Class 9

Haryana State Board HBSE 9th Class Maths Solutions Chapter 2 बहुपद Ex 2.5 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 2 बहुपद Exercise 2.5

प्रश्न 1.
उपयुक्त सर्वसमिकाओं को प्रयोग करके निम्नलिखित गुणनफल ज्ञात कीजिए
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (3 – 2x) (3 + 2x)
हल :
(i) (x + 4) (x + 10) = x² + (4 + 10)x + 4 × 10
= x² + 14x + 40 उत्तर
(ii) (x + 8 ) (x – 10) = x² + (8 – 10)x + 8 × (-10)
= x² – 2x – 80 उत्तर
(iii) (3x + 4) (3x – 5) = (3x)² + (4 – 5) × 3x + 4 × (-5)
= 9x² + (-1) × 3x – 20
= 9x² – 3x – 20 उत्तर

(iv) (y² + \(\frac {3}{2}\)) (y² – \(\frac {3}{2}\))
= (y²)² – (\(\frac {3}{2}\))²
= y⁴ – \(\frac {9}{4}\) उत्तर

(v) (3 – 2x) (3 + 2x) = (3)² – (2x)²
= 9 – 4x² उत्तर

प्रश्न 2.
सीधे गुणा किए बिना निम्नलिखित गुणनफलों के मान ज्ञात कीजिए [March, 2020]
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
हल :
(i) 103 × 107 = (100 + 3) (100 + 7)
= (100)² + (3 + 7) × 100 + 3 × 7
= 10000 + 10 × 100 + 21
= 10000 + 1000 + 21 = 11021 उत्तर

(ii) 95 × 96 = (100 – 5)(100 – 4)
= (100)² + (-5 – 4) × 100 + (-5)(-4)
= 10000 – 900 + 20
= 10020 – 900 = 9120 उत्तर

(iii) 104 × 96 = (100 + 4) (100 – 4)
= (100)² – (4)²
= 10000 – 16 = 9984 उत्तर

प्रश्न 3.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित का गुणनखंडन कीजिए
(i) 9x² + 6xy + y²
(ii) 4y² – 4y + 1
(iii) x² – \(\frac{y^2}{100}\) [B.S.E.H. 2019]
हल :
(i) 9x² + 6xy + y² = (3x)² +2 (3x) (y) + (y)²
= (3x + y)² = (3x + y) (3x + y) उत्तर
(ii) 4y² – 4y + 1 = (2y)² – 2 (2y)(1) + (1)²
= (2y – 1)² = (2y – 1) (2y- 1) उत्तर
(iii) x² – \(\frac{y^2}{100}\) = (x)² – (\(\frac {y}{10}\))²
= (x – \(\frac {y}{10}\)) (x + \(\frac {y}{10}\)) उत्तर

प्रश्न 4.
उपयुक्त सर्वसमिकाओं का प्रयोग करके निम्नलिखित में से प्रत्येक का प्रसार कीजिए
(i) (x + 2y + 4z)²
(ii) (2x – y + z)²
(iii) (-2x + 3y + 2z)²
(iv) (3a – 7b – c)²
(v) (-2x + 5y – 3z)²
(vi) [\(\frac {1}{4}\)a – \(\frac {1}{2}\)b + 1]²
हल :
(i) (x + 2y + 4z)² = x² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2 (4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 8zx उत्तर

(ii) (2x – y + z)² = [2x + (-y) + z]²
= (2x)² + (-y)² + z² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx उत्तर

(iii) (-2x + 3y + 2z) = [(-2x) + 3y + 2z²]
= (-2x)² + (3y)² + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx उत्तर

(iv) (3a – 7b – c)² = [3a + (-7b) + (-c)]² = (3a)² + (-7b)² + (-c)² + 2(3a)(-7b) + 2(-7b)(-c) + 2 (-c)(3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ac उत्तर

(v) (-2x + 5y -3z)² = [(-2x) + 5y + (-3z)]² = (-2x)² + (5y)² + (-3z)² + 2(-2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx उत्तर

प्रश्न 5.
गुणनखंडन कीजिए –
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8xz
हल:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz = (2x)² + (3y)² + (-4z)² + 2(-4z)(2x)
= (2x + 3y – 4z)²
= (2x + 3y – 4z) (2x + 3y – 4z) उत्तर

(ii) 2x² + y² + 8z² – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8xz = (-\(\sqrt{2}\)x)² + (y)² + (2\(\sqrt{2}\)z)² + 2(\(\sqrt{2}\)z) (-\(\sqrt{2}\)x)
= (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)\(\sqrt{2}\)
= (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z) (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z) उत्तर

प्रश्न 6.
निम्नलिखित घनों को प्रसारित रूप में लिखिए
(i) (2x + 1)³
(ii) (2a – 3b)³
(iii) [\(\frac {3}{2}\)x + 1]³
(iv) [x – \(\frac {2}{3}\)y]³ [B.S.E.H. March, 2018]
हल :
(i) (2x + 1)³ = (2x)³ + (1)³ + 3 × 2x × 1 (2x + 1)
[∵ (a + b)³ = a³ + b³ + 3ab(a + b)]
= 8x³ + 1 + 6x (2x + 1)
= 8x³ + 1 + 12x² + 6x
= 8x³ + 12x² + 6x + 1 उत्तर

(ii) (2a – 3b)³ = (2a)³ – (3b)³ – 3 (2a) (3b)(2a – 3b)
[∵ (a – b)³ = a³ – b³ – 3ab(a – b)]
= 8a³ – 27b³ – 18ab (2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab² उत्तर

प्रश्न 7.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित के मान ज्ञात कीजिए
(i) (99)³ (ii) (102)³ (iii) (998)³
हल :
(i) (99)³ = (100 – 1)³
= (100)³ – (1)³ – 3 × 100 × 1 (100 – 1)
= 1000000 – 1 – 30000 + 300
= 1000300 – 30001 = 970299 उत्तर

(ii) (102)³ = (100 + 2)³
= (100)³ + (2)³ + 3 × 100 × 2(100 + 2)
= 1000000 + 8 + 600 (100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208 उत्तर

(iii) (998)³ = (1000 – 2)³
= (1000)³ – (2)³ – 3 × 1000 × 2(1000 – 2)
= 1000000000 – 8 – 6000 × (1000 – 2)
= 1000000000 – 8 – 6000000 + 12000
= 1000012000 – 6000008
= 994011992 उत्तर

प्रश्न 8.
निम्नलिखित में से प्रत्येक का गुणनखंडन कीजिए
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³ – 12a²b + 6ab²
(iii) 27 – 125a³ – 135a + 225a²
(iv) 64a³ – 27b³ – 144a²b + 108ab²
(v) 27p³ – \(\frac{1}{216}-\frac{9}{2}\)p² + \(\frac {1}{4}\)p [March 2020]
हल :
(i) 8a³ + b³ + 12a²b + 6ab²
= (2a)³ + (b)³ + 3 (2a) (b) (2a + b)
= (2a + b)³
= (2a + b)(2a + b)(2a + b) उत्तर

(ii) 8a³ – b³ – 12a²b + 6ab²
= (2a)³ – (b)³ – 3 (2a)(b)(2a – b)
= (2a – b)³
= (2a – b)(2a – b)(2a – b) उत्तर

(iii) 27 – 125a³ – 135a + 225a²
= (3)³ – (5a)³ – 3(3)(5a) (3 – 5a)
= (3 – 5a)³ = (3 – 5a) (3 – 5a) (3 – 5a) उत्तर

(iv) 64a³ – 27b³ – 144a²b + 108aba²
= (4a)³ – (3b)³ – 3 (4a) (3b) (4a – 3b)
= (4a – 3b)³
= (4a – 3b) (4a – 3b) (4a – 3b) उत्तर

प्रश्न 9.
सत्यापित कीजिए
(i) x³ + y³ = (x + y) (x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)
हल:
(i) R.H.S. = (x + y) (x² – xy + y²)
= (x) (x² – xy + y²) + y (x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³
= L.H.S.
अतः सत्यापित हुआ।

(ii) R.H.S. = (x – y) (x² + xy + y²)
= x (x² + xy + y²) – y (x² + xy + y²)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³
= L.H.S.
अतः सत्यापित हुआ।

प्रश्न 10.
निम्नलिखित में से प्रत्येक का गुणनखंडन कीजिए
(i) 27y³ + 125z³ [B.S.E.H. 2019]
(ii) 64m³ – 343n³
हल :
(i) 27y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z) [(3y)² – 3y × 5z + (5z)²]
= (3y + 5z) [9y² – 15yz + 25z²] उत्तर

(ii) 64m³ – 343n³ = (4m)³ – (7n)³
= (4m – 7n) [(4m)² + 4m × 7n + (7n)²]
= (4m – 7n) [16m² + 28mn + 49n²] उत्तर

प्रश्न 11.
गुणनखंडन कीजिए
27x³ + y³ + z³ – 9xyz [B.S.E.H. March, 2017, 2018]
हल :
हम जानते हैं कि a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
अब 27x³ + y³ +z³ – 9xyz
= (3x)³ + (y)³ + (z)³ -3 (3x) (y) (z)
= (3x + y + z) [(3x)² + (y)² + (z)² – (3x)(y) – (y)(z) – (z)(3x)]
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3zx) उत्तर

प्रश्न 12.
सत्यापित कीजिए
x³ + y³ + z³ – 3xyz = \(\frac {1}{2}\)(x + y + z) [(x – y)² + (y – z)² + (z – x)²]
हल :
यहाँ पर (x – y)² + (y – z)² + (z – x)² = (x² + y² – 2xy) + (y² + z² – 2yz) + (z² + x² – 2zx)
= 2x² + 2y² + 2z² – 2xy – 2yz – 2zx
= 2 (x² + y² + z² – xy – yz – zx)
दोनों पक्षों को \(\frac {1}{2}\)(x + y + z) से गुणा करने पर,
\(\frac {1}{2}\)(x + y + z) [(x – y)² + (y – z)² + (z – x)²]
= \(\frac {1}{2}\)(x + y + z) × 2 (x² + y² + z² – xy – yz – zx)
= (x + y + z) (x² + y² + z² – xy – yz – zx) (i)
हम जानते हैं कि
x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx) (ii)
समीकरण (i) व (ii) से
x³ + y³ + z³ – 3xyz = \(\frac {1}{2}\)(x + y + z) [(x – y)² + (y – z)² + (z – x)²] |इति सिद्धम]

प्रश्न 13.
यदि x + y + z = 0 हो, तो दिखाइए कि x³ + y³ + z³ = 3xyz है।
हल :
यहाँ पर
x + y + z = 0
⇒ x + y = – z
दोनों पक्षों को घन करने पर
(x + y)³ = (-z)³
या x³ + y³ + 3xy (x + y) = – z³
या x³ + y³ – 3xyz = – z³ [∵ x + y = – z]
या x³ + y³ + z³ = 3xyz [इति सिद्धम]

प्रश्न 14.
वास्तव में पनों का परिकलन किए बिना निम्नलिखित में से प्रत्येक का मान ज्ञात कीजिए –
(i) (-12)³ + (7)³ + (5)³
(ii) (28)³ + (-15)³ + (-13)³ [B.S.E.H. March, 2020]
हल:
(i) माना x = – 12, y = 7 और z = 5
यहाँ पर
x + y + z = – 12 + 7 + 5 = 0
यदि x + y + z = 0 हो तो
x³ + y³ + z³ = 3xyz
अतः
(-12)³ + (7)³ + (5)³ = 3 × (-12) × 7 × 5 उत्तर
= – 1260 उत्तर

(ii) माना x = 28, y = – 15 और z = – 13
यहाँ पर
x + y + z = 28 – 15 – 13 = 0
यदि x + y + z = 0 हो तो
x³ + y³ + z³ = 3xyz
(28)³ + (-15)³ + (-13)³ = 3 × (28) × (-15) × (-13)
= 16380 उत्तर

प्रश्न 15.
नीचे दिए गए आयतों, जिनमें उनके क्षेत्रफल दिए गए हैं, में से प्रत्येक की लंबाई और चौड़ाई के लिए संभव व्यंजक दीजिए
(i) क्षेत्रफल : 25a² – 35a +12
(ii) क्षेत्रफल : 35y² + 13y – 12
हल :
(i) 25a² – 35a + 12 = 25a² – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
हम जानते हैं कि
आयत का क्षेत्रफल = लंबाई × चौड़ाई
⇒ 25a² – 35a + 12 = (5a – 3) × (5a – 4)
अतः आयत की लंबाई 5a-3 तथा चौड़ाई 5a-4 उत्तर

(ii) 35y² + 13y – 12 = 35y² + 28y – 15y – 12
= 7y (5y + 4) – 3 (5y + 4)
= (5y + 4)(7y – 3)
हम जानते हैं कि
आयत का क्षेत्रफल = लंबाई × चौड़ाई
⇒ 35y² + 13y – 12 = (7y – 3) × (5y + 4)
अतः आयत की लंबाई 7y – 3 तथा चौड़ाई 5y + 4 उत्तर

प्रश्न 16.
बनाभों (cuboids), जिनके आयतन नीचे दिए गए हैं, की विमाओं के लिए संभव व्यंजक क्या हैं ?
(i) आयतन : 3x² – 12x
(ii) आयतन : 12ky² + 8ky – 20k
हल :
(i) 3x² – 12x = 3x(x – 4)
= 3 × x × (x – 4)
हम जानते हैं कि
घनाभ का आयतन = लंबाई × चौड़ाई × ऊँचाई
अतः दिए गए व्यंजक की लंबाई, चौड़ाई एवं ऊँचाई क्रमशः 3, x तथा (x – 4) है। उत्तर

(ii) 12ky² + 8ky – 20k = 4k [3y² + 2y – 5]
= 4k [3y² + 5y – 3y – 5]
= 4k [y(3y + 5) – 1(3y + 5)]
= 4k (3y + 5)(y – 1)
हम जानते हैं कि
घनाभ का आयतन = लंबाई × चौड़ाई × ऊँचाई
अतः दिए गए व्यंजक की लंबाई, चौड़ाई एवं ऊँचाई क्रमशः 4k, (3y + 5) तया (y – 1) है। उत्तर

Haryana State Board HBSE 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Exercise 9.1

प्रश्न 1.
निम्नलिखित आकृतियों में से कौन-सी आकृतियां एक ही आधार और एक ही समांतर रेखाओं के बीच स्थित हैं? ऐसी स्थिति में, उभयनिष्ठ आधार और दोनों समांतर रेखाएं लिखिए।

हल :
(i) ΔPDC व समांतर चतुर्भुज ABCD एक ही आधार DC तथा समांतर रेखाओं DC व AB के मध्य में स्थित हैं।
(ii) चतुर्भुज SRNM व SRQP एक आधार SR पर हैं परंतु एक ही समांतर रेखाओं के बीच नहीं हैं।
(iii) ΔTQR व समांतर चतुर्भुज PQRS एक ही आधार QR तथा समांतर रेखाओं QR व PS के मध्य में स्थित हैं।
(iv) ΔRPQ व || चतुर्भुज ABCD एक ही आधार पर नहीं हैं परंतु एक ही समांतर रेखाओं BC व AD के मध्य में स्थित हैं।
(v) समांतर चतुर्भुज ABCD व समांतर चतुर्भुज APQD एक ही आधार AD तथा एक ही समांतर रेखाओं AD व BQ के मध्य में स्थित हैं।
(vi) चतुर्भुज PSDA, PSCB व PSRQ एक ही आधार PS पर हैं परन्तु एक ही समांतर रेखाओं के बीच नहीं हैं।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 5 युक्लिड के ज्यामिति का परिचय Ex 5.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 5 युक्लिड के ज्यामिति का परिचय Exercise 5.1

प्रश्न 1.
निम्नलिखित कथनों में से कौन-से कथन सत्य हैं और कौन-से कथन असत्य हैं ? अपने उत्तरों के लिए कारण दीजिए
(i) एक बिंदु से होकर केवल एक ही रेखा खींची जा सकती है।
(ii) दो भिन्न बिंदुओं से होकर जाने वाली असंख्य रेखाएँ हैं।
(iii) एक सांत रेखा दोनों ओर अनिश्चित रूप से बढ़ाई जा सकती है।
(iv) यदि दो वृत्त बराबर हैं, तो उनकी त्रिज्याएँ बराबर होती हैं।
(v) आकृति में, यदि AB = PQ और PQ = XY है, तो AB = XY होगा।

हल :
(i) यह कथन असत्य है, क्योंकि एक बिंद से होकर एक नहीं, बल्कि अनेक रेखाएँ खींची जा सकती हैं। जैसे बिंद O से दर्शाई गई हैं।

(ii) यह कथन असत्य है, क्योंकि दो बिंदुओं से होकर एक ओर केवल एक ही रेखा खींची जा सकती है, जैसे बिंदु X और Y से होती हुई केवल एक ही रेखा खींची जा सकती है।

(iii) यह कथन सत्य है, क्योंकि अभिगृहीत 2 अनुसार एक सांत रेखा (Terminated Line) को अनिश्चित रूप से बढ़ाया जा सकता है। जैसे AB को बढ़ाया हुआ दिखाया गया है।

(iv) यह कथन सत्य है, क्योंकि यदि एक वृत्त द्वारा घेरे गए क्षेत्र को दूसरे वृत्त पर अध्यारोपित किया जाए तो वे संपाती होंगे। अतः इनके केंद्र और परिसीमाएँ संपाती होंगी। इस कारण इनकी त्रिज्याएँ संपाती होंगी।
(v) यह कथन सत्य है, क्योंकि यूक्लिड के प्रथम अभिगृहीत अनुसार वे बस्तुएँ जो एक ही वस्तु के बराबर हों एक-दूसरे के बराबर होती हैं।

प्रश्न 2.
निम्नलिखित पदों में से प्रत्येक की परिभाषा दीजिए। क्या इनके लिए कुछ ऐसे पद हैं, जिन्हें परिभाषित करने की आवश्यकता है? वे क्या हैं और आप इन्हें कैसे परिभाषित कर पाएँगे ?
(i) समांतर रेखाएँ[B.S.E.H, 2018]
(ii) लंब रेखाएँ
(iii) रेखाखंड [B.S.E.H. 2018]
(iv) वृत्त की त्रिज्या
(v) वर्ग
हल :
उपरोक्त पदों को परिभाषित करने के लिए हमें निम्नलिखित पदों की आवश्यकता पड़ेगी
(a) बिंदु – एक बिंदु केवल कल्पना है जिसकी न कोई लंबाई, न कोई चौड़ाई और न कोई मोटाई होती है अर्थात एक बिंदु वह है जिसका कोई भाग नहीं होता।
(b) रेखा – एक रेखा चौड़ाई रहित लंबाई होती है। इसे दोनों ओर अनिश्चित रूप से बढ़ाया जा सकता है।
(c) तल – तल की लंबाई तथा चौड़ाई होती है। इसकी कोई मोटाई नहीं होती; जैसे कागज।
(d) किरण – एक निश्चित बिंदु से चलकर अंत तक जाने वाली रेखा किरण कहलाती है अर्थात इसका आरंभिक बिंदु होता है। परंतु अंत बिंदु नहीं होता; जैसे किरण AB दशाई गई है।

(e) कोण – एक उभयनिष्ठ बिंदु वाली दो किरणों का सम्मिलन कोण कहलाता है; जैसे \(\overline{AB}\) और \(\overline{AC}\) के बीच का क्षेत्र ∠BAC कहलाता है।

(f) वृत्त – किसी तल में किसी निश्चित बिंदु से उसी तल में दी गई समान अचर दूरी पर स्थित बिंदुओं का समुच्चय वृत्त कहलाता है। निश्चित बिंदु को वृत्त का केंद्र तथा निश्चित दूरी को वृत्त की त्रिज्या कहते हैं।

(i) समांतर रेखाएँ-वे दो रेखाएँ समांतर रेखाएँ कहलाती हैं, जब (a) वे प्रतिच्छेद न करें, (b) जब वे समतलीय हों।
आकृति में l₁ तथा l₂ दो समांतर रेखाएँ दर्शाई गई हैं।

(ii) लंब रेखाएँ – दो रेखाएँ AB तथा CD जो एक ही तल में स्थित हों, लंब रेखाएँ कहलाती हैं, यदि वे समकोण बनाएँ; जैसे AB ⊥ CD.

(iii) रेखाखंड – एक रेखाखंड रेखा का वह भाग होता है, जब दो विभिन्न बिंदु A तथा B एक रेखा पर दिए गए हैं, तब रेखा का भाग जिसके अंतःबिंदु A तथा B हों रेखाखंड कहलाती है।

इसे AB कहा जाता है। AB तथा BA एक ही रेखाखंड को दर्शाते हैं।

(iv) वृत्त की त्रिज्या – केंद्र से वृत्त की परिधि पर एक बिंदु की दूरी वृत्त की त्रिज्या कहलाती है।

(v) वर्ग-वह चतुर्भुज जिसकी चारों भुजाएँ बराबर हों तथा प्रत्येक कोण समकोण हो, वर्ग कहलाता है।

प्रश्न 3.
नीचे दी हुई दो अभिधारणाओं पर विचार कीजिए
(i) दो भिन्न बिंदु A और B दिए रहने पर, एक तीसरा बिंदु C ऐसा विद्यमान है, जो A और B के बीच स्थित होता है।
(ii) यहाँ कम-से-कम ऐसे तीन बिंदु विद्यमान हैं कि वे एक रेखा पर स्थित नहीं हैं।
क्या इन अभिधारणाओं में कोई अपरिभाषित शब्द हैं ? क्या ये अभिधारणाएँ अविरोधी हैं ? क्या ये यूक्लिड की अभिधारणाओं से प्राप्त होती हैं ? स्पष्ट कीजिए।
हल :
ऐसे अनेक अपरिभाषित शब्द हैं जिनकी जानकारी छात्र को होनी चाहिए। ये संगत होते हैं, क्योंकि इनमें दो अलग-अलग स्थितियों का अध्ययन किया जाता है अर्थात
(i) यदि दो बिंदु A और B दिए हुए हों, तो उनके बीच में स्थित एक बिंद होता है।
(ii) यदि A और B दिए हुए हों, तो आप एक ऐसा बिंद ले सकते हैं जो A और B से होकर जाने वाली रेखा पर स्थित नहीं होता।
ये अभिगृहीत यूक्लिड की अभिगृहीतों का अनुसरण नहीं करते। फिर भी ये अभिगृहीत 5.1 का अनुसरण करते हैं क्योंकि दो दिए विभिन्न बिंदुओं से केवल एक रेखा ही गुजर सकती है।

प्रश्न 4.
यदि दो बिंदुओं A और B के बीच एक बिंदु C ऐसा स्थित है कि AC = BC है, तो सिद्ध कीजिए कि AC = \(\frac {1}{2}\) AB है। एक आकृति खींचकर इसे स्पष्ट कीजिए। [B.S.E.H. March 2017, 2019]
हल :
यहाँ पर एक बिंदु C है जोकि दो बिंदुओं A तथा B के बीच में इस प्रकार स्थित है कि AC = BC. दोनों ओर AC जोड़ने पर,

AC + AC = AC + BC
⇒ 2AC = AB [∵ AC + CB, AB के संपाती है।]
या AC = \(\frac {1}{2}\)AB
इति सिद्धम

प्रश्न 5.
प्रश्न 4 में, C रेखाखंड AB का एक मध्य-बिंदु कहलाता है। सिद्ध कीजिए कि एक रेखाखंड का एक और केवल एक ही मध्य-बिंदु होता है।
हल :

माना AB का अन्य मध्य-बिंदु D है.
AD = DB ……..(i)
लेकिन दिया गया है C, AB का मध्य-बिंदु है।
⇒ AC = CB ……..(ii)
समीकरण (ii) में से समीकरण (i) को घटाने पर
AC – AD = CB – DB
या DC = – DC
या DC + DC = 0
या 2DC = 0
या DC = \(\frac {0}{2}\) = 0
इस प्रकार C और D संपाती बिंदु हैं।
अतः प्रत्येक रेखाखंड का एक और केवल एक ही मध्य-बिंदु होता है।

प्रश्न 6.
आकृति में, यदि AC = BD है, तो सिद्ध कीजिए कि AB = CD है। [B.S.E.H. March, 2020]

हल :
यहाँ पर दिया है
AC = BD
आकृति अनुसार
AB + BC = BC + CD
[∵ AC = AB + BC तथा BD = BC + CD]
या AB + BC – BC = CD
या AB = CD [इति सिद्धम]

प्रश्न 7.
यूक्लिड की अभिगृहीतों की सूची में दिया हुआ अभिगृहीत 5 एक सर्वव्यापी सत्य क्यों माना जाता है ? (ध्यान दीजिए कि यह प्रश्न पाँचवीं अभिधारणा से संबंधित नहीं है।)
हल :
क्योंकि यूक्लिड की अभिगृहीतों की सूची में दिया हुआ अभिगृहीत 5 ब्रह्मांड की प्रत्येक चीज के लिए सत्य है। इसलिए यह सदैव सत्य है।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 12 हीरोन का सूत्र Ex 12.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 12 हीरोन का सूत्र Exercise 12.1

प्रश्न 1.
एक यातायात संकेत बोर्ड पर ‘आगे स्कूल है’ लिखा है और यह भुजा ‘a’ वाले एक समबाहु त्रिभुज के आकार का है। हीरोन के सूत्र का प्रयोग करके इस बोर्ड का क्षेत्रफल ज्ञात कीजिए। यदि संकेत बोर्ड का परिमाप 180 cm है, तो इसका क्षेत्रफल क्या होगा ?
हल :
यहाँ समबाहु त्रिभुज की प्रत्येक भुजा = a अर्थात a = a, b = a, c = a
∴ s = \(\frac{a+a+a}{2}=\frac{3 a}{2}\)

a + a + a = 180 cm
3a = 180
a = \(\frac{180}{3}\) = 60cm
∵ वांछित क्षेत्रफल = \(\frac{\sqrt{3}}{4}\) × (60)² cm²
= \(\frac{\sqrt{3}}{4}\) × 3600 = 900\(\sqrt{3}\) cm² उत्तर

प्रश्न 2.
किसी फ्लाईओवर (flyover) की त्रिभुजाकार दीवार को विज्ञापनों के लिए प्रयोग किया जाता है। दीवार की भुजाओं की लंबाइयाँ 122 m, 22 m और 120 m हैं (देखिए आकृति)। इस विज्ञापन से प्रति वर्ष ₹ 5000 प्रति m2 की प्राप्ति होती है। एक कंपनी ने एक दीवार को विज्ञापन देने के लिए 3 महीने के लिए किराए पर लिया। उसने कुल कितना किराया दिया ?

हल :
यहाँ त्रिभुजाकार दीवार के लिए
a = 122 m
b = 22 m
c = 120 m

1 वर्ग मी० क्षेत्रफल का 1 वर्ष का विज्ञापन किराया = ₹ 5000
1320 वर्ग मी० क्षेत्रफल का \(\frac{1}{4}\) वर्ष का किराया = 5000 × 1320 × \(\frac{1}{4}\)
= ₹ 16,50,000 उत्तर

प्रश्न 3.
किसी पार्क में एक फिसल पट्टी (slide) बनी हुई है। इसकी पार्वीय दीवारों (side walls) में से एक दीवार पर किसी रंग से पेंट किया गया है और उस पर “पार्क को हरा-भरा और साफ रखिए” लिखा हुआ है (देखिए आकृति)। यदि इस दीवार की विमाएँ 15 m, 11 m और 6 m हैं, तो रंग से पेंट हुए भाग का क्षेत्रफल ज्ञात कीजिए।

हल :

आकृति यहाँ पेंट की गई दीवार की विमाएँ
a = 15 m
b = 11 m
c = 6 m

प्रश्न 4.
उस त्रिभुज का क्षेत्रफल ज्ञात कीजिए जिसकी दो भुजाएँ 18 cm और 10 cm हैं तथा उसका परिमाप 42 cm है।
हल :
यहाँ पर
a = 18 cm
b = 10 cm
a+ b + c = 42 cm
या 18 + 10 + c = 42
या c = 42 – 28 = 14 cm

प्रश्न 5.
एक त्रिभुज की भुजाओं का अनुपात 12 : 17 : 25 है और उसका परिमाप 540 cm है। इस त्रिभुज का क्षेत्रफल ज्ञात कीजिए। [B.S.E.H. March, 2019]
हल :
यहाँ पर त्रिभुज की भुजाओं का अनुपात = a : b : c = 12 : 17 : 25
अनुपाती योग = 12 + 17 + 25 = 54
∴ a = \(\frac{12}{54}\) × 540 = 120cm
b = \(\frac{17}{54}\) × 540 = 170cm
c = \(\frac{25}{54}\) × 540 = 250cm

प्रश्न 6.
एक समद्विबाहु त्रिभुज का परिमाप 30 cm है और उसकी बराबर भुजाएँ 12 cm लंबाई की हैं। इस त्रिभुज का क्षेत्रफल ज्ञात कीजिए।
हल :
यहाँ पर
a = 12 cm
b = 12 cm
c = 30 – (a + b)
= 30 – (12 + 12) = 30 – 24 = 6 cm

= \(\sqrt{15 \times 3 \times 3 \times 9}\) cm²
= \(\sqrt{15 \times 9 \times 9}\) cm²
= 9\(\sqrt{15}\) cm²

Haryana State Board HBSE 9th Class Maths Solutions Chapter 7 त्रिभुज Ex 7.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 7 त्रिभुज Exercise 7.1

प्रश्न 1.
चतुर्भुज ACBD में, AC = AD है और AB कोण A को समद्विभाजित करता है (देखिए आकृति)। दर्शाइए कि ΔABC ≅ ΔABD है।
BC और BD के बारे में आप क्या कह सकते हैं?
हल :
यहाँ पर ΔABC और ΔABD में,

AC = AD [दिया है]
∠CAB = ∠BAD [∵ AB कोण Aको समद्विभाजित करता है।]
AB = AB [उभवनिष्ठ]
अतः ΔABC ≅ ΔABD [भुजा-कोण-भुजा सर्वांगसमता]
⇒ BC = BD [∵ सर्वांगसम त्रिभुजों के संगत भाग] [इति सिद्धम]

प्रश्न 2.
ABCD एक चतुर्भुज है, जिसमें AD = BC और ∠DAB = ∠CBA है (देखिए आकृति)। सिद्ध कीजिए कि
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC

हल :
(i) यहाँ पर ΔABD और ΔBAC में,
AD = BC [दिया है]
∠DAB = ∠CBA [दिया है]
AB = AB [उभयनिष्ठ]
अतः ΔABD ≅ ΔBAC [भुजा-कोण-भुजा सर्वांगसमता]

(ii) क्योंकि ΔABD ≅ ΔBAC ∴ BD = AC [∵ सर्वांगसम त्रिभुजों के संगत भाग]

(iii) क्योंकि ΔABD ≅ ΔBAC ∴ ∠ABD = ∠BAC [∵ सर्वांगसम त्रिभुजों के संगत भाग]
[इति सिद्धम]

प्रश्न 3.
एक रेखाखंड AB पर AD और RC दो बराबर लंब रेखाखंड हैं (देखिए आकृति)। दर्शाइए कि CD, रेखाखंड AB को समद्विभाजित करता है।

हल :
यहाँ पर AB और CD, O पर प्रतिच्छेद करते हैं।
∴ ∠AOD = ∠BOC [शीर्षाभिमुख कोण]
ΔAOD और ΔBOC में,
∠AOD = ∠BOC [प्रमाणित]
∠DAO = ∠OBC
AD = BC [दिया है]
∴ ΔAOD ≅ ΔBOC [कोण-कोण-भुजा सर्वांगसमता]
⇒ OA = OB
[∵ सर्वागसम त्रिभुजों के संगत भाग]
अतः CD, रेखाखंड AB को समद्विभाजित करता है। [इति सिद्धम]

प्रश्न 4.
l और m दो समांतर रेखाएं हैं जिन्हें समांतर रेखाओं p और q का एक अन्य युग्म प्रतिच्छेदित करता है (देखिए आकृति)। दर्शाइए कि ΔABC ≅ ΔCDA है।

हल :
क्योंकि l तथा m समांतर रेखाएँ दूसरी समांतर रेखाओं p और q व द्वारा प्रतिच्छेदित की जाती हैं। अतः AD || BC तथा AB || CD
⇒ ABCD समांतर चुतर्भुज है।
⇒ AB = CD और BC = AD [∵ समांतर चतुर्भुज की सम्मुख भुजाएँ बराबर होती हैं।]
अब ΔARC और ΔCDA में,
AB = CD [प्रमाणित]
BC = AD [प्रमाणित]
AC = AC [उभयनिष्ठ]
ΔABC ≅ ΔCDA [भुजा-भुजा-भुजा सर्वांगसमता] [इति सिद्धम]

प्रश्न 5.
रेखा l कोण A को समद्विभाजित करती है और B रेखा l पर स्थित कोई बिंदु है। BP और BQ कोण A की भुजाओं पर B से डाले गए लंब हैं (देखिए आकृति)। दर्शाइए कि –
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ है अर्थात बिंदु B कोण की भुजाओं से समदूरस्थ है।

हल :
(i) यहाँ पर ΔAPB तथा ΔAQB में,
∠APB = ∠AQB [प्रत्येक = 90°]
∠PAB = ∠QAB [क्योंकि AB, ∠PAQ का समद्विभाजक है।]
AB = AB [उभयनिष्ठ]
⇒ ΔAPB ≅ ΔAQB
[कोण-कोण भुजा सर्वांगसमता]

(ii) ∵ ΔAPB ≅ ΔAQB
∴ BP = BQ [सर्वांगसम त्रिभुजों के संगत भाग]
अर्थात बिंदु B, कोण A की भुजाओं से समदूरस्थ है। [इति सिद्धम]

प्रश्न 6.
आकृति में, AC = AE, AB = AD और ∠BAD = ∠EAC है। दर्शाइए कि BC = DE है।

हल :
ΔBAD और ΔEAC में,
∠BAD = ∠EAC [दिया है।]
दोनों ओर ∠DAC जोड़ने पर,
∠BAD + ∠DAC = ∠EAC + ∠DAC
या ∠BAC = ∠EAD
अब ΔBAC और ΔEAD में,
AB = AD [दिया है]
∠BAC = ∠EAD [प्रमाणित]
AC = AE [दिया है]
अतः
ΔBAC ≅ ΔEAD [भुजा-कोण-भुजा सर्वांगसमता]
⇒ BC = DE [सर्वांगसम त्रिभुजों के संगत भाग] [इति सिद्धम]

प्रश्न 7.
AB एक रेखाखंड है और Pइसका मध्य-बिंदु है। D और E रेखाखंड AB के एक ही ओर स्थित दो बिंदु इस प्रकार हैं कि ∠BAD = ∠ABE और ∠EPA = ∠DPB है। (देखिए आकृति)। दर्शाइए कि
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE

हल :
(i) ∠EPA = ∠DPB [दिया है|
दोनों ओर ∠DPE जोड़ने पर,
∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒ ∠DPA = ∠EPB
अब, ΔEBP तथा ΔDAP में,
∠EPB = ∠DPA [प्रमाणित]
BP = AP [दिया है।]
और ∠EBP = ∠DAP [दिया है]
अतः ΔEBP ≅ ΔDAP [कोण-भुजा-कोण सर्वांगसमता]

(ii) क्योंकि ΔEBP ≅ ΔDAP
AD = BE
[∵ सर्वांगसम त्रिभुजों के संगत भाग]

प्रश्न 8.
एक समकोण त्रिभुज ABC में, जिसमें कोण समकोण है, M कर्ण AB का मध्य-बिंद्ध है। C को M से मिलाकर बिंदु D तक इस प्रकार बढ़ाया गया है कि DM = CM है। बिंदु D को बिंदु B से मिला दिया जाता है (देखिए आकृति)। दर्शाइए कि –

(i) ΔAMC ≅ ΔBMD
(ii) ∠DRC एक समकोण है।
(iii) ΔDBC ≅ ΔACB
(iv) CM = \(\frac {1}{2}\)AB
हल :
(i) ΔAMC तथा ΔBMD में,
AM = BM [∵ M, AB का मध्य-बिंदु है]
∠AMC = ∠BMD [शीर्षाभिमुख कोण]
और CM = MD [दिया गया है]
अतः ΔAMC ≅ ΔBMD [भुजा-कोण-भुजा सर्वांगसमता]
[इति सिद्धम]

(ii) क्योंकि
ΔAMC ≅ ΔBMD
∠BDM = ∠ACM [∵ सर्वांगसम त्रिभुजों के संगत भाग]
परंतु ये एकांतर कोण है, अतः BD || CA
या ∠CBD + ∠BCA = 180° [∵ तिर्यक रेखा के एक ही ओर के आंतरिक कोण]
या ∠CBD + 90° = 180° [∵ ∠BCA = 90°]
या ∠DBC = 90° [इति सिद्धम]

(iii) अब ΔDBC तथा ΔACB में,
BD = CA [सर्वांगसम ΔBMD व ΔAMC के भाग]
∠DBC = ∠ACB [प्रत्येक = 90°]
BC = BC [उभयनिष्ठ]
ΔDBC ≅ ΔACB [भुजा-कोण-भुजा सर्वांगसमता] [इति सिद्धम]

(iv) क्योंकि
CD = AB [∵ सर्वांगसम त्रिभुजों के संगत भाग]
\(\frac {1}{2}\)CD = \(\frac {1}{2}\)AB
या CM = \(\frac {1}{2}\) AB
[इति सिद्धम]

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.9 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Exercise 13.9

प्रश्न 1.
एक लकड़ी के बुकशेल्फ (book-shell) की बाहरी विमाएं निम्न हैं : ऊंचाई = 110 सें०मी०, गहराई = 25 सें०मी०, चौड़ाई = 85 सें०मी० (देखिए आकृति)। प्रत्येक स्थान पर तख्तों की मोटाई 5 सें०मी० है। इसके बाहरी फलकों पर पालिश कराई जाती है और आंतरिक फलकों पर पेंट किया जाना है। यदि पालिश कराने की दर 20 पैसे प्रति सें०मी० है और पेंट कराने की दर 10 पैसे प्रति सें०मी० है, तो इस बुक-शैल्फ पर पालिश और पेंट कराने का कुल व्यय ज्ञात कीजिए।

हल :
यहां पर, पॉलिश वाले तल का क्षेत्रफल = (110 × 85 + 2 × 85 × 25 + 2 × 25 × 110 + 4 × 75 × 5 + 2 × 110 × 5) सें०मी०²
= (9350 + 4250 + 5500 + 1500 + 1100) सें०मी०²
= 21700 सें०मी०²
इस पर पॉलिश कराने का खर्च = 21700 × \(\frac{20}{100}\) = ₹ 4340
पेंट वाले तल का क्षेत्रफल = (6 × 75 × 20 + 2 × 90 × 20 + 75 × 90) सें०मी०²
= (9000 + 3600 + 6750) सें०मी०² = 19350 सें०मी०²
इस पर पेंट कराने का खर्च = 19350 × \(\frac{10}{100}\) = ₹ 1935
पॉलिश व पेंट कराने का कुल खर्च = 4340 + 1935 = ₹ 6275 उत्तर

प्रश्न 2.
किसी घर के कंपाउंड के सामने की दीवार को 21 सें०मी० व्यास वाले लकड़ी के गोलों को छोटे आधारों पर टिका कर सजाया जाता है, जैसा कि आकृति में दिखाया गया है। इस प्रकार के आठ गोलों का प्रयोग इस कार्य के लिए किया जाना है और इन गोलों को चांदी वाले रंग में पेंट करवाना है। प्रत्येक आधार 1.5 सें०मी० त्रिज्या और ऊंचाई 7 सें०मी० का एक बेलन है तथा इन्हें काले रंग से पेंट करवाना है। यदि चांदी के रंग के पेंट करवाने की दर 25 पैसे प्रति से०मी० है तथा काले रंग के पेंट करवाने की दर 5 पैसे प्रति सें०मी० हो, तो पेंट करवाने का कुल व्यय ज्ञात कीजिए।

हल :
माना
लकड़ी के गोले का व्यास (d) = 21 सें०मी०
लकड़ी के गोले की त्रिज्या = \(\frac{21}{2}\) सें०मी०
∴ गोले का पृष्ठीय क्षेत्रफल = 4πr2
= \(4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}\) = 1386 सें०मी०²
गोले के जितने तल पर चांदी वाला रंग होगा = \(\left[1386-\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2}\right]\) सें०मी०²
= \(\left[1386-\frac{99}{14}\right]\) सें०मी०²
= [1386 – 7.07] सें०मी०²
= 1378.93 सें०मी०²
8 गोलों में जितने तल पर चांदी वाला रंग होगा = 1378.93 × 8 सें०मी०² = 11031.44 सें०मी०²
चांदी का पेंट करने की दर = 25 पैसे प्रति सें०मी०²
∴ चांदी वाले रंग पर व्यय = \(\frac{11031.44 \times 25}{100}\)
= ₹ 2757.86
बेलन की त्रिज्या (r) = 1.5 = \(\frac{15}{10}=\frac{3}{2}\) सें०मी०
बेलन की ऊंचाई (h) = 7 सें०मी०
बेलन का वक्र पृष्ठीय क्षेत्रफल = 2πrh
= \(2 \times \frac{22}{7} \times \frac{3}{2} \times 7\) = 66 सें०मी०²
8 बेलनों का वक्र पृष्ठीय क्षेत्रफल = 66 × 8 = 528 सें०मी०²
काले रंग का पेंट करवाने की दर = 5 पैसे प्रति सें०मी०²
बेलनों पर काला रंग कराने पर व्यय = \(\frac{528 \times 5}{100}\) = ₹ 26.40
दोनों रंगों के पेंट पर कुल व्यय = 2757.86 + 26.40
= ₹ 2784.26 उत्तर

प्रश्न 3.
एक गोले के व्यास में 25% की कमी हो जाती है। उसका वक्र पृष्ठीय क्षेत्रफल कितने प्रतिशत कम हो गया है?
हल :
माना,
गोले का पहला व्यास = 2x मी०
गोले की पहली त्रिज्या = x मी०
गोले का पहला पृष्ठीय क्षेत्रफल = 4πr² = 4πx² मी०²
25% कमी करने के पश्चात्

Haryana State Board HBSE 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Exercise 9.2

प्रश्न 1.
आकृति में, ABCD एक समांतर चतुर्भुज है, AE ⊥ DC और CF ⊥ AD है। यदि AB = 16 सें०मी०, AE = 8 सें०मी० और CF = 10 सें०मी० है, तो AD ज्ञात कीजिए। [B.S.E.H. March, 2018]

हल :
यहाँ पर आधार (AB) = 16 सें०मी०
शीर्षलंब (AE) = 8 सें०मी०
∴ समांतर चतुर्भुज ABCD का क्षेत्रफल = आधार × शीर्षलंब
= 16 सें०मी० × 8 सें०मी०
= 128 सें०मी०²
दूसरी अवस्था में
शीर्षलंब (CF) = 10 सें०मी०

प्रश्न 2.
यदि E, F, G और H क्रमशः समांतर चतुर्भुज ABCD की भुजाओं के मध्य-बिंदु हैं, तो दर्शाइए कि ar (EFGH) = \(\frac{1}{2}\)ar (ABCD) है।
हल :

दिया है : समांतर चतुर्भुज ABCD में E, F, G व H क्रमशः भुजाओं AB, BC, CD व DA के मध्य बिंदु हैं।
इन्हें मिलाने पर चतुर्भुज EFGH प्राप्त होता है।
सिद्ध करना है : ar (EFGH) = \(\frac{1}{2}\)ar (ABCD)
रचना : H व F को मिलाओ।
प्रमाण : ΔHGF और समांतर चतुर्भुज HDCF समान आधार HF और समान समांतर रेखाओं HF और DC के मध्य स्थित हैं।
∴ ar (ΔHGF) = \(\frac{1}{2}\)ar (HDCF) …..(i)
इसी प्रकार, ΔHEF और समांतर चतुर्भुज ABFH समान आधार HF और समान समांतर रेखाओं HF और AB के मध्य स्थित है।
∴ ar (ΔHEF) = \(\frac{1}{2}\)ar (ABFH) …..(ii)
समीकरण (i) एवं (ii) को जोड़ने पर,.
ar (ΔHGF) + ar (ΔHEF) = \(\frac{1}{2}\)[ar (HDCF) + ar (ABFH)]
⇒ ar (EFGH) = \(\frac{1}{2}\)ar (ABCD) [इति सिद्धम]

प्रश्न 3.
P और Q क्रमशः समांतर चतुर्भुज ABCD की भुजाओं DC और AD पर स्थित बिंदु हैं। दर्शाइए कि ar (APB) = ar (BQC) है।
हल :

दिया है :
समांतर चतुर्भुज ABCD की भुजाओं DC और AD पर क्रमशः P व Q दो बिंदु स्थित हैं।
सिद्ध करना है : ar (ΔAPB) = ar (ΔBQC)
प्रमाण : यहाँ पर ΔAPB तथा || चतुर्भुज ABCD एक ही आधार AB तथा समांतर रेखाओं AB व CD के मध्य में हैं।
∴ ar (ΔAPB) = \(\frac{1}{2}\)ar (|| चतुर्भुज ABCD) …..(i)
इसी प्रकार ΔBQC तथा || चतुर्भुज ABCD एक ही आधार BC तथा समांतर रेखाओं AD व BC के मध्य में हैं।
∴ ar (ΔBQC) = \(\frac{1}{2}\)ar (॥ चतुर्भुज ABCD) …..(ii)
समीकरण (i) व (ii) की तुलना में,
ar (ΔAPB) = ar (ΔBQC) [इति सिद्धम]

प्रश्न 4.
आकृति में, P समांतर चतुर्भुज ABCD के अभ्यंतर में स्थित कोई बिंदु है। दर्शाइए कि
(i) ar (APB) + ar (PCD) = \(\frac{1}{2}\)ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

हल :

दिया है : समांतर चतुर्भुज ABCD के अभ्यंतर में स्थित कोई बिंदु P है।
सिद्ध करना है : (i) ar (ΔAPB) + ar (ΔPCD) = \(\frac{1}{2}\)ar (ABCD)
(ii) ar (ΔAPD) + ar (ΔPBC) = ar (ΔAPB) + ar (ΔPCD)
रचना : P से EPF समांतर AB या DC तथा GPH समांतर AD या BC खींचिए।
प्रमाण : क्योंकि AB || EF m(रचना से)
तथा AE || BF [|| चतुर्भुज की सम्मुख भुजाएं]
∴ AEFB एक || चतुर्भुज है। इसी प्रकार EDCF भी एक || चतुर्भुज है।
ΔAPB तथा || चतुर्भुज AEFB एक ही आधार AB तथा समांतर रेखाओं AB तथा EF के मध्य में स्थित हैं।
∴ ar (ΔAPB) = \(\frac{1}{2}\)ar (|| चतुर्भुज AEFB) …..(i)
इसी प्रकार ar (ΔPCD) = \(\frac{1}{2}\)ar (|| चतुर्भुज EDCF) …..(ii)
समीकरण (i) व (ii) को जोड़ने पर,
ar (ΔAPB + ar (ΔPCD) = \(\frac{1}{2}\)ar (|| चतुर्भुज AEFB) + \(\frac{1}{2}\)ar (|| चतुर्भुज EDCF)
ar (ΔAPB) + ar (ΔPCD) = \(\frac{1}{2}\){ar (|| चतुर्भुज AEFB) + ar (|| चतुर्भुज EDCF)}
∴ ar (ΔAPB) + ar (ΔPCD) = \(\frac{1}{2}\)ar (|| चतुर्भुज ABCD) …..(iii)
इसी प्रकार हम सिद्ध कर सकते हैं कि,
ar (ΔAPD) + ar (ΔPBC) = \(\frac{1}{2}\)ar (|| चतुर्भुज ABCD) …..(iv)
समीकरण (iii) व (iv) की तुलना से,
ar (ΔAPB) + ar (ΔPCD) = ar (ΔAPD) + ar (ΔPBC) [इति सिद्धम]

प्रश्न 5.
आकृति में, PQRS और ABRS समांतर चतुर्भुज हैं तथा X भुजा BR पर स्थित कोई बिंदु है। दर्शाइए कि
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = \(\frac{1}{2}\)ar (PQRS)

हल :
दिया है : PQRS और ABRS समांतर चतुर्भुज एक ही आधार SR तथा दो समांतर रेखाओं SR व PB के बीच स्थित हैं तथा X भुजा BR पर स्थित कोई बिंदु है।
सिद्ध करना है : (i) ar (PQRS) = ar (ABRS)
(ii) ar (ΔAXS) = \(\frac{1}{2}\)ar (PQRS)
प्रमाण : (i) समांतर चतुर्भुज PQRS और समांतर चतुर्भुज ABRS समान आधार RS और एक ही समांतर रेखाओं SR तथा PB के बीच स्थित हैं।
∴ ar (PQRS) = ar (ABRS) [इति सिद्धम]
(ii) ΔAXS और समांतर चतुर्भुज ABRS समान आधार AS और एक ही समांतर रेखाओं AS बीच RB के बीच स्थित है।
∴ ar (ΔAXS) = \(\frac{1}{2}\)ar (ABRS)
या ar (ΔAXS) = \(\frac{1}{2}\)ar (PQRS) [भाग (i) से] [इति सिद्धम]

प्रश्न 6.
एक किसान के पास समांतर चतुर्भुज PQRS के रूप का एक खेत था। उसने RS पर स्थित कोई बिंदु A लिया और उसे P और Q से मिला दिया। खेत कितने भागों में विभाजित हो गया है? इन भागों के आकार क्या हैं? वह किसान खेत में गेहूँ और दालें बराबर-बराबर भागों में अलग-अलग बोना चाहता है। वह ऐसा कैसे करें?
हल :

इस प्रकार खेत तीन भागों में बँट जाता है तथा तीनों भाग त्रिभुज के आकार में हैं।
(i) ΔAPQ (ii) ΔASP (iii) ΔARQ
क्योंकि ΔAPQ तथा || चतुर्भुज PQRS एक ही आधार PQ तथा एक ही समांतर रेखाओं PQ तथा Rs के मध्य में स्थित है।
∴ ar (ΔAPQ) = \(\frac{1}{2}\)ar (|| चतुर्भुज PORS)
⇒ ar (ΔAPQ) = ar (ΔAPS) + ar (ΔAQR)
किसान को या तो गेहूँ ΔAPQ में तथा दालें अन्य दो त्रिभुजों में बोनी चाहिएं या दालें ΔAPQ में तथा गेहूँ अन्य दो त्रिभुजों में बोनी चाहिए।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Ex 6.3 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Exercise 6.3

प्रश्न 1.
आकृति में, ΔPQR की भुजाओं QP और RQ को क्रमशः बिंदुओं 5 और T तक बढ़ाया गया है। यदि ∠SPR = 135° है और ∠PQT = 110° है, तो ∠PRQ ज्ञात कीजिए।

हल :
क्योंकि QS एक रेखा है।
∴ ∠QPR + ∠SPR = 180° [रैखिक युग्म]
या ∠QPR + 135° = 180°
या ∠QPR = 180° – 135° = 45°
अब ∠TQP = ∠QPR + ∠PRQ [∵ बाह्य कोण, अंतः अभिमुख कोणों के योग के बराबर होता है]
या 110° = 45° + ∠PRQ
या ∠PRQ = 110° – 45° = 65°
अतः ∠PRQ = 65° उत्तर

प्रश्न 2.
आकृति में, ∠X = 62° और ∠XYZ = 54° है। यदि YO और ZO क्रमशः ΔXYZ के ∠XYZ और ∠XZY के समद्विभाजक हैं, तो ∠OZY और ∠YOZ ज्ञात कीजिए।

हल :
ΔXYZ में,
∠YXZ + ∠XYZ + ∠XZY = 180°
या 62° + 54° + ∠XZY = 180°
[∵ ∠YXZ = 62°, ∠XYZ = 54°]
या ∠XZY = 180° – 62° – 54° = 180° – 116° = 64°
क्योंकि YO तथा ZO, ∠XYZ तथा ∠XZY के समद्विभाजक हैं, इसलिए
∠OYZ = \(\frac {1}{2}\) × ∠XYZ = \(\frac {1}{2}\) × 54° = 27°
तथा
∠OZY = \(\frac {1}{2}\) × ∠XZY = \(\frac {1}{2}\) × 64° = 32°
अब ΔOYZ में,
∠YOZ + ∠OYZ + ∠OZY = 180° .
या ∠YOZ + 27° + 32° = 180°
या ∠YOZ + 59° = 180°
या ∠YOZ = 180° – 59° = 121°
अतः ∠OZY = 32° व ∠YOZ = 121° उत्तर

प्रश्न 3.
आकृति में, यदि AB || DE, ∠BAC = 35° और ∠CDE = 53° है, तो ∠DCE ज्ञात कीजिए।

हल :
आकृति में, AB || DE तथा तिर्यक रेखा AE इन्हें क्रमशः A तथा E पर काटती है।
∴ ∠DEA = ∠BAE [एकांतर कोण]
परंतु ∠BAE = 35°
⇒ ∠DEA = 35° या ∠DEC = 35°
ΔDEC में,
∠DCE + ∠DEC + ∠CDE = 180°
या ∠DCE + 35° + 53° = 180°
⇒ ∠DCE = 180° – 35° – 53° = 180° – 88° = 92°
अतः ∠DCE = 92° उत्तर

प्रश्न 4.
आकृति में, यदि रेखाएँ PQ और Rs बिंदु T पर इस प्रकार प्रतिच्छेद करती हैं कि ∠PRT = 40°, ∠RPT = 95° और ∠TSQ = 75° है, तो ∠SQT ज्ञात कीजिए।

हल :
ΔPRT में,
∠PRT + ∠RTP + ∠TPR = 180°
या 40° + ∠RTP + 95° = 180°
⇒ ∠RTP = 180° – 40° – 95° = 180° – 135° = 45°
परंतु ∠STQ = ∠RTP [शीर्षाभिमुख कोण]
⇒ ∠STQ = 45°

अब ΔTQS में,
∠SQT + ∠STQ + ∠TSQ = 180°
या ∠SQT + 45° + 75° = 180°
या ∠SQT = 180° – 45° – 75°
= 180° – 120° = 60°
अतः SQT = 60° उत्तर

प्रश्न 5.
आकृति में, यदि PQ ⊥ PS, PQ || SR, ∠SQR = 28° और ∠QRT = 65° है, तो x और y के मान ज्ञात कीजिए।

हल :
हम जानते हैं कि किसी त्रिभुज का बाय कोण अंतः अभिमुख कोणों के योग के बराबर होता है।
∴ ΔSRQ में,
∠QRT = ∠QSR + ∠RQS
या 65° = ∠QSR + 28°
या ∠QSR = 65° – 28° = 37°
अब क्योंकि PQ || SR तथा तिर्यक रेखा PS इन्हें क्रमशः P और S पर प्रतिच्छेदित करती है।
∴ ∠PSR + ∠SPQ = 180° [अंतः कोण युग्म]
या (∠PSQ + ∠QSR) + 90° = 180°
या y + 37° + 90° = 180°
या y + 127° = 180°
या y = 180° – 127° = 53°
समकोण ΔPQS में,
∠PQS = 180° – ∠SPQ – ∠PSQ
या x = 180° – 90° = 53°
x = 180° – 143° = 37°
अतः x = 37° व y = 53° उत्तर

प्रश्न 6.
आकृति में, ΔPQR की भुजा QR को बिंदु S तक बढ़ाया गया है। यदि ∠PQR और ∠PRS के समद्विभाजक बिंदु T पर मिलते हैं, तो सिद्ध कीजिए कि ∠QTR = \(\frac {1}{2}\)∠QPR है।

हल :
ΔPOR में, बाह्य ∠PRS = ∠P + ∠Q
दोनों ओर 2 से भाग करने पर
⇒ \(\frac {1}{2}\) बाह्य ∠PRS = \(\frac {1}{2}\)∠P + \(\frac {1}{2}\)∠Q ……(i)
क्योंकि. TR, ∠PRS को समद्विभाजित करता है।
∴ ∠PRS = 2∠TRS
इसी प्रकार TQ, ∠Q को समद्विभाजित करता है।
∴ ∠Q = 2∠TQR
∠PRS व ∠Q का मान समीकरण (i) में रखने पर
\(\frac {1}{2}\)(2∠TRS) = \(\frac {1}{2}\)∠P + \(\frac {1}{2}\)(2∠TQR)
या ∠TRS = \(\frac {1}{2}\)∠P + ∠TQR
परंतु ΔTQR में, ∠TRS = ∠TQR + ∠QTR
अतः ∠TQR + ∠QTR = \(\frac {1}{2}\)∠P + ∠TQR
या ∠QTR = \(\frac {1}{2}\)∠P
या ∠QTR = \(\frac {1}{2}\)∠QPR [इति सिद्धम]

Haryana State Board HBSE 9th Class Maths Solutions Chapter 12 हीरोन का सूत्र Ex 12.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 12 हीरोन का सूत्र Exercise 12.2

प्रश्न 1.
एक पार्क चतुर्भुज ABCD के आकार का है, जिसमें ∠C = 90°, AB = 9 m, BC = 12 m, CD =5 m और AD = 8 m है। इस पार्क का कितना क्षेत्रफल है ?
हल :
ΔBCD का क्षेत्रफल = \(\frac{1}{2}\) × BC × CD
= (\(\frac{1}{2}\) × 12 × 5) m² = 30 m²
पाइथागोरस प्रमेय से,

BD² = BC² + CD²
⇒ BD² = 12² + 5²
⇒ BD² = 144 + 25
⇒ BD² = 169
BD = \(\sqrt{169}\) = 13 m
Δ ABD के लिए
a = 13 m , b = 8 m site c = 9m
अब s = \(\frac{1}{2}\)(a + b + c)
= \(\frac{1}{2}\)(13 + 8 + 9) m
= \(\frac{1}{2}\) × 30 m = 15 m

अतः चतुर्भुजाकार पार्क Δ BCD का क्षेत्रफल = (Δ BCD + Δ ABD) का क्षेत्रफल
= (30 + 35.4) m² = 65.4 m² (लगभग) उत्तर

प्रश्न 2.
एक चतुर्भुज ABCD का क्षेत्रफल ज्ञात कीजिए, जिसमें AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5cm और AC = 5 cm है।
हल :
यहाँ पर ΔABC के लिए a = 3 cm, b = 4 cm, c = 5 cm

s = \(\frac{a+b+c}{2}=\frac{3+4+5}{2}\) cm
= \(\frac{12}{2}\) cm = 6 cm

अतः चतुर्भुज (ΔBCD) का क्षेत्रफल = (ΔABC) + (ΔADC) का क्षेत्रफल
= (6 + 9.2) cm²
= 15.2 cm² (लगभग) उत्तर

प्रश्न 3.
राधा ने एक रंगीन कागज से एक हवाई जहाज का चित्र बनाया, जैसा कि आकृति में दिखाया गया है। प्रयोग किए गए कागज का कुल क्षेत्रफल ज्ञात कीजिए।

हल :
यहाँ पर भाग I के लिए
a = 5 cm, b = 5 cm, c = 1 cm

भाग II के लिए
आयत का क्षेत्रफल = लंबाई × चौड़ाई
= 6.5 × 1 cm² = 6.5 cm²
भाग III के लिए
समलंब का क्षेत्रफल = 3 × समबाहु त्रिभुज का क्षेत्रफल
= 3 × \(\frac{\sqrt{3}}{4}\) (1)²cm²
= 3 × \(\frac{1.732}{4}\) cm²
= \(\frac{5.196}{4}\) = 1.3 cm²
भाग IV के लिए
त्रिभुज का क्षेत्रफल = \(\frac{1}{2}\) × 6 × 1.5 cm² = 4.5 cm²
भाग V के लिए
त्रिभुज का क्षेत्रफल = \(\frac{1}{2}\) × 6 x 1.5 cm² = 4.5 cm²
अतः राधा द्वारा प्रयोग किए गए कागज का कुल क्षेत्रफल
= भाग [I + II + III + IV + V] का क्षेत्रफल
= [2.5 + 6.5 + 1.3 + 4.5 + 4.5] cm²
= 19.3 cm² (लगभग) उत्तर

प्रश्न 4.
एक त्रिभुज और एक समांतर चतुर्भुज का एक ही आधार है और क्षेत्रफल भी एक ही है। यदि त्रिभुज की भुजाएँ 26 cm, 28 cm और 30 cm हैं तथा समांतर चतुर्भुज 28 cm के आधार पर स्थित है, तो उसकी संगत ऊँचाई ज्ञात कीजिए।
हल :
यहाँ त्रिभुज के लिए
a = 26 cm, b = 28 cm, c = 30 cm

समांतर चतुर्भुज का आधार = 28 cm
प्रश्नानुसार
समांतर चतुर्भुज का क्षेत्रफल = त्रिभुज का क्षेत्रफल
⇒ आधार × संगत ऊँचाई = 336
या 28 × संगत ऊँचाई = 336
या संगत ऊँचाई = \(\frac{336}{28}\) = 12 cm उत्तर

प्रश्न 5.
एक समचतुर्भुजाकार घास के खेत में 18 गायों के चरने के लिए घास है। यदि इस समचतुर्भुज की प्रत्येक भुजा 30 m है और बड़ा विकर्ण 48 m है, तो प्रत्येक गाय को चरने के लिए इस घास के खेत का कितना क्षेत्रफल प्राप्त होगा ? [B.S.E.H. March, 2019]
हल :
हम जानते हैं कि समचतुर्भुज का विकर्ण इसे दो बराबर त्रिभुजों में बाँटता है।

प्रत्येक त्रिभुज के लिए
a = 30m
b = 30m
c = 48m

अतः समचतुर्भुजाकार खेत का क्षेत्रफल = 2 × त्रिभुज का क्षेत्रफल
= 2 × 432 m² = 864 m²
18 गायों को चरने के लिए उपलब्ध क्षेत्रफल = 864 m²
1 गाय को चरने के लिए उपलब्ध क्षेत्रफल = \(\frac{864}{18}\) m²
= 48 m² उत्तर

प्रश्न 6.
दो विभिन्न रंगों के कपड़ों के 10 त्रिभुजाकार टुकड़ों को सीकर एक छाता बनाया गया है (देखिए आकृति)। प्रत्येक टुकड़े के माप 20 cm, 50 cm और 50 cm हैं। छाते में प्रत्येक रंग का कितना कपड़ा लगा है ?

हल :
यहाँ पर प्रत्येक त्रिभुजाकार टुकड़े के लिए
a = 20 cm
b = 50 cm
c = 50 cm

अतः छाते में प्रत्येक रंग के कपड़े का क्षेत्रफल = 5 × 200\(\sqrt{6}\) cm²
= 1000\(\sqrt{6}\) cm² उत्तर

प्रश्न 7.
एक पतंग तीन भिन्न-भिन्न शेडों (shades) के कागजों से बनी है। इन्हें आकृति में I, II और III से दर्शाया गया है। पतंग का ऊपरी भाग 32 cm विकर्ण का एक वर्ग है और निचला भाग 6 cm, 6 cm और 8 cm भुजाओं का एक समद्विबाहु त्रिभुज है। ज्ञात कीजिए कि प्रत्येक शेड का कितना कागज प्रयुक्त किया गया है ?

हल :
पतंग के ऊपरी भाग का विकर्ण = 32 cm
∴ ऊपरी भाग का कुल क्षेत्रफल = \(\frac{1}{2}\) × पहला विकर्ण × दूसरा विकर्ण
= \(\frac{1}{2}\) × 32 × 32 cm²
= 16 × 32 cm² = 512 cm²
512 अतः छायांकित भाग I का क्षेत्रफल = छायांकित भाग II का क्षेत्रफल = \(\frac{512}{2}\)
= 256 cm² उत्तर
अब भाग III के लिए
a = 6 cm, b = 6 cm, c = 8 cm

प्रश्न 8.
फर्श पर एक फूलों का डिज़ाइन 16 त्रिभुजाकार टाइलों से बनाया गया है, जिनमें से प्रत्येक की भुजाएँ 9 cm, 28 cm और 35 cm हैं (देखिए आकृति)। इन टाइलों को 50 पैसे प्रति cm² की दर से पॉलिश कराने का व्यय ज्ञात कीजिए।

हल :
प्रत्येक त्रिभुजाकार टाइल के लिए
a = 9 cm, b = 28 cm, c = 35 cm

= \(\sqrt{36 \times 27 \times 8 \times 1}\) cm²
= \(\sqrt{6 \times 6 \times 3 \times 3 \times 3 \times 2 \times 2 \times 2}\) cm²
= 6 × 3 × 2 × \(\sqrt{6}\) cm²
= 36\(\sqrt{6}\) cm²
अतः फर्श में लगी 16 टाइलों का कुल क्षेत्रफल = 16 × 36\(\sqrt{6}\) cm²
= 576 × 2.45 cm²
= 1411.2 cm²
1 वर्ग सें०मी० टाइल पर पॉलिश करने का व्यय = 50 पैसे = ₹ \(\frac{50}{100}\)
1411.2 वर्ग सें०मी० टाइल पर पॉलिश करने का व्यय = ₹\(\frac{50}{100}\) × 1411.2
₹ 705.60 उत्तर

प्रश्न 9.
एक खेत समलंब के आकार का है जिसकी समांतर भुजाएँ 25 m और 10 m हैं। इसकी असमांतर भुजाएँ 14 m और 13 m हैं। इस खेत का क्षेत्रफल ज्ञात कीजिए।
हल :

माना ABCD एक समलंब के आकार का खेत है जिसमें
AB = 10 m, BC = 14 m
CD = 25 m, DA = 13 m
BE||AD खींचिए जो CD को E पर काटे तथा BM⊥CD खींचे जो CD को M पर मिले। इस प्रकार ABED एक समांतर चतुर्भुज है।
AB = DE = 10 m
BE = DA = 13 m
CE = DC – DE = 25 – 10 = 15 m
अब ΔBEC में
a = 13 m
b = 15 m
c = 14 m

अतः समलंब ABCD का क्षेत्रफल = \(\frac{1}{2}\) × (AB + CD) × BM
= \(\frac{1}{2}\) × (10 + 25) × 11.2 m²
= \(\frac{1}{2}\) × 35 × 11.2 m²
= 35 × 5.6 m²
= 196 m² उत्तर

Haryana State Board HBSE 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 5 The Fundamental Unit of Life

Very Short-Answer Type Questions

Question 1.
Who discovered first cell and when ?
Answer:
A scientist named Robert Hooke discovered cell in 1665.

Question 2.
What was discovered by Leeuwenhoek?
Answer:
Leeuwenhoek discovered bacteria cell in 1674 with the help of a microscope.

Question 3.
Which scientist gave the name protoplasm to the living material inside the cell?
Answer:
J.E. Purkinje in 1939.

Question 4.
Which scientists proposed cell theory?
Answer:
M.Schleiden (1838) and T. Schwann (1839).

Question 5.
What is a cell ?
Answer:
The cell is the structural and functional unit of living organisms.

Question 6.
Which scientist expanded cell theory?
Answer:
Virchow (1855).

Question 7.
When was the electron microscope discovered?
Answer:
In 1940.

Question 8.
What is a unicellular organism?
Answer:
When a single cell of the organism performs all the life functions, it is called a unicellular organism.

Question 9.
What is a multicellular organism?
Answer:
When more than one cell group together to form an organism, it is called a multicellular organism.

Question 10.
Give some examples of unicellular organisms.
Answer:
Amoeba, Euglena, Entamoeba, Paramoecium.

Question 11.
Name one multicellular organism.
Answer:
Man.

Question 12.
Name the longest cell.
Answer:
Nerve cell.

Question 13.
Name the smallest cell.
Answer:
Mycoplasma galisepticum.

Question 14.
Which is the largest cell?
Answer:
Ostrich egg.

Question 15.
Name the smallest cell found in the human body.
Answer:
Lymphocyte.

Question 16.
Name the largest plant cell.
Answer:
Acetabularia.

Question 17.
Which cell organelle is responsible for protein synthesis?
Answer:
Ribosome.

Question 18.
In which organism is a division of labour found?
Answer:
In multicellular organisms.

Question 19.
What are cell organelles?
Answer:
In every cell, some specific components perform specific functions. They are called as cell organelles.

Question 20.
Cell organelles group together to form which unit?
Answer:
Cell.

Question 21.
Name the main functional zones oLa cell.
Answer:
The plasma membrane, nucleus cytoplasm.

Question 22.
Name one special feature of plasma membrane.
Answer:
Selective permeability.

Question 23.
What is osmosis?
Answer:
The movement of water or some other liquid substance from a region of high concentration to a region of low concentration through a semi-permeable membrane is called osmosis.

Question 24.
Give one example of osmosis.
Answer:
Absorption of water by plant roots.

Question 25.
Why does kishmish (dried raisin) get swollen in kheer?
Answer:
As the protoplasm of the kishmish is more concentrated than the kheer, internal osmosis takes place due to which it gets swollen.

Question 26.
What is diffusion?
Answer:
The movement of a substance from a region of high concentration to a medium of low concentration is called diffusion.

Question 27.
What happens when a cell is kept in a hypertonic solution?
Answer:
The cell will lose water by osmosis.

Question 28.
What is the membrane called which allows only specific substances to pass through it?
Answer:
Semi-permeable membrane.

Question 29.
What is plant cell wall made of?
Answer:
Cellulose.

Question 30.
Which cell organelle controls the activities of the cell?
Answer:
Nucleus.

Question 31.
Write the full form of RNA.
Answer:
Ribonucleic acid.

Question 32.
Write the full form of DNA.
Answer:
Deoxyribonucleic acid.

Question 33.
What is a gene?
Answer:
The functional unit of a chromosome is called a gene.

Question 34.
Which is the carrier of hereditary characters ?
Answer:
Chromosome.

Question 35.
How many chromosomes are found in man ?
Answer:
23 pairs or 46 chromosomes.

Question 36.
Name two cell organelles.
Answer:
Mitochondria, Golgi apparatus.

Question 37.
Which cell organelle is knows as suicide bag?
Answer:
Lysosome.

Question 38.
Name one secretive organelle of cell.
Answer:
Golgi apparatus.

Question 39.
Name the powerhouse of cell.
Answer:
Mitochondria.

Question 40.
What is ATP?
Answer:
ATP is the energy of the cell.

Question 41.
Name an organelle which is found only in plant tissue.
Answer:
Plastids.

Question 42.
Name the organelle which is helpful in cell division.
Answer:
Centrosome.

Question 43.
What are prokaryotic cells ?
Answer:
The cells which are small, have no cell organelle and have only one chromosome, are called prokaryotic cells.

Question 44.
What are eukaryotic cells ?
Answer:
Cells which are large in size, have cell organelle and more than one chromosome are called eukaryotic cells.

Question 45.
What is the green colour of plants due to ?
Answer:
Due to green pigment.

Question 46.
Name a prokaryotic cell.
Answer:
Bacteria.

Question 47.
Name a eukaryotic cell.
Answer:
Animal or plant cell.

Question 48.
What is cytoplasm ?
Answer:
The substance enclosed by the plasma membrane is called cytoplasm.

Question 49.
What is protoplasm ?
Answer:
Cytoplasm and nucleus together constitute protoplasm.

Question 50.
How many types of endoplasmic reticulum are there ?
Answer:
Two types:
(i) Rough Endoplasmic Reticulum (RER),
(ii) Smooth Endoplasmic Reticulum (SER).

Question 51.
What is the main function of SER ?
Answer:
To Synthesis lipid molecules.

Question 52.
What does ER denote ?
Answer:
Endoplasmic Reticulum.

Question 53.
Which organelle detoxifies toxic substances and drugs in the liver cells ?
Answer:
SER.

Question 54.
Name coloured plastids.
Answer:
Chromoplast.

Question 55.
Name a white and colourless plastid.
Answer:
Leucoplast.

Question 56.
Which plastid contains chlorophyll ?
Answer:
Chloroplast.

Question 57.
Are the vacuoles present in plants smaller or larger (in size)?
Answer:
Larger.

Short-Answer Type Questions

Question 1.
What is a cell ?
Answer:
Just as a house is made up of small bricks, living organisms are formed of small units called cells. Cell is the structural and functional unit of each living organism cell has its own organisation.

Question 2.
Write a note on the size of the cell.
Answer:
The size of the cell is so small that it is difficult to see it. It can be seen under a microscope, but to see and measure its structural organisation, a special unit called micron (p) on a micrometre is used. A micron is the l/000th part of a millimetre. Generally cells are around 30 microns in size. The smallest cell is mycoplasmas galisepticum and the largest cell is the egg of ostrich which is 170 x 135 mm in size. The size of pleuropneumonia is 0.1 pm. The size of the cell depends on the function it performs.

Question 3.
What is cell theory (principal) ?
Answer:
Cell is the basic unit of life. This is the cell theory. According to this theory, all living organisms are made up of cells. Cell is the functional and structural unit of life. Cell is a body containing protoplasm. Cell theory was proposed by M. Schleiden (1838) and T. Schwann (1839) after studying plant and animal cells. Afterwards, in 1855, Virchow expanded the cell theory by saying, ‘The new cells are formed from division of pre-existing cell.’ This is the cell theory.

Question 4.
What are unicellular and multicellular organisms ?
Answer:
1. Unicellular organisms:
The organisms which are formed of single cell are called unicellular organisms; e.g. amoeba, fungi, chlamydomonas, paramoecium, euglena etc.

2. Multicellular organisms:
The organisms which are formed of many cells grouped together are called multicellular organisms; e.g. hydra, whale, sponge, ferm, man, tagetus, fish, etc.

Question 5.
Draw a labelled diagram of the cells of onion peel as seen under a microscope.
Answer:
Cells of onion peel:

Question 6.
Draw a labelled diagram of various cells found in human body.
Answer:
Different cells of human body

Question 7.
What are the major differences between the cells of unicellular and multicellular organisms ?
Answer:
The main differences between the cells of unicellular and multicellular organisms are :
(1) The cell of a unicellular organism is generally circular or oval, while the cells of multicellular organisms are cylindrical.

(2) As a unicellular organism is made up of a single cell, all the life activities are performed by this single cell. On the other hand in multicellular organisms, all cells perform certain specific functions.

(3) As all the life activities are performed by a single cell in unicellular organisms, these activities are not performed efficiently. But division of labour is found in the cells multicellular organisms. Different cells perform specific functions, which they do efficiently.

(4) Amoeba, Paramoecium, etc. are unicellular organisms while dog, frog, man, etc. are multicellular organisms.

Question 8.
What are the advantages of multicellularity ?
Answer:
Multicellularity has a lot of advantages. Because of multicellularity, division of labour is found in higher animals. Because of difference in the types of cells, various cells perform specific functions in these organisms. All the cells co ordinate with each other. Thus, all the activities relating to growth and development depend upon the co ordination between all the cells.

Question 9.
What are the differences between plasma membrane and cell wall ? Give their functions also.
Answer:
Differentiate between Plasma Membrane and cell wall are as follows :
Plasma Membrane:
1. It is living.
2. It is formed of proteins, and lipid peptides.
3. It is smooth and flexible.
4. It is semi-permeable.

Cell Wall:
1. It is non-living.
2. It is formed of cellulose, lignin and pectin.
3. It is hard.
4. It is permeable.

Functions:
Plasma Membrane: Only molecules of definite size can pass through it. It is semi-permeable. It is a protective covering of the cell.

Cell Wall: It provides shape, size and protection to the cell.

Question 10.
What happens when we put an animal cell or a plant cell into a sugar solution ?
Answer:
The following can happen if we put a living cell in sugar solutions having different concentration:
(1) When put in a hypotonic solution, water will enter the cell by osmosis.
(2) When put in an isotonic solution,there will be no net movement of water through the cell membrane.
(3) When put in a hypertonic solution, the water will come out of the cell through osmosis.

Question 11.
How is diffusion important for a cell ?
Answer:
Diffusion plays an important role for the entry and exit of water and gases in and out of the cell. It helps the cell to get nutrition from the outside environment. The entry and exit of various molecules and in and out of the cell takes place by diffusion only.

Question 12.
What is cell wall ?
Answer:
Cell Wall: Cell wall is found only in plant cells. Some bacteria also possess cell wall. As the cell wall is made up of cellulose, it is permeable to water. Because of hardening of cellulose, it is non-living.

Question 13.
Write the advantages and disadvantages of cell wall.
Answer:

Advantages:
(1) It provides a definite shape to the cell.
(2) The cell remains structurally strong and protected.
(3) The plasma membrane remains protected.
(4) Because of its permeability, the cell wall maintains balance of different materials inside the cell.
(5) It prevents the cell from the attack of disease causing molecules.
(6) It provides a definite structure to the cell.

Disadvantages:
(1) The pace of cell division gets slowed down because of cell wall.
(2) It reduce the expanding power of cell.

Question 14.
Write a note on protoplasm shrinkage.
Answer:
If a plant cell is kept in a hypertonic solution, then exosmosis starts, i.e. water starts coming out of cytoplasm, due to which the cell wall loosens. If the hypertonic solution is highly concentrated, then the shrinkage of protoplasm continues. This process is called protoplasm shrinkage.

Question 15.
What are the main functional zones of a cell ?
Answer:
The main functional zones of a cell are plasma membrane, cysoplasm and nucleus which perform the following functions : ‘
1. Plasma membrane: It is the outer covering of every cell. It is made of lipids and proteins. It protects the cell and controls the movement of molecules inside and outside the cell.

2. Cytoplasm: Apart from the nucleus it constitutes the remaining part of the cell. It has various cell organelles like Golgibodies, mitochondria, endoplasmic, reticulum, ribosomes, lysosomes and cnloroplasts which perform various functions.

3. Nucleus: It is the main part of the cell, circular of oval-shaped. It controls all the activities of the cell. It contains nucleoplasm, nucleolus and chromosomes. Nucleus is covered by nuclear membrane.

Question 16.
Name the main cell organelles.
Answer:
The main cell organelles are:
(1) Endoplasmic Reticulum
(2) Ribosomes
(3) Lysosomes
(4) Mitochondria
(5) Nucleus
(6) Golgibodies
(7) Cilia and Plagella
(8) Vacuoles
(9) Plastids
(10) Centrosomes
(11) Perotisomes.

Question 17.
Write a note on nucleus.
Answer:
Nucleus is circular and the largest cell organelle found in the centre of cytoplasm. It controls and co ordinates all the activities of a cell. Nucleus is surrounded by nuclear membrane which is filled by nucleoplasm. Nucleolus and chromosomes are present in the nucleoplasm. There are some cells which possess many nuclei. These are called coenocytes.

Question 18.
Give four functions of nucleus.
Answer:
Functions of nucleus :
(1) It controls all the activities of the cell.
(2) It controls cell division.
(3) It takes part in reproduction.
(4) It maintains the functional ability of the cell.

Question 19.
Name both the nucleic acids present in the cell. What functions do they perform ?
Answer:
(1) DNA (Deoxyribose nucleic acid),
(2) RNA (Ribose nucleic acid).

(i) Functions of DNA: It contains genes. DNA stores the hereditary characters and passes these characters to the next generation. So, it is the hereditary material.
(ii) Functions of RNA: It synthesises proteins. In many types of viruses, RNA also acts as the hereditaiy material.

Question 20.
Write a note on Gene.
Answer:
Small granular structures present on the chromosomes are called genes. Genes are always found in pail’s on the chromosomes. One pair of genes controls one character. For example, our height is decided by a pair of two genes present in our body. Similarly, other characters/features of our body, like skin colour, colour of the eyes, etc. develop from a pair of genes each.

Question 21.
What are genes ? What is their function ?
Answer:
Gene:
Gene is a part of deoxyribonucleic acid (DNA). They are present on the chromosome, one above the other, in a specific order. Gene is the hereditary unit of life. In living organisms one character is controlled by one pair of genes. Genes hold the chloroplasts together.

Question 22.
Describe prokaryotic and eukaryotic cells.
Answer:
Prokaryotic and eukaryotic ceils are of following types:
1. Prokaryotic ceils: Pro means primitive (primary) and karyon means nucleus. These cells do not possess nuclear membrane. Only one chromosome is found in these ceils. The cell organelles are not enclosed by a membrane in these cells, like blue-green algae, bacterial cell.

2. Eukaryotic cells: Eu means continuous and karyon means nucleus. These cells have a well-defined nucleus which is enclosed by a nuclear membrane. More than one chromosomes are present in these cells, like plant and animal cells.

Question 23.
Define the following :
Protoplasm, cytoplasm, nucleoplasm.
Answer:
(i) Protoplasm : All the substances present inside the cell membrane constitute protoplasm. Protoplasm is the very basis of life. It performs all the activities of life.
It has two parts:
(a) Nucleus
(b) Cytoplasm.

(ii) Cytoplasm : Apart from the nucleus, the remaining part of the protoplasm is called cytoplasm. It contains living as well as non-living material. Many cell organelles are present in the cytoplasm i.e. Mitochondria, ribosomes, Golgibodies, lysosomes, vacuoler, endoplasmic reticulum etc.

(iii) Nucleoplasm : It is found inside the nucleus in which proteins, nucleolus, enzymes and chromosomes are present.

Question 24.
Write a note on Endoplasmic Reticulum (E.R.) Or White down the functions of Endoplasmic Reticulum (E.R.).
Answer:
Near the nucleus, a large network of membrane-bound tubes is found, which is called endoplasmic reticulum. These look like long tubules or round or oblong bags (vesicles). ER is of two types:
(i) Rough Endoplasmic Reticulum (RER),
(ii) Smooth Endoplasmic Reticulum (SER).
ER helps in intracellular transport of materials. It also acts as a cytoskeleton for some bio-chemical reactions. In the liver cells of vertebrates it plays a crucial role in detoxifying many poisons and drugs.

Question 25.
Differentiate RER and SER.
Answer:
Differences between RER and SER as below:
RER (Rough Endoplasmic Reticulum)
1. They are present on the surface of the ribosomes.
2. They, lie near the nucleus.
3. They help in photosynthesis.

SER (Smooth Endoplasmic Reticulum):
1. They don’t have any ribosomes.
2. They are present near the plasma membrane.
3. They help in lipid synthesis and in the formation of plasma membrane.

Question 26.
Write a note on Golgibodies.
Answer:
Golgibodies were first described by Camillo Golgi. The substance secreted by Golgibodies is accumulated in the vacuoles. Golgibodies form the cell wall. They are made up of vesicles, vacuoles and tubules. They are called golgi complex or golgi apparatus. It packages the materials produced by the ER and despatches it outside the cell.

Question 27.
Which cell organelle is called digestive bag and why ?
Answer:
Lysosome functions as a digestive bag in the cell.
Lysosomes occur in all the cells. These are spherical structures enclosed by a green coloured membrane. These contain many membrane bound small vesicles which are filled with catabolic substances lenzymes, also which are called as digestive bags which contain enzymes. These vesicles digest organic substances in the cell. These work as intercellular digestive system.

Lysosomes also function in digestion of food particles as well as worn-out cell organisms as a result of which old cells die. Therefore, they are also called ‘suicidal bags’ because during disturbances in cellular metabolism, when the cell gets damaged, lysosomes burst and the enzymes digest their own cell. RER make the digestive enzymes of the lysosomes.

Question 28.
Write a note on lysosomes.
Answer:
Lysosomes: Lysosomes are sac like structures which are also called as suicidal bags. They are found in all cells. In each cells, there are over one hundred lysosomes. They were discovered by D. Dueve, a scientist, in 1955. Their function is the elimination of some disturbances in the cell as well as destruction of foreign particles. Many digestive enzymes are found in them due to which they are also called digestive enzymes.

Question 29.
What is a centrosome ?
Answer:
Centrosome: Centrosome is a spherical structure found near the nucleus. It is found only in animal cells and is totally absent in plant cells. It contains one or two centrioles. In plant cells, polar caps are present instead of centrosomes. Centrosome play an important role in cell division.

Question 30.
Write a note on ribosomes.
Answer:
They are made up of RNA and proteins. These extremely small bodies are either scattered freely in the cytoplasm or are attached on the outer surface membrane of the rough endoplasmic reticulum. In the cell, ribosomes systhesise proteins with the help of amino acids.

Question 31.
Write a note on mitochondria.
Answer:
Mitochondria is a spherical or rod-shaped important organelle of the cell bounded by two membranes. The outer membrane is smooth but the inner layer contains folds. These folds are called cristae. In between the two membranes, a fluid called matrix is present. Mitochondria produces respiratory enzymes. It also oxidises food to produce energy, so it is also called as energy house or power house of the cell.

Question 32.
Differentiate the structural organisation of lysosomes and mitochondria.
Answer:
Differences between the structural organisation of lysosomes and mitochondria are as follows :
Lysosomes:
1. These are sac like structures
2. These are enclosed by a single membrane.
3. They do not bear any folds on the inner surface.
4. They contain digestive enzymes, which digest the food material.
5. They do not contain DNA.

Mitochondria:
1. These are spherical or rod-shaped organelles.
2. These are enclosed by a double membrane.
3. They bear folds on the inner surface, which are called cristae.
4. They contain some specific enzymes, which help in respiration.
5. They contain DNA.

Question 33.
Give differences between the functions of lysosomes and mitochondria.
Answer:
Differences between the functions of lysosomes and mitochondria as follows :
Lysosomes
1. They produce energy by digesting stored food in the times of emergency.
2. They do not contain ATP.
3. The destroy viruses and bacteria.
4. They function to destroy old and damaged cell organelles.
5. Through destructive process, they destroy the unwanted materials and are also called as suicidal bags.

Mitochondria
1. They produce energy through respiration.
2. They store energy in ATP and release it according to the requirement.
3. They do not do so.
4. They are not capable to perform such functions.
5. They do not perform any protective functions.

Question 34.
Write any five differences between ribosomes and centrosomes.
Answer:
Differences between ribosomes and centrosomes are as follows:
Ribosomes:
1. These are granular and spherical structure.
2. These are found both in plant and animal cells.
3. They synthesise proteins from the amino acids.
4. They do not take part in cell division.
5. They are found either scattered freely in the cytoplasm or attached to the Rough Endoplasmic Reticulum.

Centrosomes:
1. These are small star-like structure present outside the nuclear Wall.
2. These are found only in the animal cells.
3. They do not take part in protein synthesise.
4. They form Aster Rays at the time of cell division.
5. They occur in pairs near the nucleus.

Question 35.
Write a brief note on chloroplasts.
Answer:
Chloroplast is found in all green-coloured plants. It a double membrane cell organelle on the inner membrane of which are found flat structures, thylakoids, the number of which is innumerable in a chloroplast. Thylakoids are held together in a granum. Chloroplast plays an important role in photosynthesis. Green coloured pigment called chlorophyll found in chloroplast. Its ground matter is called stroma.

Question 36.
Write a note on plastids.
Answer:
Plastids are present only in plant cells. They are of two types:
(1) Chromoplast (coloured plastid).
(2) Leucoplast (white and colourless plastid).
The plastid having green pigment is called chloroplast. It is essential for photosynthesis. Yellow and purple coloured pigments are also found in chloroplasts. Plastids also contain their own DNA and ribosomes.

Question 37.
Why does papaya look yellow, spinach look green and watermelon look red from inside ?
Answer:
Different pigments present in plants impart different colours in them. The pigment named carotene gives yellow colour to papaya, green pigment chlorophyll in spinach gives it green colour; xanthomycin pigment gives red colour to the inside of the watermelon. So, different colours found in plants are due to different pigments.

Question 38.
Write a note on vacuoles.
Answer:
Vacuoles are the storage sacs for solid or liquid materials. Vacuoles are small in size in animal cells whereas plant cells have larger vacuoles. In plant cell, vacuoles provide turgidity and rigidity to the cells. They play an important role in the disposal of waste products from the body.

Essay Type Questions

Question 1.
Explain the microscopic structure of an animal cell.
Answer:
Microscopic structure of an animal cell: The basic unit of life in living organism is called a cell. It is a small sized structure filled with protoplasm. The body of every animals is formed by grouping together of many cells. The shape ofthe cell varies and their number also varies (from organism to organism). Cell has two main parts:
(a) Cell membrane,
(b) Protoplasm.

(a) Cell membrane:
A cell is enclosed by a thin membrane which is made up of plasma. It is very thin, soft and smooth. Cell membrane is semi-permeable. It gives shape to the cell and protects the protoplasm in it. It controls the movement of material inside and outside the cell.

(b) Protoplasm:
All the material present inside the cell membrane constitutes protoplasm. In fact, protoplasm is the basis of life. All life activities are completed in the protoplasm. Protoplasm has two main parts:
(1) Nucleus,
(2) Cytoplasm.

1. Nucleus:
Every cell has a spherical structure in its centre known as nucleus. Nucleus controls all the activities of the cell. Nucleus has four main parts nuclear membrane, nucleoplasm, chromatin reticulum, nucleolus. Nucleus is enclosed by a double membrane which is attached to endoplasmic reticulum. Nucleus contains DNA, RNA and substance chromatin which is made up of proteins. Chromosomes are found on the thread-like structures which constitute chromatin. Chromosomes are composed of DNA and protein. Chromosomes carry the hereditary traits from parents to offspring through genes. In the centre of the nucleus is present a dense, dark straining organelle called nucleolus. It stores RNA in it and helps in cell division.

2. Cytoplasm: Protoplasmic material of the cell other than nucleus is called cytoplasm. It contains living as well as non-living particles.

Following cell organelles are present in the cytoplasm:

(i) Mitochondria: These are double membraned spherical or rod-shaped structures. Mitochondria contains DNA, RNA and respirator)’ enzymes. These are helpful in respiration. Mitochondria are also called energy houses or powerhouses of the cell.

(ii) Ribosomes: These are spherical structures composed of RNA and protein. They are found scattered freely throughout the cell cytoplasm in prokaryots. However, in eukaryotic cells they are found either in free state in the cytoplasm or attached on the outer surface membrane of RER.

(iii) Endoplasmic Reticulum:
It is a large network of membrane-bound tubules. ER communicates both with plasma-membrane as well as nuclear membrane. Ribosomes are found on its outer surface membrane (in case of RER). It is a network formed of proteins and lipids. It helps in the transport of materials within the cell. It establishes contact between cytoplasm and nucleus.

(iv) Golgibodies:
These are membrane-bound structures found near the nucleus. Its function is not so clear or distint, but it helps in the formation of cell wall and despatches the material synthesised in ER outside the cell.

(v) Centrosome:
It is a small spherical structure found near the nucleus. It contains one or two granules, the centrioles. It helps in cell division.

(vi) Lysosomes:
These are spherical or oval structures present inside the cell. They contain digestive enzymes. Lysosomes help in digestive activity and destruction of damaged cell organelles.

(vii) Vacuoles:
These are a few spherical or egg shaped regions in the cell where there is very little cytoplasm. These are called vacuoles. Vacuoles help in the disposal of poisonous material from the cell as well as in the storage of food particles. In some unicellular organisms, specialised vacuoles also help in expelling excess water outside the cell.

(viii) Cilia and Flagella:
These are some thin, hair-like structures which help in movement of the organism in a fluid medium. Cilia are more in number but smaller in size whereas flagella are one or two in number. Their length is comparatively more.

Question 2.
What is a nucleus ? Write its functions.
Answer:
Every cell has a nucleus present in the centre. It is spherical or oval-shaped. The cell which have more than one nucleus is called coenocyte.

Following are the main parts of a nucleus :
1. Nuclear membrane: It is a thin and elastic membrane which keeps the nucleus separate from the cytoplasm.
2. Nucleoplasm: The fluid present inside the nucleus is called nucleoplasm.
3. Nucleolus: It is present inside the nucleus and controls the synthesis of protein.
4. Chromosomes: Every nucleus contains thin and long structures called chromosomes. Chromosomes are composed of a substance called nucleoprotein. This substance is formed by the joining of nucleic acids and proteins. The number of chromosomes for one type‘of organisms is fixed.

Functions of Nucleus: Following are the functions of a nucleus:
(i) The main function of the nucleus is to control and coordinate the activities of the cell.
(ii) A cell cannot remain alive without a nucleus.
(iii) It controls the hereditary characters with the help of genes.

Question 3.
What are the main functions of:
(a) Plasma membrane
(b) Mitochondria
(c) Chromosomes
(d) Nucleolus
(e) Lysosomes
(f) Cell wall
(g) Ribosome
(h) Golgi apparatus
(j) Peroxisomes ?
Answer:
The main functions of each cell constituent are as follows :
(a) Plasma membrane: Being semi-permeable, it controls the movement of materials inside and outside the cell.
(b) Mitochondria: It is the powerhouse of the cell. It oxidises carbohydrates to produce energy.
(c) Chromosomes: They are the carriers of hereditary traits and DNA is present in them.
(d) Nucleolus: It synthesises ribosomes.
(e) Lysosomes: It functions as a digestive sac,/bag because it plays the role of intercellular digestion.
(f) Cell wall: It protects the internal contents of the cell.
(g) Ribosomes: These help in protein synthesis.
(h) Chloroplast: Its green pigment produces food during photosynthesis.
(i) Golgi apparatus: It forms lysosomes and peroxisomes.
(j) Peroxisomes: They are helpful in the oxidation reactions.

Practical Work

Experiment 1:
To observe osmosis with dried raisin.

Procedure:
(1) Take 3-4 dried raisins and put them in water. After 30 minutes observe them. Raisins swell because water enters them by osmosis.
(2) Place the swollen raisins in concentrated sugar solution and observe after half an hour. Raisins shrink because water in them moves to the solution because of osmosis.

Experiment 2:
To prepare a temporary slide of onion peel.

Procedure:
(1) Take onion peel from its concave side and place it in the watch glass containing water.
(2) Put a drop of water on a glass slide and with the help of a brush transfer the onion peel from the watch glass to the slide (on the drop of water) and put a drop of iodine and place a cover slip on it. Take care that there is no air bubble beneath the coverslip.
(3) Observe this under a compound microscope and make a diagram of the cells as per your observation.

Quick Review of the Chapter

1. Who coined the term ‘cell’?
(A) Robert Brown
(B) Robert Hooke
(C) M. Schielden
(D) Virchow
Answer:
(B) Robert Hooke

2. Which substance ¡s used by Robert Hooke in discovery of cell?
(A) membrane of an onion
(B) cork
(C) a leaf
(D) Toot of a plant
Answer:
(B) cork

3. Which scientist discovered living cells first of all?
(A) Robert Hooke
(B) Leeuwenhoek
(C) Virchow
(D) Robert Brown
Answer:
(B) Leeuwenhoek

4. Who discovered the nucleus of a cell?
(A) Purkinje
(B) Virchow
(C) Robert Brown
(D) Robert Hooke
Answer:
(C) Robert Brown

5. Who gave the name protoplasm to the living material inside the cell?
(A) Purkinje
(B) Robert Brown
(C) M. Schleiden
(D) Virchow
Answer:
(A) Purkinje

6. Who proposed ‘Cell Theory’?
(A) M. Schleiden
(B) T. Schwann
(C) Virchow
(D) both (A) and (B)
Answer:
(D) both (A) and (B)

7. …………… is not a unicellular organism.
(A) Amoeba
(B) Euglena
(C) Ferns
(D) Paramoecium
Answer:
(C) Ferns

8. When was electron microscope discovered?
(A) in 1936
(B) in 1940
(C) in 1942
(D) in 1944
Ans.
(B) in 1940

9. Which organelle is not found in all organisms?
(A) plasma membrane
(B) cell wall
(C) nucleus
(D) cytoplasm
Answer:
(B) cell wall

10. Which organelle allows in and out of materials and stops them?
(A) plasma membrane
(B) cytoplasm
(C) nuclei’s
(D) centrosome
Answer:
(A) plasma membrane

11. The process of osmosis is related to:
(A) plasma membrane
(B) cell wall
(C) selective permeable membrane
(D) nucleus membrane
Answer:
(C) selective permeable membrane

12. The use of osmosis process ¡s carried out in the plants as:
(A) absorption of water by roots
(B) exchange of gases in leaves
(C) intake of nutrition from outside environment
(D) all of the above
Answer:
(D) all of the above

13. Which is not a feature of plasma membrane?
(A) flexible
(B) formed of lipid, protein
(C) study possible under simple microscope
(D) absence of selective permeable feature
Answer:
(D) absence of selective permeable feature

14. Plant cell wall is composed of which substance?
(A) minerals
(B) fats
(C) cellulose
(D) glucose
Answer:
(C) cellulose

15. The longest cell of human body is:
(A) RBC
(B) WBC
(C) neuron cell
(D) skin cell
Answer:
(C) neuron cell

16. Which is non-living?
(A) plasma membrane
(B) cell wall
(C) golgibodies
(D) ribosome
Answer:
(B) cell wall

17. Which waste require to be excreted out by the cell?
(A) CO₂
(B) O
(C) water
(D) all of the above
Answer:
(A) CO₂

18. controls all functions of a cell.
(A) centrosome
(B) golgi apparatus
(C) nucleus
(D) chromosome
Answer:
(C) nucleus

19. Where are chromosomes found?
(A) in nucleus
(B) in nucleolus
(C) in protoplasm
(D) in cytoplasm
Answer:
(A) in nucleus

20. Hereditary characters are found in
(A) nucleus
(B) chromosome
(C) nucleolus
(D) mitochondria
Answer:
(B) chromosome

21. Which substance forms chromosome?
(A) DNA
(B) Protein
(C) fat
(D) both (A) and (B)
Answer:
(D) both (A) and (B)

22. The functional unit of a chromosome is called
(A) chioroplast
(B) chromosome
(C) gene
(D) lamella
Answer:
(C) gene

23. The feature of prokaryotic is not:
(A) small size
(B) nucleus present
(C) only one chromosome
(D) nucleus is not surrounded by nucleus membrane
Answer:
(B) nucleus present

24. The feature of eukaryotic ¡s not:
(A) small size
(B) nucleus region fully developed
(C) nucleus is found
(D) more than one chromosome
Answer:
(A) smaLl size

25. Ribosomes synthesise ………..
(A) fats
(B) proteins
(C) carbohydrates
(D) all three
Answer:
(B) proteins

26. What is formed by endoplasmic reticulum (ER)?
(A) respiratory system
(B) food synthesis system
(C) reticulum system
(D) all three
Answer:
(C) reticulum system

27. Which organelle detoxifies toxic substances and drugs in the liver cells?
(A) ER
(B) goigi apparatus
(C) lysosome
(D) vacuoles
Answer:
(A) ER

28. Which scientist told about Golgi apparatus first of all?
(A) Virchow
(B) Purkinje
(C) Camino
(D) T. Schwann
Answer:
(C) Camillo

29. Lysosome is formed by
(A) ER
(B) golgi apparatus
(C) vacuole
(D) mitochondria
Answer:
(B) Golgi apparatus

30. Which is not an organ of an animal cell? ‘
(A) plasma membrane
(B) cytoplasm
(C) vacuole
(D) cell wall
Answer:
(D) cell wall

31. How many m are there in ijtm?
(A) 106
(B)
(C) l0
(D) 106
Answer:
(A) l06

32. What is the name of system of disposal of waste of a cell?
(A) golgi apparatus
(B) vacuoles
(C) lysosome
(D) centrioles
Answer:
(C) lysosome

33. ………. is called a suicidal bag.
(A) golgi apparatus
(B) lysosome
(C) ER
(D) plastid
Answer:
(B) lysosome

34. ………….. is called powerhouse of a cell.
(A) centriole
(B) Golgi bodies
(C) mitochondria
(D) plastid
Answer:
(C) mitochondria

35. plays an important role in photosynthesis.
(A) m mitochondria
(B) ch chloroplast
(C) vacuole
(D) Golgi bodies
Answer:
(B) chloroplast

36. pigments are found in the chloroplast.
(A) green
(B) yellow
(C) pink
(D) all above mentioned
Answer:
(A) green

37. are storage sacs of solid or liquid substances.
(A) lysosome
(B) golgi apparatus
(C) vacuoles
(D) ER
Answer:
(C) vacuoles

38. In which ‚ear cell was discovered?
(A) 1565
(8) 1605
(C) 1600
(D) 1665
Answer:
(D) 1665

39. The fundamental organizational unit of life is:
(A) Chromoplasts
(B) Cell
(C) Leucoplasts
(D) Plasma
Answer:
(B) Cell

Haryana State Board HBSE 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 7 Diversity in Living Organisms

Very Short-Answer Type Questions

Question 1.
Define Taxonomy.
Answer:
Science of classification of organism is known as Taxonomy.

Question 2.
Who classified the animals on the basis of their habitat ?
Answer:
Greek thinker Aristotle classified the animals into different groups based upon their habitat and habits.

Question 3.
Which type of cells have nucleus, cell organelles and cell wall ?
Answer:
Eukaryotic cells.

Question 4.
In which cell have the capacity to produce multicellular organisms ?
Answer:
Cell with nucleus.

Question 5.
Name the principle due to which some cell group together to form a single organism.
Answer:
Division of labour.

Question 6.
What are plants ?
Answer:
The organisms which produce their own food (autotrophs) are called plants.

Question 7.
What are animals ?
Answer:
The organisms which are heterotrophic known as animals.

Question 8.
Who proposed the concept of evolution ?
Answer:
Charles Darwin in his book “The Origin of Species” in 1859.

Question 9.
Which are primitive or lower organisms ?
Answer:
The group of organisms still possess their ancient body on earth is known as primitive or lower organisms.

Question 10.
Which are advanced or higher organisms ?
Answer:
The group of organisms who have undergone lot of changes in their body are called advanced or higher organisms.

Question 11.
Name the biologists who classified kingdom of organisms into broad categories.
Answer:
Ernst Haeckel (1894), Robert Whittaker (1959) and Carl Woese (1977).

Question 12.
What is the name of classification proposed by Whittaker ?
Answer:
Five kingdoms.

Question 13.
Which biologist classified the Monera Kingdom into Archaebacteria and Eubacteria ?
Answer:
Carl Woese.

Question 14.
Define species.
Answer:
Species is a group of organisms that are capable of breeding amongst themselves so as to produce fertile young ones. It is the smallest unit of classification.

Question 15.
What is the basic unit of classification ?
Answer:
Species.

Question 16.
What is nomenclature ?
Answer:
The system of naming animals and plants is termed as nomenclature.

Question 17.
Who is the father of Taxonomy ?
Answer:
Carolus Linnaeus.

Question 18.
Why does system of writing scientific names called bionomial system ?
Answer:
In this system every organism is given two names. One name is generic and the other is specific name.

Question 19.
What is the scientific name of human ?
Answer:
Homo sapiens.

Question 20.
What is the International System of nomenclature?
Answer:
Binomial system.

Question 21.
What are different categories of classification ?
Answer:
Species, Genus, Family, Order, Class, Phylum and Kingdom.

Question 22.
What is Genus ?
Answer:
Similar species together form genus.

Question 23.
How many kinds of kingdoms classified the living world in ancient time ?
Answer:
Two kingdoms:
(i) Plantae
(ii) Animalia.

Question 24.
What is classification ?
Answer:
Arranging of organism into groups on the basis of similarities and dissimilarities, which also show their relationship is called classification.

Question 25.
Write an importance of classification.
Answer:
It makes the study of organisms easy.

Question 26.
Which scientist classified the all organisms of the world into plantae and animalia kingdoms?
Answer:
Carolus Linnaeus in 1758.

Question 27.
Who raised the third kingdom Protista?
Answer:
E.H. Haeckel in 1866.

Question 28.
Who proposed the fourth kingdom Monera and fifty kingdom Animalia ?
Answer:
Robert H. Whittaker in 1969.

Question 29.
Which is the living organisms grow in whole life ?
Answer:
Plants.

Question 30.
How many sub-kingdoms of plant kingdom divided by Eichler ?
Answer:
Two: Cryptogamae and Phanerogamae.

Question 31.
Fungi cannot prepare their own food. Why ?
Answer:
Because chlorophyll is not present in them.

Question 32.
How many types of Angiosperms ?
Answer:
They are of two types:
(i) Dicot plants
(ii) Monocot plants.

Question 33.
Give two examples of Gymnosperms.
Answer:
(i) Cycas
(ii) Pinus.

Question 34.
Give one difference between algae and fungi.
Answer:
Chlorophyll present in algae but absent in fungi.

Question 35.
What is lichen?
Answer:
The symbiotic relationship of algae with fungi is called lichen.

Question 36.
Which sub-group of Cryptogams related to fungi ?
Answer:
Thallophyta.

Question 37.
Give one example of bryophyta.
Answer:
Mosses.

Question 38.
Give one example of pteridophyta.
Answer:
Ferns.

Question 39.
What are cryptogams ?
Answer:
The plants which do not bear seeds and spores located in their sex orgAnswer:

Question 40.
What is phanerogams ?
Answer:
Those plants in which seeds are produced after reproduction are called phanerogams.

Question 41.
What is meant by gymnosperms ?
Answer:
The plants bear naked seeds are called gymnosperms.

Question 42.
What is meant by angiosperms ?
Answer:
The plants which bear seeds into the fruit is called angiosperms. .

Question 43.
What is the main feature of plantae ?
Answer:
They are autotrophic.

Question 44.
What is the main feature of Animalia ?
Answer:
Heterotrophic.

Question 45.
What is porifera ?
Answer:
Those animals which bear pores all over their body are known as porifera.

Question 46.
Give two examples of coelenterata.
Answer:
Hydra and sea anemone.

Question 47.
What is the name of flatworm ?
Answer:
Platyhelminthes.

Question 48.
What is the scientific name of tapeworm ?
Answer:
Taenia solium.

Question 49.
Which group of animals is heterotrophic ?
Answer:
Nematoda.

Question 50.
Give one example of annelida.
Answer:
Earthworm.

Short-Answer Type Questions

Question 1.
What is the science of Taxonomy and Classification ?
Answer:
Taxonomy:
The branch of biology in which organisms are classified is called Taxonomy. First of all D. Candolli used this word Taxonomy involves identification, nomenclature and classification of organisms.

Classification:
It arranges the organisms in groups or set on the basis of their relations. Animals and plants are ranked in different categories as follow: Phylum, Class, Order, Family, Genus and Species. First of all, Carolus Linnaeus (1707-1778) gave idea about it in his book ‘Systema Naturae Therefore, Carolus Linnaeus is known as father of Taxonomy.

Question 2.
Why is it necessary to classify the living organisms ? Explain in your own words.
Answer:
The number of organisms in the world is very large. These organisms possess some similarities and some dissimilarties. The structure of some organisms is simple and that of others is complex. The organisms differ in their structure, nature, life-cycle, nutrition, respiration and reproduction, etc. Therefore, so as to make their study convenient, their classification is essential. The organisms are classified on the basis of similarities and dissimilarities. The science of classifiation provides a base for other branches of biology.

Question 3.
What is classification ? Explain its importance.
Answer:
Grouping organisms on the basis of similarities, dissimilarities and their inter-relationships is called classification.

Importance:
(1) Classification gives a base to the other branches of biology.
(2) Classification makes the study of organisms easy and convenient.
(3) Classification provides a clear picture of organisms.
(4) It provides information about inter-relationship of organisms.
(5) The study of geography is completely based on classification of plants and animals.
(6) The classification of organisms also contributes to the development of knowledge in other subjects.
(7) The branches of biology like ecology, cell biology, etc. have developed due to science of classification.

Question 4.
What is binomial nomenclature ? Explain with an example.
Answer:
Binomial Nomenclature: This system of nomenclature was given by Carolus Linnaeus. According to this system 4200 species of animals were named and were published in a book ‘SystemaNaturae’in 1758. In this system each plant or animal is given two names : The first name indicates the genera and is known as generic name whereas the second name indicates the specific name or species. Linnaeus is known as the father of Taxonomy.

For Example:
the scientific name of mango is Mangifere indica, that of mustard is Brassica compestris and that of man is Homo sapiens.
In all these the first name is generic name and the second name is that of species. All these names are universal and are used throughout the world.

Question 5.
What are the salient features of two kingdom system of classification ? Why are bacteria and fungi classified along with plants ?
Answer:
The living organisms are generally classified into two kingdoms : kingdom Animalia and kingdom Plantae. In both these kingdoms the characteristic features of organisms are same but they still differ in some characters. The organisms belonging to kingdom animalia are motile and cannot prepare their own food as they lack chlorophyll.

On the other hand the organisms belonging to kingdom plantae are (generally) autotrophs and prepare their own food. They are non-motile and remain confined to the same place. Their cell wall is made up of cellulose. They contain chlorophyll and are green. Bacteria and fungi do not contain green coloured pigment called chlorophyll, so they are not autotrophs. These obtain food/nourishment from dead or living organisms.

Still they are classified along with plants (kingdom Plantae) because:
(1) Their cell wall is made up of cellulose.
(2) They obtain their food/nourishment in soluble form.

Question 6.
Describe the different categories of classification ?
Answer:
Categories of the classification are the following:

Question 7.
What is Cryptogamae and Phanerogamae ?
Answer:
1. Cryptogamae: They are lowest group of plants. They lack flowers and seeds plants. They have bidden reproductive sex orgAnswer: Asexual reproduction takes place by the formation of spores.

2. Phanerogamae: They are highest group of plants. They bear roots, stem, leaves, flower, fruit and ’ seeds. They reproduce sexually by means of seeds. The seed contains embryo and nourishing material.

Question 8.
How is cryptogamae classified ?
Answer:
They are flowerless plants. Reproductive organs are hidden in them. There is no external flower and seed consists in them. According to classification, by Lindale and Eichler, cryptogamae is divided into three main groups:
(i) Thallophyta; e.g. Ulothrix, Penicillium, Lichens, etc.
(ii) Bryophyta; e.g. Funaria, Marchantia, etc.
(iii) Pteridophyta; e.g. Ferns, Selaginella, etc.

Question 9.
What is the difference between monocot and dicot plants ?
Answer:
These are the following differences between monocot and dicot plants :
Monocots:
1. Seeds with one cotyledon.
2. The leaves of these plants have parallel veination.
3. The root system is fibrous in them.
4. There is no secondary growth.
5. Stem divided into nodes.
Example: Wheat, maize, bajra, etc.

Dicots:
1. Seeds with two cotyledons.
2. Veination in their leaves is reticulate with a network of veins.
3. Tap root system is present in them.
4. Secondary growth is present.
5. Stem is not divided into nodes.
Example: Mustard (sarson), pea, gram, etc.

Question 10.
What is animalia ? Write names of main phylum and important features of this group. Answer: Animalia : The multicellular organisms with lack cell wall are called animalia. Their important features are following :
(i) They are heterotrophic and autotropic.
(ii) Their growth is definite.
(iii) These are free moving organisms.

They can be classified in the following :
(i) Porifera
(ii) Coelenterata
(iii) Platyhelminthes
(iv) Nematoda
(v) Annelida
(vi) Arthropoda
(vii) Mollusca
(viii) Echinodermata
(ix) Protochordata
(x) Vertebrata

Question 11.
Describe main features of porifera phylum.
Answer:
The main features of porifera are following:
(i) They mostly occur in saline water although some are found in fresh water.
(ii) They are multicellular.
(iii) Cells in them are loosely held together. Thus, they have cellular level of organisation.
(iv) Skeleton is not found in them.
(v) The shape of their body is the vase or sac like.
(vi) They bear pores all over body, with simple large opening known as osculum on the top.
(vii) They possess a unique canal system.
(viii) They reproduce by asexual and sexual methods.
Example:
Sycon, Spongia, Euplectella, Spongilla, etc.

Question 12.
Describe main features of Coelenterata.
Answer:
The main features of Coelenterata is following:
(i) They are solitary or colonial.
(ii) Anus is absent in them.
(iii) They possess radial symmetry.
(iv) They possess coelenterata.
(v) They bear tentacles.
(vi) Their body have stinging cells called cnidoblasts.
(vii) Polyp and medusa come into their life cycle one by one.
(viii) Reproduction is usually asexual in polyp form and sexual in the medusa form Examples : Hydra, Obelia, Sea anemone, Coral, etc.

Question 13.
Describe the main features of platyhelminthes.
Answer:
Platyhelminthes are characterised by the following main features:
(i) They are often dorso ventrally flattened.
(ii) They are usually endoparasites.
(iii) They are triploblastic, the body develops from three layers.
(iv) Their body having bilateral symmetry.
(v) Sexual reproduction takes place in them.
(vi) Only one hole is present in the alimentary canal.
(vii’) They have dorsoventrally flat and leaf like or ribbon like body.
Examples:
Tapeworm, planaria, liver- fluke, etc.

Question 14.
Write the main features of the nematoda phylum.
Answer:
Following are the main features of nematoda phylum:
(i) They are endoparasites.
(ii) They possess bilateral symmetry.
(iii) Alimentary canal is complete.
(iv) They are pseudocoelomates. Body cavity is not a true coelom in them.
(v) They are unisexual.
(vi) They are triploblastic.
(vii) Body size is microscopic and other are several centimetres in length.
Examples: Ascaris, Roundworm, Pinworm, etc.

Question 15.
Write main features of annelida phylum.
Answer:
Annelida phylum are characterised by the following main features:
(i) They live in moist soil, in freshwater or saline water.
(ii) They have long and palpented body.
(iii) They are first animals with true body- cavity or coelom.
(iv) Nephridia is present in them for the purpose of excretion.
(v) Animals have bilateral symmetry.
(vi) They are unisexual or bisexual.
(vii) Some forms have unjointed locomotary appendages, the parapodia. Others have setae for locomotion.
Examples: Earthworm, leech, nereis, sea mouse, etc.

Question 16.
What are the main features of arthropoda ?
Answer:
Arthropoda is characterised by the following main features :
(i) They occur everywhere, on land, in soil, in fresh and saline water.
(ii) They are heterotrophic.
(iii) They possess jointed legs.
(iv) Their body is segmented and divided into three regions: head, thorax and abdomen.
(v) Interior part of the body forms the distinct head, which bears well-developed sense orgAnswer:
(vi) Their body is covered externally by a chitinous exoskeleton.
(vii) Circulatory system is of open type.
(viii) Haemocoel present in them.
(ix) Male and female reproductive organs are separate.
Examples: Cockroach, crab, scorpion, bee, butterfly, mosquito, etc.

Question 17.
Write features of hemichordata phylum.
Answer:
Features of hemichordata are the following :
(i) They are worm-like unsegmented organisms.
(ii) They are all marine animals.
(iii) They may be solitary or colonial.
(iv) Their body is divisible into proboscis, collar, and trunk.
(v) They are bilaterally symmetrical.
(vi) Respiration takes place by gill slits.
(vii) Sexes are mostly separate.
Examples: Balanoglossus, Cephalodiscus, etc.

Question 18.
Write major differences of non-chordata and chordata.
Answer:
Non-chordata and chordata have following differences :
Non-Chordates:
1. There is no tail behind the anus.
2. Haemogloblin plasma dissolved in their blood.
3. Heart, if present, is dorsal.
4. Nervous system is ventral and solid.
5. Pharynx gill slits are absent

Chordates:
1. Tail is present behind their anus.
2. Haemoglobin is present in their red blood cells.
3. Heart is ventral and in a pericardial cavity.
4. Nervous system is dorsal and hollow.
5. Pharynx gill slits are present.

Question 19.
Write important features of chordata phylum and classified this phylum. Answer: Important features of chordata phylum are the following :
(i) Presence of nerve chord and hollow nervous system.
(ii) They all have paired gill slits.
(iii) Solid, elastic notochord occurs in all chordata and in any stage of life.
(iv) Presence of tail behind the anus stage.

Classification:
Chordata is divided into three subphyla:
(i) Urochordata
(ii) Cephalochordata
(iii) Vertebrata.

Essay Type Questions

Question 1.
Explain the classification of cryptogamae.
Answer:
Cryptogamae:
These are flowerless and seedless plants. They have hidden reproductive orgAnswer: They do not bear flowers and seeds externally.

They are divided into three divisions:

1. Thailophyta: They do not possess a stem, root or leaf and so called as thallus. Embryo formation after fertilization is absent. Their sex organs are single-celled. There is three clear group of plants in this division :
(A) Algae: Ulothrix, Cladophora, Ulva.
(B) Fungi: Aspergillus, Penicillium, Agaricus.
(C) Lichens: like Lichen perennail, Fruticose lichen.

2. Bryophyta:
The body of these plants are divided into structures like roots, stem or leaf. They grow on moist shady placed. Vascular tissue are absent in them. Their reproductive organs are multicellular. Embryo formation takes place after fertilisation.
Examples: Riccia, Funaria, Marchantia etc.

3. Pteridophyta:
The plant body is differentiated into root, stem and leaves. Vascular tissues are present in them. Their sex organs are multicellular. Embryo formation takes place after fertilization in them.
Examples: Fern (Pteris, selaginella).

Question 2.
Give the names of major phylurii with one or two features with examples.
Answer:
Major features with examples of the animal’s phylum are following:

Practical Work

Experiment 1.
To differentiate between monocots and dicots.

Procedure:
Take some seeds of wheat, gram, pea, maize, rice, tamarind and put into water. When they swollen completely, then classified and note in table. Those have one cotyledon (wheat, maize, rice) are monocots and those have two cotyledons (gram, pea, tamarind) are dicots. Make table to see other difference in them.

Quick Review of the Chapter

1. Father of Taxonomy is:
(A) Aristotle
(B) Robert Hook
(C) De-Candoli
(D) Carolus Linnaeus
Answer:
(D) Carolus Linnaeus

2. Which is not the base for the classification of organisms like Monera?
(A) cell structure
(B) nutrition
(C) body organization
(D) habitat
Answer:
(D) habitat

3. Which animal is not Monera?
(A) diatom
(B) bacteria
(C) microplasma
(D) blue-green algae
Answer:
(A) diatom

4. Respiration in reptiles takes place:
(A) by gills
(B) by skin
(C) by lungs
(D) by skin and slits
Answer:
(C) by lungs

5. Mammalia animal is:
(A) seahorse
(B) chameleon
(C) ostrich
(D) whale
Answer:
(D) whale

6. Skeleton is hollow in:
(A) pisces
(B) amphibians
(C) reptiles
(D) aves
Answer:
(D) aves

7. Who does not contribute in the classification of animals in a large kingdom ?
(A) Ernst Haeckel
(B) Robert Whittaker
(C) Carl Woese
(D) Darwin
Answer:
(D) Darwin

8. Which scientist divided the Monera Kingdom into Archaebacteria and Eubacteria?
(A) Eichler
(B) Carl Woese
(C) Haeckel
(D) Carolus
Answer:
(B) Car! Woese

9. Which of the following is a gymnosperm plant?
(A) mango
(B) pea
(C) cycas
(D) guava
Answer:
(C) cycas

10. Which of the follow ing has outer covering?
(A) unio
(B) pila
(C) chiton
(D) all of the above
Answer:
(D) all of the above

11. Which of the following group have mammary glands?
(A) aves
(B) mammals
(C) reptiles
(D) none of the above
Answer:
(B) mammals