordinates of point P = \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)
= \(\left(\frac{2 \times \frac{7}{2}+1 \times 4}{2+1}, \frac{2 \times \frac{9}{2}+1 \times 2}{2+1}\right)\)
= \(\left(\frac{7+4}{3}, \frac{9+2}{3}\right)=\left(\frac{11}{3}, \frac{11}{3}\right)\)

(iii) E is the mid-point of AC.
Co-ordinates of point E = \(\left(\frac{4+1}{2}, \frac{2+4}{2}\right)=\left(\frac{5}{2}, 3\right)\)

Q divides BE in the.ratio 2 : 1.
Co-ordinates of point Q.
= \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1^{\prime} y_2+m_2 y_1}{m_1+m_2}\right)\)
= \(\left(\frac{2 \times \frac{5}{2}+1 \times 6}{2+1}, \frac{2 \times 3+1 \times 5}{2+1}\right)\)
= \(\left(\frac{5+6}{3}, \frac{6+5}{3}\right)=\left(\frac{11}{3}, \frac{11}{3}\right)\)
F is the mid-point of AB.
Co-ordinates of point F = \(\left(\frac{4+6}{2}, \frac{2+5}{2}\right)=\left(5, \frac{7}{2}\right)\)

Point R divides the CF in the ratio 2 : 1.
Co-ordinates of point R = \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)
= \(\left(\frac{2 \times 5+1 \times 1}{2+1}, \frac{2 \times \frac{7}{2}+1 \times 4}{2+1}\right)\)
= \(\left(\frac{10+1}{3}, \frac{7+4}{3}\right)=\left(\frac{11}{3}, \frac{11}{3}\right)\)

(iv) We observe that the co-ordinates of point P, Q, R, are same. Therefore, the point P, Q, R, are same point.
(v) See solution in Basic concepts; co-ordinates of the centroid.

Question 8.
ABCD is a rectangle formed by the points A(- 1, – 1), B(- 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle ? or a rhombus ? Justify your answer.
Solution :
Since P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.
∴ Co-ordinates of point P are \(\left[\frac{(-1-1)}{2}, \frac{(-1+4)}{2}\right]\) i.e., (- 1, \(\frac{3}{2}\))
Co-ordinates of point Q are \(\left[\left(\frac{-1+5}{2}\right),\left(\frac{4+4}{2}\right)\right]\) i.e., (2, 4)
Co-ordinates of point R are \(\left[\left(\frac{5+5}{2}\right),\left(\frac{4-1}{2}\right)\right]\) i.e., (5, \(\frac{3}{2}\))
Co-ordinates of point R are \(\left[\left(\frac{5-1}{2}\right),\left(\frac{-1-1}{2}\right)\right]\) i.e., (2, – 1).
Now,

Thus PQ = QR = RS = SP, but diagonal PR ≠ diagonal SQ.
Hence quadrilateral PQRS is a rhombus.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are :
(i) (2, 3), (- 1, 0), (2, – 4)
(ii) (- 5, – 1), (3, – 5), (5, 2)
Solution:
(i) Let A (2, 3), B (- 1, 0) and C(2, – 4) be the vertices of the given triangle.
Here, x = 2, y = 3, x = -1, y = 0, x = 2, y = – 4
∴ Area of ∆ABC = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [2(0 + 4)+ (- 1)(- 4 – 3) + 2(3 – 0)]
= \(\frac{1}{2}\) [8 + 7 + 6]
= \(\frac{1}{2}\) × 21 = 10.52
Hence, Area of ∆ABC = 10.5 square units.

(ii) Let A(- 5, – 1), B(3, – 5) and C(5, 2) be the vertices of given triangle.
Here x₁ = – 5, y₁ = – 1, x₂ = 3, y₂ = 5, x₃ = 5, y₃ = 2
∴ Area of ∆ABC = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [- 5(- 5 – 2) + 3(2 + 1) + 5 (- 1+ 5)]
= \(\frac{1}{2}\) [- 5 × – 7 + 3 × 3 + 5 × 4]
= \(\frac{1}{2}\) (35 + 9 + 20)
= \(\frac{1}{2}\) (64) = 32
Hence, Area of ∆ABC = 32 square units.

Question 2.
In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, – 2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, – 5)
Solution :
(i) Let the given points be A(7, – 2), B(5, 1) and C(3, k)
Here, x₁ = 7, y₁ = – 2, x₂ = 5, y₂ = 1, x₃ = 3, y₃ = k
Since, the given points are collinear.
Area of ∆ABC = 0
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac{1}{2}\) [7(1 – k) + 5(k+ 2) + 3(- 2 – 1)] = 0
⇒ \(\frac{1}{2}\) [7 – 7k + 5k + 10 + 3 × (- 3)] = 0
⇒ \(\frac{1}{2}\) (- 2k + 17 – 9)= 0
⇒ \(\frac{1}{2}\) (-2k + 8) = 0
⇒ – 2k + 8 = 0
⇒ – 2k = – 8
⇒ k = \(\frac{-8}{-2}\)
⇒ k = 4
⇒ Hence, k = 4.

(ii) Let the given points be A(8, 1), B(k, – 4) and C(2, – 5)
Here, x₁ = 8, y₁ = 1, x₂ = k, y₂ = – 4, x₃ = 2, y₃ = – 5
Since, the given points are collinear.
Area of ∆ABC = 0
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac{1}{2}\) [8(- 4 + 5) + k(- 5 – 1) + 2(1 + 4)] = 0
⇒ \(\frac{1}{2}\) [8 × 1 + k × (- 6) + 2 × 5] = 0
⇒ \(\frac{1}{2}\) [8 – 6k + 10] = 0
⇒ (- 6k + 18) = 0
⇒ – 6k + 18 = 0
⇒ 6k = 18
⇒ k = \(\frac{18}{6}\)
⇒ k = 3
Hence, k = 3.

Question 3.
Find the area of the triangle formed by the joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution :
Let A (0, – 1), B (2, 1) and C (0, 3) be the vertices of given triangle.
Let D, E and F are the mid points of sides of the triangle AB, BC and CA respectively. Then,
The co-ordinates of D are \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)\) i.e., (1, 0)
The co-ordinates of E are \(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)\) i.e., (1, 2)
The co-ordinates of F are \(\left(\frac{0+0}{2}, \frac{-1+3}{2}\right)\) i.e., (0, 1)

Thus, co-ordinates of the mid-points are D(1, 0), E(1, 2), F(0, 1)
∴ Area of ∆DEF = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [1(2 – 1) + 1(1 – 0) + 0(0 – 2)]
= \(\frac{1}{2}\) (1 × 1 + 1 × 1 + 0)
= \(\frac{1}{2}\) (1 + 1)
= \(\frac{1}{2}\) × 2
= 1 square unit.
Now, Area of ∆ABC = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [0(1 – 3) + 2(3 + 1) + 0(- 1- 1)]
= \(\frac{1}{2}\) [0(- 2) + 2 × 4 + 0 (- 2)]
= \(\frac{1}{2}\) (0 + 8 + 0)
= \(\frac{1}{2}\) × 8
= 4 square units.
Area of ∆DEF : Area of ∆ABC = 1 : 4
Hence, area of ∆DEF = 1 square unit and
Area of ∆DEF : Area of ∆ABC =1 : 4.

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are(- 4, – 2), (- 3, – 5), (3, – 2) and (2, 3).
Solution :
Let A(- 4, – 2), B(- 3, – 5), C(3, – 2) and D(2, 3) be the vertices of a quadrilateral. Join A to C.
Area of quadrilateral ABCD = Area(∆ABC) + Area(∆ACD)
Now, Area of ∆ABC = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [- 4(- 5 + 2) + (- 3) (- 2 + 2) + 3 (- 2 + 5)]
= \(\frac{1}{2}\) [- 4 × – 3 – 3 × 0 + 3 × 3]
= \(\frac{1}{2}\) (12 – 0 + 9)

= \(\frac{21}{2}\) square units.
and, Area of ∆ACD = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [- 4(- 2 – 3)+ 3(3 + 2) + 2(- 2 + 2)]
= \(\frac{1}{2}\) [- 4 × (- 5) + 3 × 5 + 2 × 0]
= \(\frac{1}{2}\) (20 + 15 + 0)
= \(\frac{1}{2}\) × 35
= \(\frac{35}{2}\)
Hence, Area of quadrilateral ABCD = \(\frac{21}{2}\) + \(\frac{35}{2}\)
= \(\frac{56}{2}\) = 28 square units.

Question 5.
You have studied in class IX, that a median of a triangle divides it into two triangles of equal areas.Verify this result for ∆ABC whose vertices are A(4, – 6), B(3, – 2) and C(5, 2).
Solution :
Since AD is the median
∴ D is the mid point of BC. Then co-ordinates of D are \(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\) i.e., (4, 0)

Now Area of ∆ABD = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [4(- 2 – 0) + 3(0 + 6) + 4(- 6 + 2)]
= \(\frac{1}{2}\) [4 × (- 2) + 3 × 6 + 4 × (- 4)]
= \(\frac{1}{2}\) [- 8 + 18 – 16]
= \(\frac{1}{2}\) (- 24 + 18)
= \(\frac{1}{2}\) × – 6 = – 3
Since, area cannot be negative, we will take numerical value of – 3 i.e. 3 square units and
area of ∆ADC = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [4(0 – 2) + 4(2 + 6) + 5(- 6 – 0)]
= \(\frac{1}{2}\) [4 × (- 2) + 4 × 8 + 5 (- 6)]
= \(\frac{1}{2}\) (- 8 + 32 – 30)
= \(\frac{1}{2}\) (- 6)
= – 3 i.e., 3 square units.
Hence Area of ∆ABD = Area of ∆ADC.

Haryana State Board HBSE 6th Class Social Science Solutions Civics Chapter 9 Urban Livelihoods Textbook Exercise Questions and Answers.

Haryana Board 6th Class Social Science Solutions Civics Chapter 9 Urban Livelihoods

HBSE 6th Class Civics Urban Livelihoods Textbook Questions and Answers

Bachchu Manjhi – A Cycle-Rickshaw Puller
I come from a village in Bihar where I worked as a mason. My wife and three children live in the village. We don’t own land. In the village I did not get masonry work regularly. The income that I earned was not enough for our family.

After I reached this city, I bought an old cycle rickshaw and paid for it in instalments. This was many years ago. I come to the bus stop every morning and take the customers wherever they want to go. I work till 8.30 pm in the evening. I take rides of up to 6 kilometres in the surrounding area. Each customer gives me Rs. 5-10 per trip depending on the distance. When I’m ill I can’t do this work, so on those days I don’t earn anything.

I stay with my friends in a rented room. They work in a nearby factory. I earn between? 80¬100 every day, out of which I spend? 50-60 on food and rent. The rest I save for my family. I visit my village two or three times a year to see my family. Though my family survives on the money I send, my wife also earns from agricultural work that she gets once in a while.

Class 6th Civics Chapter 9 HBSE Question 1.
Why did Bachchu Manjhi come to the city?
Answer:
Bachchu Manjhi did not get regular masonry work in the village. So he came to the city to find a regular job so he could earn enough for the family.

Urban Livelihoods Question Answer HBSE Class 6th Question 2.
Why can’t Bachchu Manjhi stay with his family?
Answer:
Bachchu Manjhi could not stay with his family as he did not have any room to stay in and not enough earning for 81 the whole family. He himself stayed with his friends in a rented room.

Class 6 Civics Chapter 7 HBSE Urban Livelihoods Question 3.
Talk to a vegetable vendor or hawker and find out how do they organise their work, their way of preparing, purchasing, selling, etc.
Answer:
The vegetable vendors or hawkers organise their work in the following ways:
(а) They purchase vegetables or groceries from the wholesale market.
(b) They arrange them on the trolley cycle.
(c) They give loud calls on the roads, streets to sell their goods.
(d) Sometimes they sell their goods in weekly markets.

Class 6 Urban Livelihoods HBSE Civics Question 4.
Bachchu Manjhi has to think twice before taking a day off from work?
Answer:
Bachchu Manjhi earns a small daily income out of which he saves only a very small portion. He has to send something for his family too. So, Bachchu Manjhi has to think twice before taking a day off from work.

Question 5.
Why did Harpreet and Vandana start a showroom? What do they have to do to face difficulties?
Answer:
Harpreet and Vandana started a showroom since:
1. Harpreet started working in the shop of his father and uncle after she completed her college.
2. Vandana is a dress designer. She realised that people prefer to buy readymade clothes, rather than have them stitched.
Vandana also felt the need for an attractive display of readymade products for their showroom:
1. They buy things from different places like Mumbai, Ahmedabad, Ludhiana, etc.
2. They had to put them in show cases.
3. They also had to advertise their products.

Question 6.
Talk to a shop owner in a market and find out how he plans his work. Have there been any changes in his business in the past twenty years?
Answer:
A shop owner in the market had to take a showroom on rent earlier. Now they had to purchase it. He buys things on installments and sells them on cash. There have been changes in the business in the past twenty years.
There have been changes in the business since the past.
(а) There is more competition. So, shopkeepers have to sell quality goods.
(b) They also have to give more stress on the advertisement.
(c) To attract customers, shopkeepers have to provide them with other service like after sales service, free-home delivery, etc.

Question 7.
What are the difference between those who sell on the street and those in the market?
Answer:
Those who sell things in the street buy their things from the local market. Those who sell things in the market, purchase their things in bulk. They need to adhere to government regulations and pay taxes also.

Question 8.
Why do you think small workshops and factories employ casual workers?
Answer:
Small workshops and factories employ casual workers as they do not have regular work.

Question 9.
Describe the working conditions of people like Nirmala keeping in mind the following: working hours, condition in the working place, earnings and the days of work available.
Answer:
Working conditions of people like t Nirmala are as follows:
(а) Working Hours: Workers have to work for long hours. A normal working day begins at 9 A.M. and finishes by 10 P.M., sometimes even later. They have to generally work for 6 days a week. When the work has to be done urgently, she worked on Sundays too.
(b) Conditions in the workplace: The conditions in the working are not very healthy.
(c) Earnings: They are paid only f 80 per day for eight hours and extra 40 for working late hours. They have no bargaining power.
(d) They are all casual workers as they work only for some months of the year. After this period, they have to move to other work.

Question 10.
Would you say that domestic workers like housemaids are also casual workers? Why? Describe the workday of one such woman detailing the work she does in other peoples’ houses.
Answer:
Yes, the domestic workers like housemaids are also casual workers. They work as long as their employer wants them.
The working of one woman in other people’s houses:
1. She washes utensils, cleans the house, washes the clothes, cuts vegetables daily needed for meals.
2. Her work starts at 9 A.M. in the morning and finishes at 7 P.M.
3. Her wages are 100 a day.
4. She is also provided with some eatables during the day.

EXERCISE

Question 1.
Read and discuss description of the living conditions of workers who come to the labour chowk.
Most workers that we find at the labour chowk cannot afford permanent accommodation and sleep on pavements near the chowk, or they pay ₹ 6/- a night for a bed at a nearby night shelter run by the Municipal Corporation. To compensate for the lack of security, local tea and cigarette shops function as banks, moneylenders and safety lockers, all rolled into one. Most workers leave their tools at these shops for the night and safe keeping and pass on any extra money to them. The shopkeepers keep the money safety and also offer loans to labourers in need.
Source: Aman Sethi, Hindu on-line
Answer:
Labourers those who generally do not find regular work go to labour chowk. If they get any hope of source of income. At labour chowk; workers get work available at that time, but they cannot afford permanent job. Their job is for only one or two days. They spend their night on pavements or in the ‘Rein Vasera’ built by Municipal Corporation. Their daily earning is not sure. Such workers leave their tools in tea shops for safety.

Their economic position is imbalanced. They eat what they earn and spend on their families. Even they take money from money-lenders to pass their time. (Note: Everybody has different view, so you can write your own view about the life of chowk labourers, their living standard and way of earning).

Question 2.
Complete the following table and discuss how their work is different:

Answer:
1. Name: Bachchu Manjhi.
Place of Work: A cycle-rickshaw puller in the city bus stop.
Earning: ₹ 100/- per day.
Security of work: No
Benefits received: Saving send for expenditure of the family.
Work on their own or employed: Work on their own.

2. Name: Harpreet Vandana.
Place of Work: Readymade showroom.
Earning: Good
Security of Work: Yes
Benefits received: Increase in income.
Work on their or own or employed: Their own work.

3. Name: Nirmala
Place of Work: Garment Factory.
Earning: ₹ 80 per day + ₹ 40 overtime.
Security of Work: No
Benefits received: He gets benefit during summer season, when business is on full swing.
Work on their own or employed: She was employed.

4. Name: Sudha.
Place of Work: Company.
Earning: ₹ 30,000/- p.m.
Security of Work: Yes.
Benefits received: She receives benefits such as:
(i) Provident Fund
(ii) Medical facilities
(iii) Holidays: Sundays and also can get some annual leaves.

Their works are different. Bachchu was a rickshaw puller, Harpreet Vandana is owner of a showroom. Nirmala does work in a garment factory and Sudha is working in a company. Among them Sudha’s work is safe and secure because she has a permanent job and facilities as generally enjoyed by a government servant.

Question 3.
In what ways is a permanent and regular job different from a casual job? Discuss.
Answer:

Question 4.
What benefits does Sudha get along with her salary?
Answer:
Sudha enjoys the following benefits with her salary:
(i) She deposits her Provident Fund as a saving for her old age.
(ii) She gets off on every Sunday, national festivals and also gets some annual leaves.
(iii) Her company provides her medical facilities.
(iv) She can get medical leave when she feels ill.

Question 5.
Fill in the following table to show the services provided by people in the markets which you visit frequently.

Answer:

HBSE 6th Class Civics Urban Livelihoods Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Name some activities which are carried out on the streets.
Answer:
Hair-cutting, shoe-repairing, flower-selling, craft-making.

Question 2.
How was the street vending earlier looked upon by people?
Answer:
Street vending was earlier looked upon by people as an obstruction to traffic and to people walking.

Question 3.
Where does Nirmala work?
Answer:
Nirmala works as a tailor in an export garment unit.

Question 4.
Why are the casual workers asked to leave by the employers?
Answer:
Casual workers are asked to leave when they complain about their pay or working conditions.

Short Answer Type Questions

Question 1.
What suggestions have been made; so that street vendors have a place to work and that there is also a free flow of traffic to the people?
Answer:
(a) Hawking zones have been suggested for towns and cities.
(b) Mobile vendors should be allowed to move around freely.
(c) Hawkers need to be part of committees that are set up to take these and other decisions relating to them.

Question 2.
What extra benefits does Sudha get?
Answer:
Sudha gets the following extra benefits:
(i) A part of her salary is kept in a fund with the government. She will earn interest on these savings.
(ii) She gets medical facilities for her family.
(iii) She gets medical leave if she falls ill and her salary is not cut if she takes this leave.

Question 3.
Who are casual workers?
Answer:
Casual workers are required to come as and when the employer needs them. They are employed when the employer gets a big order or during certain seasons. At other times of the year they have to find some other work.

Long Answer Type Questions

Question 1.
Name different categories of people who do work on daily wages.
Answer:
The people who work on daily wages are
(i) People in the labour chowk are daily wage earners who wait with their tools for people to come and take them to work.
(ii) People who work in factories and small workshops are daily wage earners. They are employed on casual basis i.e., they are required to come when there they are needed.

Question 2.
What is a Call Centre? Write about the work done in the Call Centre. Why are they set up in India?
Answer:
A Call Centre is a centralised office that deals with problems and questions that consumers/customers have regarding goods purchased and services like banking, ticket booking etc. It is a new form of employment in big cities. Call centres are generally set up as large rooms with work stations that include a computer, a telephone set, and supervisors stations. India has become a major centre not only for Indian companies but also for foreign companies.

Call Centres are set up in India because they can get people who can speak English and will work for lower wages.

Urban Livelihoods Class 6 HBSE Notes

• Vendor: One who sells things of daily use by going door to door.
• Urban areas: Towns and cities.
• Business person: One who earns his livelihood by engaging himself in some business.
• Employer: One who gives job to someone.
• Casual worker: One who is engaged in temporary work.
• Labour chowk: A place where daily wage labourers gather together with their tools and wait for people to come and hire them for work.
• Call centre: It gives a new form of employment to the people of big cities. It is a centralised office that deals with problems and questions that consumers/customers have regarding goods purchased and services like banking, ticket booking, etc.
• Hawker: One who sells things by going from one place to place asking people to buy them.

Haryana State Board HBSE 6th Class Social Science Solutions Civics Chapter 8 Rural Livelihoods Textbook Exercise Questions and Answers.

Haryana Board 6th Class Social Science Solutions Civics Chapter 8 Rural Livelihoods

HBSE 6th Class Civics Rural Livelihoods Textbook Questions and Answers

Class 6 Civics Chapter 8 HBSE Rural Livelihoods Question 1.
Based on the above diagram would you say the Thulasi earns money throughout the year?
Answer:
No. Thulasi works in seasonal job which is agriculture she earns only during those seasons when she has work on her fields. According to the given data, it can be important that from June to December, she earns.

Class 6 Civics Chapter 8 Question Answer HBSE  Question 2.
Describe the work that Thulasi does. How is it different from the work that Raman does?
Answer:
Thulasi transplants paddy in the fields. Thulasi’s husband, is also a labourer. They don’t own any land. While Thulsi transplants, Raman sprays pesticides.

Rural Livelihoods Class 6 Civics Chapter 8 Question Answer Question 3.
Thulasi gets paid very little money for the work. She does, why do you think agricultural labourers like her are forced to accept low wages.
Answer:
1. Agricultural labourers are unskilled and do not have the knowledge or access to higher paid jobs.
2. They are under the debt of landlords or moneylenders who employ them on low wages.

Question Answer Class 6 Civics Chapter 8 HBSE Rural Livelihoods Question 3.
In what ways would her way of earning a living have been different if Thulasi own some farm land? Discuss.
Answer:
1. If Thulasi would have owned a farm land she would have been the possessor of all the process.
2. She would have an increased income by selling all her-crop to the market.
3. This would have an impact on her livelihood.

Question 4.
What are the crops grown in your region or nearby rural are? What kinds of works do agricultural labourers do?
Answer:
Maize, Wheat and Rice is planted in the neighbourhood village.
Agricultural labourers do the following works:
(a) Ploughing fields
(b) Preparing saplings
(c) Transplanting saplings
(d) Irrigating crops
(e) Weeding
(f) Harvesting
(g) Storage of crops.

Question 5.
What work does Sekar’s family do. Why do you think Sekar does not usually employ labourers for doing farm work?
Answer:
Sekar’s family works in the field, harvests crop and does all sort of agricultural work.
Sekar might not be employing labourers for farm work to cut cost on manual labour. Money has to be given to labour for doing work.

Question 6.
Why does Sekar not go to the town market to get a better price for his paddy?
Answer:
Sekar earns very less to satisfy his needs. He cannot earn a higher income. There is very little produce lift. Some is used in loans, some so satisfy their home needs.

Question 7.
Sekar’s sister Mina had also taken a loan from the trader. She does not loan to sell her paddy to him but will pay back her loan. Write an imaginary conversation between Mina and the trader’s agent and the arguments given by each person.
Answer:
AGENT: Mina/when will you repay me loan? How long can I excuse you?
MINA : Just fifteen days more, Sir/ I will repay it as soon as I sell my paddy.
AGENT : But I have already offered you excuse for one month.
MINA : Sorry, Sir.
AGENT : Next time, I’ll not take off.

Question 8.
What are the similarities between Sekar’s and Thulasi lives. Your answer could be based on the land they have, the need to work on the land that belong to others or the loans that they need and their earning.
Answer:
Both Sekar and Thulasi are agricultural labourers who accept loans (even at higher interest to meet then daily needs, they have low wage difference.
Thulasi owns no land while Sekar owns two acres of field on which he works.
Thulasi has a definite source of income paid by Ramalingam while Sekar’s income depends upon his crop produce.

Read again Sekar’s and Thulasi’s accounts. What do they say about Ramalingam, the large farmer? Together with what you have read fill in the details below.

Question i.
How much land does he have?
Answer:
20 acres.

Question ii.
What does Ramalingam do with the paddy grown in his field?
Answer:
The paddy is sold to traders in nearby towns.

Question iii.
A part from farming how else does he earn?
Answer:
His family owns a rice mill and shop selling seeds, pesticides. This given them extra substantial income.

Question 9.
Why do both Sekar’s and Aruna’s family have to borrow? What similarities and differences do you find?
Answer:
Sekar’s family had to borrow loan to as keep their produce while Aruna buys loan for the maintenance of their requirement. Sekar buys loan from an informal source of credit while Aruna busy loans from formal source of credit.

Question 10.
Have you heard of Tsunami? What is this and what damage do you think it might have done to the life of fishing families like Aruna’s.
Answer:
Do yourself.

EXERCISE

Short Answer Type Questions

Question 1.
You have probably noticed that people in Kalpattu are engaged in a variety of non-farm work. List five of these.
Answer:
(i) Making Baskets,
(ii) Making pots
(iii) Making bullocks carts
(iv) Making utensil
(v) Making bricks. .

Question 2.
List the different types of people you read about in Kalpattu who depend on farming. Who is the poorest among them and why?
Answer:
In Kalpattu, there are people who provides service such as blacksmiths, nurses, washerman, weavers, barbers, cycle repair mechanics, people whose work depends on farming are shopkeepers, traders, peasants etc.
Labour is the poorest of all sense it works only when in demand.

Question 3.
Imagine you are a member of a fishing family and you are discussing whether to take a loan from the bank for an engine. What would you say?
Answer:
If I were a member of the fishing family, I would say we must take loan from a bank or cooperative to improve the condition of our occupation because:
(a) Catamaran engine is useful because with their help, fishes can be caught in deep waters.
(b) It will help increase income, reduce time and help in overall development.

Question 4.
Poor rural labourers like Thulasi often do not have access to good medical facilities, good schools, and other resources. You have read about inequality in the first unit of this text. The difference between her and Ramalingam is one of inequality. Do you think this is a fair situation? What do you think can be done?
Discuss in class.
Answer:
Thulasi is a labourer while Ramalingam is a rich landlord.
1. Thulasi has a poor living standard, Ramalingam has a higher living standard.
2. Thulasi leads a sad life while Ramalingam leads happy life.

Question 5.
What do you think the government can do to help farmers like Sekar when they get into debt? Discuss.
Answer:
Debts-trap is a common problem for farmers in India.
(i) Decrease the rate of interest at which loan is given.
(ii) Give compensation in case of a natural calamity.
(iii) Additional time can be given.
(iv) Formal sources of credit should be increase.

Question 6.
Compare the situation of Sekar and Ramalingam by filling in the following table:

Answer:

HBSE 6th Class Civics Rural Livelihoods Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is the main work of the people in the village of Kalpattu?
Answer:
Paddy is the main crop that is grown in irrigated lands. Most of the families earn a living through agriculture.

Question 2.
What is the type of cultivation in Nagaland?
Answer:
Terrace farming.

Question 3.
What are catamarans?
Answer:
A catamarans is a boat with turn hull, side by side or a raft made of logs of wood or pieces of timber lashed together.

Question 4.
Who are seasonal workers?
Answer:
Most small farmers, agricultural labourers, fishing families, crafts persons in the villages are seasonal

Question 5.
What operations do farming activities involve?
Answer:
Working on farms involve operations such as preparing the lands, sowing, weeding, harvesting of crops.

Question 6.
Name the main crop grown in irrigated lands.
Answer:
Paddy.

Short Answer Questions

Question 1.
Why is agriculture a seasonal job?
Answer:
Process of agricultures take place only during some seasons. Not being able to earn money forces people in many rural areas to travel long distances in search of work. This travel or migration takes place during some seasons.

Question 2.
List the different sources of income of rural areas?
Answer:
(a) Many people in rural areas depend upon forest, animal husbandry, dairy produce, fishing, etc.
(b) Collecting mahua, tendu leave, honey to be sold to traders is an important source of additional income.
(c) Selling .milk, to cooperative is another source.

Question 3.
What is the involvement coasts in crop production?
Answer:
Equipments that are related to agriculture, seeds, pesticides, fertilizers, minerals water, irrigational facilities.

Question 4.
Name the three category of farmers.
Answer:
The three categories of farmers are:
(i) Big land owners, who employ other farmers to work on their fields.
(ii) Farmers who have very small plots of land. They have to do other work, when there is a lean season on the farm. The produce is not sufficient to meet the needs of their family.
(iii) Landless peasants, who work on the fields of big land owners and are paid for the work done by them.

Question 5.
How do money-lenders and traders exploit the poor farmers and others? Give examples.
Answer:
The traders and money-lenders exploit the poor farmers and others by taking a part of their produce at a lower price, e.g., Sekar has to; sell his paddy at a lower price than the market rate. Aruna has to sell her fish catch to the trader and cannot auction it.

Long Answer Questions

Question 1.
Write about the everyday work of the village fisherman.
Answer:
Most of the fisherman live close to the sea. Their work starts at 7 o’clock in the morning, when there is lot of activity on the beach. This is the time when catamarans return with their catch and women gather to buy and sell fish.

The fisherman go in groups to the sea in catamaran. Some have engines fitted in the catamaran, so that they can go far into the sea. The money is equally divided in the group, keeping one part for the equipment. The women folk sell the fish till afternoon. In the evening the fishermen untangle and repair the nets. Next day at 2 A.M. they will set out to sea again.

Rural Livelihoods Class 6 HBSE Notes

• Blacksmith: A man engaged in making iron goods, tools, instruments etc.
• Villager: A citizen of India living in a village and occupied with agriculture.
• Chakhesang Community: People of Chizami Village in Nagaland.
• Community: The people living at one place, district or country are combined known with this term.
• Family: The smallest unit of society consisting of parents and their children.
• Heterogeneous: Consisting of many different kinds of people.
• Catamaran: Catamaran is a small boat used for catching fish.
• Migration: Movement of people from one place to another to find jobs is termed as migration.
• Fisherman: A person engaged in the work of catching fish.

Haryana State Board HBSE 6th Class Social Science Solutions Civics Chapter 7 Urban Administration Textbook Exercise Questions and Answers.

Haryana Board 6th Class Social Science Solutions Civics Chapter 7 Urban Administration

HBSE 6th Class Civics Urban Administration Textbook Questions and Answers

Urban Administration Class 6 Question Answers HBSE Question 1.
List at least four different tasks that the Municipal does.
Answer:
1. It manages to collect garbage and disposes at proper place.
2. It takes care of the situation of the town.
3. It builds roads and repair them whenever required.
4. It takes care of water supply and keeps the streets and roads clean.

Class 6 Civics Chapter 7 Fill In The Blanks HBSE Question 2.
Fill in the blanks in the sentences below.
1. In a panchayat the elected members are called ———— .
2. The city is divided into several ———— .
3. In a municipal corporation the elected members are called ———— .
4. Group of councillors deal with issues that affect the ———— .
5. Elections are held once every ———— years for the panchayat as well as for the municipality.
6. While the councillors make decisions the administrative staff led by the Commissioner ———— these.
Answer:
(1) Panchs
(2) Wards
(3) Councillors
(4) entire city
(5) five
(6) implement

Urban Administration Question Answer HBSE Question 3.
What was Gangabai protesting about?
Answer:
Gangabai was protesting about the filthy environment of her locality.

Class 6 Civics Chapter 7 HBSE Urban Administration Question 4.
Why do you think Gangabai decided to approach the Ward Councillor?
Answer:
Gangabai decided to approach the Ward Councillor because it was Ward Councillor’s duty to look after the sanitation in his locality.

Class 6 Urban Administration HBSE Question 5.
What did Gangabai say when the Commissioner told that there were not enough trucks in the city?
Answer:
When the Commissioner said that there were not enough trucks in the city, Gangabai said. But you seem to have enough trucks to, collect garbage from the rich localities.

EXERCISE

Question 1.
Why did the children go to Yasmin Khala’s house?
Answer:
The children had broken street lights while playing cricket. They had no knowledge how the light was to be replaced and to whom they should pay the compensation. They first went to Rehana’s mother who gave them knowledge that it was the duty of municipal corporation to replace the lights. She directed them to Yasmin Khala’s house as she had just retired from the Municipal Corporation. So, children went to Yasmin Khala’s house to get their problems solved.

Question 2.
List four ways in which the work of the Municipal Corporation affects the life of city dweller.
Answer:
The four ways in which the Municipal Corporation affects the life of a city dweller are:
(i) It takes care of street lights, garbage collection, water supply, keeping the streets and the market clean.
(ii) It ensures that diseases do not break out in the city.
(iii) It runs school, hospitals and dispensaries.
(iv) It makes garbage around the city and maintains them.

Question 3.
Who is a Municipal Councillor?
Answer:
Municipal Councillor is an elected representative of a ward. He is also known as the ward councilor.

Question 4.
What did Gangabai do and why?
Answer:
Gangabai and women in the locality were upset over the garbage lying in the streets and unhygienic condition of the society. Gangabai proposed that it was the Ward Councillor who they should go to. Gangabai was confident that they had the right to do so as they were ones who elected him.

Question 5.
How does the Municipal Corporation earn the money to do its work?
Answer:
The Municipal Corporation is responsible for all round development of the society. It provides many services which requires a lot of money. The Municipal Corporation earn the money by collecting taxes from the people. It levied taxes on property, water and other services. Taxes are levied on the measurement and consumption of services. Taxes are also levied on education and other amenities like owning a hotel, entertainment etc.

Question 6.
Discuss:


In the two photograph you see different ways of collecting disposing garbage.
1. Which way do you think safety to person disposing garbage?
2. What are the dangers of collecting in the manner shown in the first photograph?
3. Why do you think that the proper ways of disposing garbage are not available to those who work in municipalities?
Answer:
1. In photograph-2, safer means for disposal of garbage are being adopted.

2. Dangers of collecting garbage in the manner shown in the photograph-1 are:
(i) The animals roaring around garbage dump may cause injury/harm/ hurt to them personally.
(ii) Always a fear of being get seriously ill or fall in the grip of serious diseases.

3. Municipalities have less facilities and transportation available to dispose garbage as compared to the corporations. In other words corporations are meant for big cities or Metropolitian cities. Due to growing population and slums a number of steps were taken by corporations to over come these problems. So, ways of disposing garbage are less available to those who work in municipalities.

Question 7.
Several poor people in the work as domestic servants as well as work for the Corporation, keeping the city clean. Yet the slums in which they live are quite filthy. This is because these slums seldom have any water and sanitation facilities. The reason often given by the Municipal Corporation is that the land in which the poor have set up their homes does not belong to them and that slum-dwellers do not pay taxes. However, people living in middle class neighbourhood, pay very little in taxes compared to the amount of money the corporation spends on them in setting up parks, street lighting facilities, regular garbage collection etc. Also as you read in this chapter, the property taxes collected by the Municipality makes up only 25-30 per cent of its money. Why do you think it is important that the Corporation should spend more money on slum localities? Why is it important that the Municipal Corporation provide the poor in the city with the same facilities that the rich get?
Answer:
The Corporation should spend of money on slum localities because:
(i) The share of general public in corporations earning stand between 70-75 per cent in terms of collection of various axes excluding only property tax.
(ii) Sanitation and light fall in the category of basic needs mandatory to provide by corporation like authority in a city.
(iii) These effects part of corporation would in long terms, bring buoyancy to its earning through taxes when parks, schools, water and electricity etc. facilities are provided with the poor in slums. Again it would inspire them to prepare for better in future.

HBSE 6th Class Civics Urban Administration Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is a Ward?
Answer:
Ward is a division of city for the purpose of municipal or corporation election.

Question 2.
What is a property tax?
Answer:
People who own homes pay proper tax. The larger the house the more the tax.

Question 3.
What is a Metropolitan city?
Answer:
A city in which people from different, religion, state and caste live together is called a metropolitan city.

Question 4.
Whose job is to replace the street light?
Answer:
It is the job of the Municipal Corpo-ration of the city to replace the street lights.

Short Answer Type Questions

Question 1.
What is sub-contracting? What did Yasmin Khala tell about this in case of garbage collection?
Answer:
Subcontracting means that the work that was earlier being done by government workers is how being done by a private company. Yasmin Khala told that in order to save money the Commissioners of several munici-palities across the country had hired private contractors to collect and process garbage.

Question 2.
What are hazards and dis-advantages associated with the job of collecting garbage?
Answer:
(i) The contract workers get paid less.
(ii) Their jobs are temporary.
(iii) It is quite a dangerous job.
(iv) The contract’ workers do not have any access to safety measures.
(v) They are not taken care of if they are injured while working.

Question 3.
How does the state government help the Municipalities?
Answer:
The sources of income of the Municipalities are very limited while the expenditures are more then the income. The state government helps the Municipalities by giving grants to them.

Question 4.
How are complicated decisions taken?
Answer:
Complicated decisions usually affect the entire city and therefore these decision are taken by groups of councillor who form committees to decide and debts issue. For example, if bus stands need to be improved or a crowded market place needs to have its garbage cleared more regularly.

Question 5.
What are major features and facilities available in a city?
Answer:
Major features and facilities available in a city are:
(a) Crowded markets
(b) Bigger in size then villages
(c) Many buses and cars
(d) Water facilities
(e) Electricity facilities
(f) Hospitals and schools

Long Answer Type Questions

Question 1.
What are the three main functions of the district administration?
Answer:
The three functions of the district administration are:
(i) To maintain law and order: To perform this task, all officers work under the supervision of the Collector.

(ii) To maintain land record and realise land revenue: For this purpose. Tehsildar, Naib Tehsildar, Kanungo and Lekhpal (Patwari) assist the Collector. He also provides relief when unforeseen situations as epidemics, famine are befallen.

(iii) To provide civic amenities: The district administration provide civic amenities as health facilities, education, means of transport, repair of roads etc.

Question 2.
What are the main functions performed by the Municipality or Municipal Corporation?
Answer:
The functions of the Municipality or a Municipal Corporation are generally similar. These are.
1. Functions for maintaining public health:
(i) They arrange for the disposal wastes causes numerous diseases.
(ii) They construct sewage, drain out dirty water from our cities and villages.
(iii) They protect the people from epidemics and other diseases like small-pox, cholera. They make suffice arrangements for vaccination and inoculation of the people.
(iv) They construct public latrines and urinals.
(v) They maintain dispensaries and hospitals to treat the sick.

2. Functions of Public Convenience:
(i) They arrange for water and electricity.
(ii) They maintain and repair roads, bridges and streets.
(iii) They plant trees along the roads for the convenience of travellers.
(iv) They make arrangements for the primary education of the children. For this, schools are maintained with proper basic facilities.
(v) They also manage libraries, museums and zoological parks etc.
(vi) It makes arrangements for playgrounds, parks and gardens.

3. Maintaining optional or miscellaneous departments:
(i) It maintains fire engines for extinguishing fire.
(ii) It checks food adulteration.
(iii) They remove obstacles from the public roads.
(iv) They maintain records of births and deaths.
(iv) It approves the plans of houses to be built in its areas.
(vi) They make arrangement for cremation and burial grounds.
(vii) Corporation issue permission to set-up petrol pumps, cinema houses etc.
(viii) Corporation also set-up homes for helpless women and children.

4. Tax Collection:
It impose and collect taxes like house tax, vehicles tax etc.

Urban Administration Class 6 HBSE Notes

• Municipal Corporation: Civic body in big cities is called Municipal Corporation.
• Municipal Council (Municipality): The civic body in small towns and cities is termed as municipal Council.
• Ward Councillor: Elected representative of a ward in a corporation is called Ward Councillor.
• Municipal Commissioner: The chief executive and administrator of the Municipal Corporation is called Municipal Commissioner.
• Public Defence: The maintenance of fire engines for extinguishing fire, checking food adulteration etc. .
• Mayor: Presiding officer or Chairman of a municipal corporation.
• Aldermen: Members of municipal bodies choosen for their experience and distinguished services.
• Democracy i A government of the people, for the people and by the people.
• Collector: The highest officer of the district.
• Tehsildar: He is in-charge of everything relating to land and land disputes of district.
• The Sessions Judge: He conducts/administers the highest criminal court in the district

Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.4

Question 1.
Let ∆ABC ~ ∆DEF and their areas be respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution :
We have, ∆ABC ~ ∆DEF

Hence, BC = 11.2 cm.

Question 2.
Diagonals of trapezium ABCD with AB || CD intersects each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution:
We have, AB || CD and AB = 2CD
In triangles AOB and COD,
∠ABO = ∠CDO (alternate interior ∠s)
∠AOB = ∠COD (vertically opposite ∠s)
∆AOB ~ ∆COD (By AA similarity criterion)
⇒ \(\frac{\ {ar}(\Delta \mathrm{AOB})}{\ {ar}(\Delta \mathrm{COD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2}\)
⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{(2 \mathrm{CD})^2}{\mathrm{CD}^2}\)
⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{4 \mathrm{CD}^2}{\mathrm{CD}^2}\)
⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{4}{1}\)
Hence, ar(∆AOB) : ar(∆COD) = 4 : 1.

Question 3.
In given figure ABC and DBC are two triangles on the same base BC.. If AD intersects BC at O, show that \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}\).

Solution :
Given : ∆ABC and ∆DBC are two triangles on the same base BC
To Prove: \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}\)
Construction : Draw AL ⊥ BC and DM ⊥ BC
Proof : In ∆ALO and ∆DMO

∠ALO = ∠DMO = 90° (By construction)
∠AOL = ∠DOM (vertically opposite ∠s)
∆ALO ~ ∆DMO (By AA similarity criterion)
⇒ \(\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}\)
[corresponding sides of similar triangles are proportional]

Hence proved.

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Given : Let two triangles ABC and PQR such that ar(ABC) = ar(APQR)
To Prove: ∆ABC ∆PQR
Proof: ar(∆ABC) ≅ ar(∆PQR)
⇒ ∆ABC ~ ∆PQR
⇒ ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
and \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)

⇒ AB = PQ, BC = QR, CA = RP
∴ ∆ABC ≅ ∆PQR (By SSS congruence criteriah)
Hence Proved.

Question 5.
D, E and F are respectively the mid points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution:
We have,
∵ D, E and F, are the mid points of sides AB, BC and CA respectively.
∴ DF = \(\frac{1}{2}\)BC, DE = \(\frac{1}{2}\)AC and EF = \(\frac{1}{2}\)AB
[Mid point Theorem]

Hence, ar(∆DEF) : ar(∆ABC) = 1 : 4.

Question 6.
Prove that ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding medians.
Solution:
Given: Two triangles ABC and PQR such that AABC – APQR and AD, PM are their medians respectively.

Hence proved.

Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution :
Given : A square ABCD and two equilateral ∆s BCE and ACF have been described on side BC and diagonal AC respectively.
To Prove: Area (∆BCE) = \(\frac{1}{2}\) Area (∆ACF)
Proof: We have ∆BCE and ∆ACF are equilateral triangles.

∠B = ∠A (each = 60°)
∠C = ∠C (each = 60°)
∠E = ∠F (each = 60°)
∆ BCE ~ ∆ ACF (By AAA similarity criterion)
⇒ \(\frac{\ {ar}(\triangle \mathrm{BCE})}{\ {ar}(\triangle \mathrm{ACF})}=\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\)
[By theorem 6.6] ………………..(1)
But AB = BC (sides squares)
and AC² = AB² + BC² [By Pythagoras theorem]
⇒ AC² = 2BC² ……………..(2)
From (1) and (2), we get
\(\frac{\ {ar}(\Delta B C E)}{\ {ar}(\Delta A C F)}=\frac{B^2}{2 B C^2}\)
⇒ \(\frac{\ {ar}(\triangle B C E)}{\ {ar}(\triangle A C F)}=\frac{1}{2}\)
⇒ ar(∆BCE) = \(\frac{1}{2}\) ar(∆ACF)
Hence Proved.

In question 8 and 9, tick the correct answer and justify:
Question 8.
ABC and BDE are two equilateral triangles such that D is the mid point of BC. Ratio of the areas of ∆ABC and ∆BDE is :
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Solution :
Correct answer is (C).
Justification : ∆ABC and ∆BDE are two equilateral triangles.
Let AB = BC = CA = a unit
Then BD = \(\frac{a}{2}\) unit
[∵ D is the mid point of BC]
∵ ∠CAB = ∠DBE, ∠ABC = ∠BDE, ∠ACB = ∠BED [each is 60°]
∆ABC ~ ∆BDE (By AAA similarity criterion)
⇒ \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\Delta \mathrm{BDE})}=\frac{\mathrm{AB}^2}{\mathrm{BD}^2}\) [By theorem (6.6)]

⇒ ar(∆ABC) : ar(∆BDE) = 4 : 1.

Question 9.
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Solution :
Correct answer is (D).
Justification : We have,
\(\frac{\text { Side of } 1^{\text {st }} \text { triangle }}{\text { Side of } 2^{\text {nd }} \text { triangle }}=\frac{4}{9}\)
We know that areas of two similar triangles are in the ratio of the squares of their corresponding sides.
∴ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{(\text { side of } 1 \text { st } \Delta)^2}{(\text { side of } 2 \text { nd } \Delta)^2}\)
⇒ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{4^2}{9^2}\)
⇒ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{16}{81}\)
⇒ Area of 1st triangle : Area of IInd triangle = 16 : 81.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.3

Question 1.
State which pairs of triangles in below figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

Solution:
(i) In ∆ABC and ∆PQR
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
By AAA similarity criterion.
∆ABC ~ ∆PQR.

(ii) In ∆ABC and ∆QRP
\(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{2}{4}=\frac{1}{2}\)
\(\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{2 \cdot 5}{5}=\frac{1}{2}\)
\(\frac{\mathrm{AC}}{\mathrm{PQ}}=\frac{3}{6}=\frac{1}{2}\)
∵ \(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{AC}}{\mathrm{PQ}}\)
∴ By SSS similarity criterion
∆ABC ~ ∆QRP

(iii) In ∆LMP and ∆DEF
\(\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2 \cdot 7}{4}\) = 0.675
\(\frac{\mathrm{MP}}{\mathrm{EF}}=\frac{2}{5}\) = 0.4
\(\frac{\mathrm{LP}}{\mathrm{DF}}=\frac{3}{6}\) = 0.5
∵ \(\frac{\mathrm{LM}}{\mathrm{DF}} \neq \frac{\mathrm{MP}}{\mathrm{EF}} \neq \frac{\mathrm{LP}}{\mathrm{DF}}\)
∴ These two triangles are not similar because they do not satisfy the SSS similarity criterion.

(iv) In ∆MNL and ∆QPR
∠M = ∠Q = 70°
\(\frac{\mathrm{MN}}{\mathrm{PQ}}=\frac{2 \cdot 5}{6}\) (approx)
\(\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}\) = 0.5
∵ \(\frac{\mathrm{MN}}{\mathrm{PQ}} \neq \frac{\mathrm{ML}}{\mathrm{QR}}\)
∴ These two triangles are not similar because they do not satisfy the SAS similarity criterion.

(v) In ∆ABC and ∆FDE
∠A = ∠F = 80°
\(\frac{\mathrm{AB}}{\mathrm{DF}} \neq \frac{\mathrm{AC}}{\mathrm{EF}}\) [∵ AC is not given]
∴ These two triangles are not similar because they do not satisfy the SAS similarity criterion.

(vi) In ∆DEF and ∆PQR
∠P = 180° (80° + 30°)
∠P = 180° – 110°
∠P = 70°
Now, ∠D = ∠P = 70°
∠E = ∠Q = 80°
By AA similarity criterion ∆DEF ~ ∆PQR.

Question 2.
In given figure ∆ODC – ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:
We have, ∠CDO = 70° and ∠COB = 125°
∵ BD is a line and CO ray on it
∠DOC + ∠COB = 180°
∠DOC + 125° = 1800
∠DOC = 180° – 125°
∠DOC = 55°
In ∆COD, we have
∠DOC + ∠DCO + ∠CDO = 180° [Sum of ∠s of a triangle is 180°]
55° + ∠DCO + 70° = 180°
125° + ∠DCO = 180°
∠DCO = 180° – 125°
∠DCO = 55°
Now ∆ODC ~ ∆OBA (given)
∴ ∠OAB = ∠DCO
(Corresponding ∠s of similar ∆s)
∠OAB = 55°
Hence, ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°.

Question 3.
Diagonals AC and BD of a trapezium ABCD with AB DC ntersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\).
Solution:
Given : ABCD is a trapezium in which AB || DC. Its diagonals AC and BD meet at O.
To Prove: \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
∠OAB = ∠OCD [∵ AB || DC]
∴ Alternate ∠s are equal.
∠AOB = ∠COD (vertically opposite ∠s)

∴ ∆OAB ~ ∆OCD (By AA similarity criterion)
∴ \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
Hence Proved.

Question 4.
In given figure \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Solution:
Given: \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2.
To Prove: ∆PQS ~ ∆TQR
Proof: We have

∴ ∆PQS ~ ∆TQR [By SAS similarity criterion]
Hence Proved.

Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ – ∆RTS.
Solution :
Given : S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS
To Prove: ∆RPQ ~ ∆RTS.

Proof: In ∆RPQ and ∆RTS.
∠P = ∠RTS (given)
∠R = ∠R (common)
∴ ∆RPQ ~ ∆RTS (By AA similarity criterion)
Hence Proved.

Question 6.
In given figure, If ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.

Solution:
Given: ∆ABE ≅ ∆ACD
ToProve: ∆ADE ~ ∆ABC
Proof: ∆ABE ~ ∆ACD
∴ AB = AC
and AD = AE [CPCT]
⇒ \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = 1 and \(\frac{\mathrm{AD}}{\mathrm{AE}}\) = 1
⇒ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)
Now, In ∆ADE and ∆ABC, we have
⇒ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)
⇒ \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)
and ∠DAE = ∠BAC (each = ∠A)
∆ADE ~∆ABC. [By AA similarity criterion]
Hence Proved .

Question 7.
In given figure altitudes AD and CE of AABC intersect each other at the point P. Show that :
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC.

Solution :
Given : CE ⊥ AB and AD ⊥ BC.
To Prove : (i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC.
Proof : (i) In ∆AEP and ∆CDP
∠AEP = ∠CDP (each = 90°)
∠APE = ∠CPD (vertically opposite ∠s)
∴ ∆AEP ~ ∆CDP
(By AA similarity criterion) Hence Proved.

(ii) In ∆ABD and ∆CBE
∠ADB =∠CEB (each = 90°)
∠B = ∠B (common)
∴ ∆ABD ~ ∆CBE
(By AA similarity criterion) Hence Proved.

(iii) In ∆AEP and ∆ADB
∠AEP = ∠ADB (each = 90°)
∠PAE = ∠DAB (same angle)
∴ ∆AEP ~ ∆ADB
(By AA similarity criterion) Hence Proved.

(iv) In ∆PDC and ∆BEC
∠PDC = ∠BEC (each = 90°)
∠PCD = ∠BCE (same angle)
∆PDC ~ ∆BEC
(By AA similarity criterion) Hence Proved.

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.

Solution :
Given : E is a point on side AD produced of a ||gm ABCD and BE intersects CD at F.
To Prove : ∆ABE ~ ∆CFB.
Proof : In ∆ABE and ∆CFB.
∠AEB = ∠CBF (alternate interior ∠s)
∠A = ∠C (opposite ∠s of a ||gm )
∴ ∆ABE ~ ∆CFB
(By AA similarity criterion) Hence Proved.

Question 9.
In given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that :
(i) ∆ABC ~ ∆AMP.
(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\).

Solution:
Given: ∆ABC and ∆AMP are two right angled triangles right angles at B and M respectively.
To Prove :(i) ∆ABC ~ ∆AMP.
(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\).
Proof : (i) In ∆ABC and ∆AMP.
∠ABC = ∠AMP (each = 90°)
∠A = ∠A (common)
∴ ∆ABC ~ ∆AMP (By AA similarity criterion)
Hence Proved.

(ii) ∵ ∆ABC ~ ∆AMP
∴ \(\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)
(corresponding sides of similar triangles)
Hence Proved.

Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that:
(i) \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF
Solution:
Given : CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on the sides AB and EF of ∆ABC and ∆EFG and ∆ABC ~ ∆FEG.
To Prove :
(i) ∆DCB ~ ∆HGE
(ii) ∆DCA ~ ∆HGF
(iii) \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)
Proof : (i) ∆ABC ~ ∆FEG (given)
∠ACB = ∠EGF (corresponding ∠s of similar ∆s)
\(\frac{1}{2}\) ∠ACB = \(\frac{1}{2}\) ∠EGF
[CD and GH are bisectors of ∠C and ∠G.]

∠DCB = ∠HGE ………………..(1)
and ∠B = ∠E
(corresponding ∠s of similar ∆s)
⇒ ∠DBC = ∠HEG ……………….(2)
From (1) and (2), we have
∆DCB ~ ∆HGE
Hence Proved.

(ii) In ∆DCA and ∆HGF
∠C = ∠G
(corresponding ∠s of similar ∆s)
∠ACD = ∠HGF [CD and GH are bisectors of ∠C and ∠G] ………………..(1)
∠A = ∠F
(corresponding ∠s of similar ∆s) ……………….(2)
∠ADC = ∠FHG
(third ∠s of two triangles) …………………(3)
∴ ∆DCA ~ ∆HGF. (By AAA similarity criterion)
Hence Proved.

(iii) ∵ ∆DCA ~ ∆HGF.
∴ \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)
(corresponding sides of similar ∆s)
Hence Proved.

Question 11.
In given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.

Solution :
Given : AB = AC, AD ⊥ BC and EF ⊥ AC.
To Prove : ∆ABD ~ ∆ECF
Proof : AB = AC (given)
∠B = ∠C [∵ ∠s opposite to equal side are equal]
Now, In ∆ABD and ∆ECF,
∠ABD = ∠ECF [∵ ∠B = ∠C]
∠ADB = ∠EFC = 90°
[∵ AD ⊥ BC and EF ⊥ AC]
∴ ∠ABD ~ ∆ECF
(By AA similarity criterion)
Hence Proved.

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see figure below). Show that ∆ABC ~ ∆PQR.

Solution:

(By SSS similarity criterion)
∠B = ∠Q
(Corresponding ∠s of Similar As)
Now, in ∆ABC and ∆PQR.
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\) (given)
∠B = ∠Q (Proved above)
∴ ∆ABC ~ ∆PQR [By SAS similarity criterion]
Hence Proved.

Question 13.
D is a point on the side BC of a trianlge ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD
Solution:
Given : ∠ADC = ∠BAC
To prove : CA² = CB.CD
Proof : In ∆ABC and ∆DAC
∠ADC = ∠BAC (given)
∠C = ∠C (common)
∆ABC ~ ∆DAC (By AA similarity criterion)

⇒ CA² = CB × CD.
Hence Proved.

Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution :
Given : In ∆ABC and ∆PQR
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
To Prove : ∆ABC ~ ∆PQR
Construction : Produce AD to E such that AD = DE and PM to N such that PM = MN Join CE and NR.
Proof : In ∆ADB and ∆EDC
AD = ED (By construction)
BD = CD (AD is median)
∠ADB = ∠CDE (vertically opposite ∠s)
∆ADB ≅ ∆EDC (By SAS Congruence criterion)
⇒ AB = EC (By CPCT) …………….(1)

Similarly, ∆PMQ ≅ ∆NMR (By SAS congruence criterion)
⇒ PQ = NR (By CPCT) …………..(2)
Now, \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)(given)
⇒ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
[From (1) and (2), AB = EC and PQ = NR]
⇒ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}\)
⇒ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
[By construction AD = DE and PM = MN)
∴ ∆ACE ~ ∆PRN (By SSS similarity criterion)
∠1 = ∠3 ……………….(3)
(Corresponding ∠s of similar triangles)
Similarly ∠2 = ∠4 ………………(4)
Adding (3) and (4) we get
∠1 + ∠2 = ∠3 + ∠4.
⇒ ∠A = ∠P.
Now ∆ABC and ∆PQR.
∠A = ∠P (Proved above)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}\) (given)
∴ ∆ABC ~ ∆PQR.

Question 15.
A vertical pole of length 6m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28m long. Find the height of the tower.

Solution:
Let AB be the vertical pole of length 6m casts a shadow BC and PQ be the tower casts a shadow QR. Let height of tower berm. Join AC and PR.
In ∆ABC and ∆PQR
∠B = ∠Q (each = 90°)
∠C = ∠R (Angle of elevation of the Sun)
∴ ∆ABC ~ ∆PQR (By AA similarity criterion)
∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)
[Corresponding sides of similar triangles]
⇒ \(\frac{6}{x}=\frac{4}{28}\)
⇒ x = \(\frac{6 \times 28}{4}\)
x = 42.
Hence, height of tower = 42 m.

Question 16.
If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\).

Solution :
Given : AD and PM are medians of ∆ABC and ∆PQR respectively and ∆ABC ~ ∆PQR.
To Prove: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
Proof : ∆ABC ~ ∆PQR
∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}}\) ………………(1)
(corresponding sides of similar triangles)
⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)
⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}\)
[∵ AD and PM are medians
∴ BD = CD and QM = MR]
⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)
and ∠B = ∠Q
[Corresponding ∠s of similar ∆s ABC and PQR]
∴ ∆ADB ~ ∆PMQ [By SAS similarity criterion]
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) (Corresponding sides of similar ∆s.)
Hence Proved.

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.7

Assume π = \(\frac{22}{7}\), unless stated otherwise,

Question 1.
Find the volume of the right circular cone with :
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm.
Solution:
(i) We have, Radius of cone (r) = 6 cm
Height of the cone (h) = 7 cm
∴ Volume e the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 6^2 \times 7\)
= 264 cm³.
(ii) We have,
Radius of the cone (r) = 3.5 cm
Height of the cone (h) = 12 cm
∴ Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times 12\)
= 154 cm³
Hence, volume of the cone (i) 264 cm³, (ii) 154 cm³.

Question 2.
Find the capacity in litres of a conical vessel with :
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm.
Solution:
(i) We have, Radius of the conical vessel (r) = 7 cm
Slant height of the conical vessel (l) = 25 cm
Let the height of the conical vessel be h cm.
∴ l² = h² + r²
⇒ 25² = h² + 7²
⇒ h² = 25² – 7²
⇒ h² = (25 + 7) (25 – 7)
⇒ h² = 32 × 18
⇒ h² = 576
⇒ h= \(\sqrt{576}\)
⇒ h = 24 cm
∴ Volume of the conical vessel = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24\)
= 1232 cm³
= \(\frac{1232}{1000}\) litres
= 1.232 litres.

(ii) We have, Slant height of the conical vessel (l) = 13 cm
Height of the conical vessel (h) = 12 cm
Let the radius of the conical vessel be r cm.
∴ l² = h² + r²
⇒ 13² = 12² + r²
⇒ r² = 13² – 12²
⇒ r² = (13 + 12) (13 – 12)
⇒ r² = 25
⇒ r = \(\sqrt{25}\)
⇒ r = 5 cm
∴ Volume of the conical vessel = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 5^2 \times 12\)
= \(\frac{2200}{7} \mathrm{~cm}^3\)
= \(\frac{2200}{7 \times 1000} \text { litres }\)
= \(\frac{11}{35} \text { litres }\)
Hence, volume of the cone (i) 1.232 litres, (ii) \(\frac{11}{35} \text { litres }\).

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)
Solution:
Let the radius of base of a cone be r сm.
We have, Height of the cone (h) = 15 cm
volume of the cone = 1570 cm³
⇒ \(\frac{1}{3}\)πr²h = 1570
⇒ \(\frac{1}{3}\) × 3.14 × r² × 15 = 1570
⇒ 15.7 × r² = 1570
⇒ r² = \(\frac{1570}{157}\)
⇒ r² = 100
⇒ r = \(\sqrt{100}\)
⇒ r = 10 cm
Hence, radius of the base of the cone = 10 cm.

Question 4.
If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Solution:
Let the radius of the basse of the cone be r сm.
We have, Height of the cone (h) = 9 cm
Volume of the cone = 48π cm³
⇒ \(\frac{1}{3}\)πr²h = 48π
⇒ \(\frac{1}{3}\)π × r² × 9 = 48π
⇒ 3πr² = 48π
⇒ r² = \(\frac{48 \pi}{3 \pi}\)
⇒ r² = 16
⇒ r = \(\sqrt{16}\)
⇒ r = 4 cm
∴ Diameter of the base of the cone = 2 × 4
= 8 cm
Hence,diameter of the base of cone = 8 cm

Question 5.
A conical pit of top diameter 3.5 m is 12m deep. What is its capacity in kilolitres ?
Solution:
We have,
Depth of conical pit (h) = 12 m
Diameter of conical pit (d) = 3.5 m
∴ Radius of conical pit (r) = \(\frac{3.5}{2}\) = 1.75 m
∴ Capacity of the conical pit = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times(1.75)^2 \times 12\)
= 38.5 m³
= 38.5 kilolitres
(∵ 1 m³ = 1 kilolitres)
Hence,capacity of the conical pit = 38.5 kilolitres.

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find :
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.
Solution:
We have,
The volume of the cone = 9856 cm³
The diameter of the base of the cone (d) = 28 cm
∴ Radius of the base of the cone (r) = \(\frac{28}{2}\) cm = 14 cm
(i) Let the height of the cone be h cm.
Volume of the cone = 9856 cm³
⇒ \(\frac{1}{3}\)πr²h = 9856
⇒ \(\frac{1}{3} \times \frac{22}{7} \times 14^2 \times h\) = 9856
⇒ \(\frac{616}{3}\) × h = 9856
⇒ h = \(\frac{9856 \times 3}{616}\)
⇒ h = 480cm

(ii) Let the slant height of the cone be 1 cm, then
⇒ l² = h² + r²
⇒ l² = 48² + 14²
⇒ l² = 2304 + 196
⇒ l2 = 2500
⇒ l = \(\sqrt{2500}\)
⇒ l = 50 cm.

(iii) Curved surface of the cone = πrl
= \(\frac{22}{7}\) × 14 × 50
= 2200 cm².
Hence, (i) Height of the cone = 48 cm, (ii) Slant height of the cone = 50 cm, (iii) Curved surface area of the cone = 2200 cm².

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
We have,
Sides of the triangle ABC are AB = 12 cm, BC = 5 cm and AC = 13 cm
Since ΔABC is revolved about the side AB(= 12 cm)
∴ We get a cone (AC’C) as shown in the figure.

Then,radius of the base of the cone (r) = 5 cm
and height of the cone (h) = 12 cm
∴ Volume of the cone (solid) = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × π × 5² × 12
= 100π cm³
Hence, volume of the cone so obtained = 100π cm³.

Question 8.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
Since ΔABC is revolved about BC(= 5 cm).
∴ We get a cone (CAA’) as shown in the figure.

Then,radius of the base of the cone (r) = 12 cm
and height of the cone (h) = 5 cm
Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × π × 12² × 5
= 240π cm³
Ratio of the volumes of the two solids obtained in Q. 7 and Q. 8
= 100π : 240π = \(\frac{100 \pi}{240 \pi}\)
= \(\frac{5}{2}\) = 5 : 12
Hence,volume of the cone (solid) = 240π cm³,
and required ratio of their volumes = 5 : 12.

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution:
We have,
Height of the conical heap (h) = 3 m
Diameter of the base of the conical heap (d) = 10.5 m³

∴ Radius of the base of the conical heap (d) = \(\frac{10.5}{2}\) = 5.25 m
∴ Slant height of a conical heap (l) = \(\sqrt{h^2+r^2}\)
= \(\sqrt{3^2+(5.25)^2}\)
= \(\sqrt{9+27.56}\)
= \(\sqrt{36.56}\)
= 6.05 m
∴ Volume of the conical heap = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times(5.25)^2 \times 3\)
= \(\frac{606.375}{7}\)
= 86.625 m³
Required area of canvas = Curved surface
area of the conical heap = πrl
= \(\frac{22}{7}\) × 5.25 × 6.05
= 16.5 × 6.05
= 99.825 m²
Hence,volume of the heap = 86.625 m³ and required area of canvas = 99.825 m².

Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.2

Question 1.
In given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution :
(i) we have, AD = 1.5 cm, AE = 1 cm, DB = 3 cm
DE || BC
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[By theorem 6.1 (BPT)]
\(\frac{1 \cdot 5}{3 \cdot 0}=\frac{1}{\mathrm{EC}}\) = 2
EC = \(\frac{3.0}{1.5}\) = 2
Hence, EC = 2 cm.

(ii) We have, DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm and DE || BC
∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) [By Theorem 6.1 (BPT)]
\(\frac{\mathrm{AD}}{7 \cdot 2}=\frac{1 \cdot 8}{5 \cdot 4}\)
AD = \(\frac{7.2 \times 1.8}{5.4}\) = 2.4
Hence, AD = 2.4 cm.

Question 2.
E and F are points on the sides PQ and PR respectively of a APQR. For each of the following cases, state whether EF || QR :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 018 cm and PF = 0.36 cm.

Solution :
(i) we have PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{3.0}\) = 1.3
and \(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3 \cdot 6}{2 \cdot 4}\) = 1.5
∵ \(\frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF is not parallel to QR.

(ii) We have,
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{4}{4 \cdot 5}=\frac{8}{9}\) and
\(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{8}{9}\)
∵ \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF || QR

(iii) We have,
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm,
EQ = PQ – PE = 1.28 – 0.18 = 11 cm,
PR = PF = 2.56 – 0.36 = 2.2 cm.
Now, \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{0 \cdot 18}{1 \cdot 1}=\frac{9}{55}\)
and \(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{0.36}{2 \cdot 2}=\frac{9}{55}\)
∵ \(\frac{P E}{E Q}=\frac{P F}{F R}\)
∴ EF || QR

Question 3.
In given figure, if LM || CB and LN || CD, prove that \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\).

Solution:
Given: In □ABCD, LM || CB and LN || CD
To Prove: \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)
Solution: In ∆ABC LM || CB (given)

Question 4.
In given figure DE || AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\).

Solution:
Given : In ∆ABC DE || AC and DF || AE
To prove: \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\).
Proof: In ∆BAE,
DF || AE (given)
∴ \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BF}}{\mathrm{FE}}\) ………………..(1)
[By Theorem 6.1 (BPT)]
In ∆ABC, DE || AC (given)
∴ \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}}\) [By Theorem 6.1 (BPT)]
From(1) and (2) we get
\(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\)
Hence Proved.

Question 5.
In given figure DE || OQ and DF || OR. Show that EF || QR.

Solution:
Given: In ∆POQ, DE || OQ and in POR, DF || OR
To prove: EF || QR
Proof: In ∆POQ DE || OQ (given)
∴ \(\frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PE}}{\mathbf{E Q}}\) ……………..(1)
[By Theorem 6.1 (BPT)]
In ∆POR DF || OR
∴ \(\frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PF}}{\mathrm{FR}}\) ……………..(2)
[By Theorem 6.1 (BPT)]
From (1) and (2), we get
\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF | | QR [By converse of BPT]

Question 6.
In given figure, A. B, and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:
Given: In figure A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
To prove: BC || QR
Proof: In ∆OPQ
AB || PQ (given)
\(\frac{O A}{A P}=\frac{O B}{B Q}\) …………………(1)
[By theorem 6.1 (BPT)]
In ∆OPR, AC || PR (given)
\(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}}\) …………….(2)
[By theorem 6.1 (BPT)]
From (1) and (2) we get
\(\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)
Thus B and C are respectively points on sides OQ and OR of ∆OQR, such that \(\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\).
⇒ BC || QR [By converse of BPT]
Hence Proved.

Question 7.
Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in class IX).
Solution:
Given : A ∆ABC in which D is the mid-point of AB and DE || BC and meeting AC at E.

To Prove : AE = EC
Proof : Since, DE || BC
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) ……………..(1)
[By Theorem 6.1 (BPT)]
But AD = DB
[∵ D is the mid point of AB]
∴ \(\frac{\mathrm{AD}}{\mathrm{AD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) ………………(2)
⇒ \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1
⇒ AE = EC.

Question 8.
Using theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in class IX).
Solution:
Given : A ∆ABC in which D is the mid-point of AB and E is the mid-point of AC
To prove : DE || BC
Proof : Since D and E are the mid-points of AB and AC respectively
AD = DB
⇒ \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = 1
and AE = EC
⇒ \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1

∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
⇒ DE || BC
[By Theorem of 6.2 (converse of BPT)]
Hence Proved.

Question 9.
ABCD is a trapezium in which AB | DC and its diagonals intersect each other at the point O. Show that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\).
OR
PQRS is a trapezum in which PQ || RS and its diagonals intersect each other at the point O. Prove \(\frac{\mathrm{PO}}{\mathrm{QO}}=\frac{\mathrm{RO}}{\mathrm{SO}}\)
Solution:
Given : A trapezium ABCD in which AB || CD and its diagonals AC and BD intersect at O.

To Prove: \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
Construction: Draw OE || AB through O, which meets AD at E.
Proof : We have,
EO || AB (By construction) …………(1)
DC || AB (given) …………(2)
From (1) and (2), we get
EO || DC
∴ \(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}}\) ………………..(3)
[By Theorem 6.1 (BPT)]
In ∆ DAB EO || AB
\(\frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{DO}}{\mathrm{OB}}\)
∴ \(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{OD}}\) ………………(4)
[By Theorem 6.1 (BPT)]
From (3), (4), we get
\(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{OD}}\)
⇒ \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
Hence Proved.

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\). Show that ABCD is a trapezium.
OR
The diagonals of a quadrilateral PQRS intersect each other at the point O, such that \(\frac{\mathrm{PO}}{\mathrm{QO}}=\frac{\mathrm{RO}}{\mathrm{SO}}\). Show that PQRS is a trapezium.
Solution:
Given : A quadrilateral ABCD whose diagonals AC and BD intersect at a point O such that

\(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
To Prove : ABCD is a trapezium i.e. AB || DC.
Construction : Draw OE || DC which meets AD at E.
Proof: In ∆ACD,
0E || DC (By construction)
∴ \(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{AE}}{\mathrm{ED}}\) ……………..(1)
[By Theorem 6.1 (BPT)]
But, \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\) (given)
⇒ \(\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{DO}}\) ………………..(2)
From (1) and (2), we get
\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{DO}}\)
⇒ EO || AB
[By Theorem 6.2 (converse of BPT)1
But, OE || DC (By construction)
From (3) and (4), we get
AB || CD
Hence, ABCD is a trapezium. Hence Proved.

Haryana State Board HBSE 6th Class English Solutions A Pact with the Sun Chapter 2 The Friendly Mongoose Textbook Exercise Questions and Answers.

Haryana Board 6th Class English Solutions A Pact with the Sun Chapter 2 The Friendly Mongoose

HBSE 6th Class English The Friendly Mongoose Textbook Questions and Answers

The Friendly Mongoose HBSE 6th Class English Question 1.
Why did the farmer bring a baby mongoose into the house?
Answer:
The farmer brought a baby mongoose into the house to provide a companion to his son.

The Friendly Mongoose Question Answer HBSE 6th Class English Question 2.
Why didn’t the farmer’s wife want to leave the baby alone with the mongoose?
Answer:
The farmer’s wife didn’t want to have the baby alone with the mongoose because she thought the mongoose could harm her son.

The Friendly Mongoose Class 6 Questions And Answers Question 3.
What was the farmer’s comment on his wife’s fears?
Answer:
The farmer commented that the mongoose was a friendly animal and that he was the best of his son’s friend.

The Friendly Mongoose Summary HBSE 6th Class English Question 4.
What did the farmer’s wife strike the mongoose with her basket?
Answer:
The farmer’s wife struck the mongoose with the basket because she thought the mongoose had killed her son.

The Friendly Mongoose Questions And Answers HBSE 6th Class English Question 5.
Did she repent her hasty action? How does she show her repentance?
Ans.
Yes, she repented her hasty actions as she shed tears later.

The Friendly Mongoose Question Answers 6th Class HBSE English  Question 6.
Do you have a pet a cat or a dog? If not, would you like one? How would you look after it? Are you for or against keeping birds in a cage as pets ?
Ans.
Do it yourself.

HBSE 6th Class English The Friendly Mongoose Important Questions and Answers

Friendly Mongoose Question Answer HBSE 6th Class English Question 1.
Who said this to whom and when? ‘You need n’t be a fraid.’
Answer:
The farmer said this to his wife. He said this when she was going to the market and she told that she did not want her baby to stay alone with mongoose.

Summary Of The Friendly Mongoose HBSE 6th Class English Question 2.
Why did the farmer not return for quite some time?
Answer:
The farmer did not return for quite some time because he met some friends on the way back from fields.

Question 3.
What was customary for the mongoose?
Answer:
It was customary for the mongoose to welcome the farmer’s wife when she returned home.

Question 4.
How did the mongoose prove himself to be a true friend of the baby?
Answer:
When the baby’s parents were not at home the mongoose saw a snake near the baby. He attacked the snake and tore it into pieces. Thus he saved the child. Hence, he proved himself to be a true friend of the baby.

Question 5.
What is the moral of the story?
Answer:
The moral of the story is ‘Don’t act hastily’, and ‘Think twice before you act’.

The Friendly Mongoose Word Meanings

village (विलेज) = गाँव, pet (पेट) = पालतू जानवर, companion (कम्पेनियन) = साथी, tiny (टाइनी) = सूक्ष्म, mongoose (मेन्गूस) = नेवला, fuilly grown (फूलीग्रोन) = पूर्ण विकसित, lovely (लवली) = सुन्दर, रम्य, shining (शाइनी) = चमकदार, bushy (बुशी)- झाड़ीदार, tail (टेल) = पूँछ। Cradle (क्रेडल) = पालना, alternately (आल्टरनेटली) = बारी-बारी से, rocked (रॉक्ड) = हिलाया, basket (बास्केट) टोकरी, friendly (फ्रेन्डली) = मित्रतापूर्ण तरीके से, fields (फिल्डस) = खेत, return (रिटन) = वापसी, finished (फिनिश्ड) = खत्म किया, shooping (शोपिंग) = खरीददारी, groceries (ग्रोसरीज) = घरेलू सामान, welcome (वेलकम) = स्वागत, customary (कास्टमरी) = रिवाज के अनुसार, screamed (स्क्रीम्ड) = चीखी, blood (ब्लड) = खून, paws (पॉज) = पंजे, smeared (स्मीअड) = लेप किया हुआ, सना हुआ, wicked (विकेड) = दुष्ट, screamed (स्क्रीम्ड) = चीखी, hysterically (हिस्ट्रीकली) = उन्मत्त (पागल) जैसा, blind (ब्लाइन्ड) = अंधी, rage (रेग) = नाराज, गुस्सा, strength (स्ट्रेन्थ) = ताकत, askeep (एसलीप) = सोया हुआ, torn (टान) = कटा-फटा होना, bleeding (ब्लीडिंग) = खून बहते हुए, saved (सेव्ड) = बचाया, unaware (अनवेयर) = अनजान था, sobbing (सॉबिंग) = सुबकते हुए, hastily (हेस्टली) = जल्दी से, rashly (रेशली) = लापरवाही से, stared (स्टेयड) = ताका, wiping (वाईपिंग) = पोंडते हुए, tears (टीअर्स) = आँसू, feed (फीड) = पिलाना।

The Friendly Mongoose Summary in English

There lived a farmer, his wife and their small son in a village. The farmer and his wife brought a mongoose to give their son a companion. One day the farmer and his wife had to go out of their house leaving their son alone with the mongoose. When the farmer’s wife returned home, she found the mongoose at the entrance with blood smeared on his face and paws. She thought her son had been killed by the mongoose and as such she killed the mongoose then and there. But when she went to her son, she learnt the truth. In fact the mongoose had guarded her son from the attack of a snake and that’s why her face and paws were smeared with blood. She went to the mongoose and repented a lot. But it was of no use.

The Friendly Mongoose Summary in Hindi

एक गाँव में एक किसान, उसकी पत्नी और उनका छोटा लड़का रहते थे। किसान और उसकी पत्नी एक नेवले को अपने पुत्र को एक साथी की संगति देने के लिए ले आये। एक दिन किसान और उसकी पत्नी को अपने बच्चे को अकेला छोड़कर घर से बाहर जाना पड़ा। जब किसान की पत्नी घर वापस लौटी, उसने नेवले को प्रवेश द्वार पर उसके चेहरे व पंजे को खून से सना हुआ पाया। उसने सोचा नेवले ने उसके पुत्र को मार दिया ऐसा सोचकर उसने भी उस नेवले को मार दिया। लेकन जब वह अपने पुत्र के पास गई तो उसे सच का पता चला। वास्तव में नेवले ने सांप के हमले से उसके लड़के की रक्षा की थी और इसी वजह से उसका चेहरा और पंजे खून से सन गये थे। वह नेवले के पास गई और काफी पश्चात्ताप किया। लेकिन इसका कोई फायदा नहीं था।