Shreya

Bihar Board Class 7 Maths Solutions Chapter 6 त्रिभुज और उसके गुण Text Book Questions and Answers.

BSEB Bihar Board Class 7 Maths Solutions Chapter 6 त्रिभुज और उसके गुण

Bihar Board Class 7 Maths त्रिभुज और उसके गुण Ex 6.1

Bihar Board Class 7 Math Solution In Hindi प्रश्न 1.
x का मान ज्ञात कीजिए-

हल :
∠1 + ∠2 + ∠3 = 180°
∠80° + ∠40° + ∠x = 180°
∠x = 180° – (80° + 40°)
∠x = 180° – 120°
∠x = 60°

हल :
∠1 + ∠2 + ∠x = 180°
∠15° + ∠35° + ∠x = 180°
∠x = 180° – (15° + 35°)
∠x = 180° – 50°
∠x = 130°

हल :
∠1 + ∠2 + ∠3 = 180°
∠x + ∠50° + ∠40° = 180°
∠x = 180° – (50° + 40°)
∠x = 180° – 90°
∠x = 90°

Bihar Board Class 7 Math प्रश्न 2.
∠1 + ∠2 + ∠3 = 180° [∵ समकोण ∆ का एक कोण = 90°]
∠1 = 35°, ∠2 = 90°
∠3 = 180° – (35° + 90°)
∠3 = 180° – 125°
∠3 = 55°

Bihar Board 7th Class Math Solution प्रश्न 3.

समबाहु त्रिभुज = जिसकी तीनों भुजाएँ और कोण बराबर हों।
∠1 + ∠2 + ∠3 = 180°
∠60° + ∠60° + ∠60° = 180°
तीनों कोणों की माप ∠60° + ∠60° + ∠60° = 180° होगी।

Class 7 Bihar Board Math Solution प्रश्न 4.

(i) ∠1 + ∠2 + ∠3 = 180°
∠140° + ∠y + ∠y = 180°
∠y + ∠y = 180°
∠y + ∠y = 180° – 140°
∠y + ∠y = 40°
2∠y = 40°
∠y = 20°
y = 20°
(ii) समद्विबाहु, अधिककोण त्रिभुज

Class 7 Maths Bihar Board प्रश्न 5.
माना पहला कोण = ∠1
दूसरा कोण = ∠2
तीसरा = ∠3
प्रश्नानुसार, ∠1 = ∠2 + ∠3
तथा ∠2 = ∠3
इस त्रिभुज के दो कोण बराबर हैं।
यह एक समद्विबाहु त्रिभुज है।

Bihar Board Solution Class 7 Math प्रश्न 6.

∠A = 2y
∠B = 90° (समकोण)
∠C = y°
∠A + ∠B + ∠C = 180°
2y + 90° + y = 180
3y = 180° – 90°
y = 30°
∠A = 2y = 2 × 30 = 60°
∠C = y = 30°

त्रिभुज और उसके गुण कक्षा 7 Bihar Board प्रश्न 7.

∠A = 2x
∠B = 3x
∠C = 4x
∠A + ∠B + ∠C = 180°
2x + 3x + 4x = 180°
9x = 180°
x = 20°
∠A = 2x = 2 × 20 = 40°
∠B = 3x = 3 × 20 = 60°
∠C = 4x = 4 × 20 = 80°

Class 7 Math Bihar Board प्रश्न 8.

AB = AC
∠C = ∠B = 55°
∠A + ∠B + ∠C = 180°
∠A + 55° + 55° = 180°
∠A = (180°) – (55° + 55°)
∠A = 180° – 110°
∠A = 70°

Bihar Board Class 7 Maths Solutions प्रश्न 9.

∠A + ∠B + ∠C = 180°
40° + 100° + ∠C = 180°
∠C = 180° – (40° + 100°)
∠C = 180° – 140°
∠C = 40°
x = 180° – ∠C = 180° – 40° = 140°

Bihar Board Class 7 Math Solution प्रश्न 10.

∠C = 180° – 130° = 50°
∠A + ∠B + ∠C = 180°
∠A = 180° – (85° + 50°)
∠A = 180° – 135°
∠A = 45°

Math Class 7 Bihar Board प्रश्न 11.
रेखा का नाम : कोण समद्विभाजक
कारण : क्योंकि यह कोण को दो बराबर भागों में बाँटता है।

Class 7th Math Bihar Board प्रश्न 12.
रेखा का नाम : आधार पर लम्ब
कारण : क्योंकि यह आधार को दो भागों में बाँटता है तथा आधार पर लम्ब है।

Bihar Board Math Class 7 प्रश्न 13.
रेखा का नाम : आधार पर लम्ब । क्योंकि यह लम्ब पर 90° डिग्री का कोण बनाता है।

Class 7 Math Solution Bihar Board प्रश्न 14.
अधिककोण त्रिभुज

Bihar Board Class 7th Math Solution प्रश्न 15.


(i) ये सभी रेखा एक ही हैं।
(ii) विशेषताएँ-
(क) यह कोण A को दो भागों में बाँटती है।
(ख) यह सामने वाली भुजा को दो भागों में बाँटती है ।

Class 7 Maths Bihar Board Solution प्रश्न 16.

Bihar Board Class 7th Math प्रश्न 17.
खाली स्थान भरिए-

Bihar Board Class 7 Maths त्रिभुज और उसके गुण Ex 6.2

Bihar Board Math Solution Class 7 प्रश्न 1.
पाइथागोरस प्रमेय के अनुसार D में a² + b² = c²
(i) (3, 4, 5)
हल :
a² + b² = c²
3² + 4² = 5²
9 + 16 = 25
25 = 25
यह त्रिभुज की भुजाओं को प्रदर्शित करता है।
(ii) (2, 3, 4)
हल :
a² + b² = c²
2² + 3² = 4²
4 + 9 = 16
13 = 16
यह त्रिभुज की भुजाओं को प्रदर्शित नहीं करता।
(iii) (1, 2, 3)
हल :
a² + b² = c²
1² + 2² = 3²
1 + 4 = 9
5 = 9
यह त्रिभुज की भुजाओं को प्रदर्शित नहीं करता।
(iv) (1, 3, 5)
हल :
a² + b² = c²
1² + 3² = 5²
1 + 9 = 25
10 = 25
यह त्रिभुज की भुजाओं को प्रदर्शित नहीं करता है।

प्रश्न 2.
सत्य/असत्य बताएँ-
(i) AO + OB < AB
हल :
असत्य
(ii) AO + OC > AC
हल :
सत्य
(iii) BO + OC = BC
हल :
असत्य

प्रश्न 4.
एक त्रिभुज को दो भुजाओं की माप 10 cm. और 14 cm. है तो त्रिभुज की तीसरी भुजा की-
न्यूनतम सीमा = a – b = 14 – 10 = 4 cm
अधिकतम सीमा = a + b = 14 + 10 = 24 cm
त्रिभुज की न्यूनतम सीमा 4 cm. ज्यादा और अधिकतम सीमा 24 cm. से कम होनी चाहिए।

प्रश्न 5.

प्रश्न 6.
समकोण त्रिभुज में,
a² + b² = c²
माना = 6 cm, b = 8 cm, c = 10 cm.
a² + b² = c²
6² + 8² = 10²
36 + 64 = 100
100 = 100
यह एक समकोण त्रिभुज है।

Bihar Board Class 10 Maths Solutions Chapter 5 समांतर श्रेढ़ियाँ Ex 5.3 Text Book Questions and Answers.

BSEB Bihar Board Class 10 Maths Solutions Chapter 5 समांतर श्रेढ़ियाँ Ex 5.3

Bihar Board Class 10 Maths समांतर श्रेढ़ियाँ Ex 5.3

Bihar Board Class 10 Math Book Solution In Hindi प्रश्न 1.
निम्नलिखित समान्तर श्रेढ़ियों का योग ज्ञात कीजिए :
(i) 2, 7, 12, ……., 10 पदों तक
(ii) -37, -33, -29, ….., 12 पदों तक
(iii) 0.6, 1.7, 2.8, ……, 100 पदों तक
(iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}\)….., 11 पदों तक
हल
(i) दी गई समान्तर श्रेढ़ी : 2, 7, 12, …….., 10 पदों तक
पहला पद (a) = 2, सार्वान्तर (d) = 7 – 2 = 5, पदों की संख्या (n) = 10
n पदों का योग, S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
10 पदों तक योग, S₁₀ = \(\frac{10}{2}\) [2 × 2 + (10 – 1)5]
= 5[4 + (9 × 5)]
= 5[4 + 45]
= 5 × 49
= 245
अत: 10 पदों तक का योग = 245

(ii) दी गई समान्तर श्रेढ़ी : -37, -33, -29, ….., 12 पदों तक
पहला पद (a) = -37, सार्वान्तर (d) = (-33) – (-37) = -33 + 37 = 4,
पदों की संख्या (n) = 12
पदों का योग, S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
12 पदों का योग, S₁₂ = \(\frac{12}{2}\) [(2 × -37) + (12 – 1) × 4]
= 6[-74 + (11 × 4)]
= 6[-74 + 44]
= 6 × (-30)
= -180
अत: 12 पदों तक का योग = -180

(iii) दी गई समान्तर श्रेढ़ी : 0.6, 1.7, 2.8, …… , 100 पदों तक
पहला पद (a) = 0.6, सार्वान्तर (d) = 1.7 – 0.6 = 1.1, पदों की संख्या (n) = 100
पदों तक योग, S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
100 पदों तक योग, S₁₀₀ = \(\frac{100}{2}\) [(2 × 0.6) + (100 – 1) × 1.1]
= 50[1.2 + 99 × 1.1]
= 50[1.2 + 108.9]
= 50 × 110.1
= 5505
अत: 100 पदों तक का योग = 5505

Bihar Board Class 10 Math Book Solution In Hindi Pdf Download प्रश्न 2.
नीचे दिए हुए योगफलों को ज्ञात कीजिए :
(i) 7 + 10\(\frac{1}{2}\) + 14 +…..+ 84
(ii) 34 + 32 + 30 +………+10
(iii) -5 + (-8) + (-11) + ….. + (-230)
हल

Bihar Board Math Solution प्रश्न 3.
एक A.P. में,
(i) a = 5, d = 3 और a_n = 50 दिया है। n और S_n ज्ञात कीजिए।
(ii) a = 7 और a₁₃ = 35 दिया है। d और S₁₃ ज्ञात कीजिए।
(iii) a₁₂ = 37 और d = 3 दिया है। n और S₁₂ ज्ञात कीजिए।
(iv) a₃ = 15 और S₁₀ = 125 दिया है। d और a₁₀ ज्ञात कीजिए।
(v) d = 5 और S₉ = 75 दिया है। a और a₉ ज्ञात कीजिए।
(vi) a = 2, d = 8 और S_n = 90 दिया है। n और a_n ज्ञात कीजिए।
(vii) a = 8, a_n = 62 और S_n = 210 दिया है। n और d ज्ञात कीजिए।
(viii) a_n = 4, d = 2 और S_n = -14 दिया है। n और a ज्ञात कीजिए।
(ix) a = 3, n = 8 और S = 192 दिया है। d ज्ञात कीजिए।
(x) l = 28, S = 144 और कुल 9 पद हैं। a ज्ञात कीजिए।
हल
(i) दिया है, a = 5, d = 3 और a_n = 50
अनुक्रम A.P. में है और a_n = 50
a + (n – 1)d = 50
⇒ 5 + (n – 1) 3 = 50
⇒ 5 + 3n – 3 = 50
⇒ 3n = 50 + 3 – 5
⇒ 3n = 48
⇒ n = 16
सूत्र S_n = \(\frac{n}{2}\) [2a + (n – 1) d] से,
S₁₆ = \(\frac{16}{2}\) [(2 × 5) + (16 – 1) × 3]
= 8 [10 + (15 × 3)]
= 8 [10 + 45]
= 8 × 55
= 440
अत: n = 16 तथा S_n = 440

(ii) दिया है, a = 7 और a₁₃ = 35
यहाँ, a₁₃ = 35

= \(\frac {13}{2}\) × 42
= 13 × 21
= 273
अत: d = \(\frac{7}{3}\) तथा S₁₃ = 273

(iii) दिया है, a₁₂ = 37 और d = 3
यहाँ, a₁₂ = 37
⇒ a + (12 – 1)d = 37
⇒ a + 11d = 37
⇒ a + 11 x 3 = 37
⇒ a + 33 = 37
⇒ a = 4
तब, S₁₂ = \(\frac{12}{2}\) [2a + (12 – 1)d]
= 6 [(2 × 4) + 11 × 3]
= 6[8 + 33]
= 6 × 41
= 246
अत: a = 4 तथा S₁₂ = 246

(iv) दिया है, a₃ = 15 और S₁₀ = 125
a₃ = 15
a + (3 – 1)d = 15
a + 2d = 15 …… (1)
और S₁₀ = 125
\(\frac{10}{2}\) [2a + (10 – 1)d] = 125
2a + 9d = \(\frac{125 \times 2}{10}\) = 25
2a + 9d = 25 …….(2)
समीकरण (1) को 2 से गुणा करके समीकरण (2) में से घटाने पर,
(2a + 9d) – (2a + 4d) = 25 – 30
5d = -5
d = -1
समीकरण (1) में d का मान रखने पर,
a + 2(-1) = 15
a = 15 + 2 = 17
a10 = a + (10 – 1)d
= 17 + 9 × (-1)
= 17 – 9
= 8
a₁₀ = 8
अतः d = -1 और a₁₀ = 8

(v) दिया है, d = 5 और S₉ = 75
S₉ = \(\frac{9}{2}\) [2a + (9 – 1)d]
= \(\frac{9}{2}\) [2a + 8d]
= 9a + 36d
= 9(a + 4d)
परन्तु S₉ = 75 दिया है
9(a + 4d) = 75

(viii) दिया है, a_n = 4, d = 2 और S_n = -14
यहाँ, a_n = 4
⇒ a + (n – 1)d = 4
⇒ a + (n – 1)2 = 4
⇒ a + 2n – 2 = 4
⇒ a + 2n = 6 ……..(1)
S_n = -14
\(\frac{n}{2}\) [2a + (n – 1) 2] = -14
⇒ n[a + n – 1] = -14 ……..(2)
समीकरण (1) से, a = 6 – 2n
तब, समीकरण (2) में a का मान रखने पर,
n(6 – 2n + n – 1) = -14
⇒ n(5 – n) = -14
⇒ 5n – n² = -14
⇒ n² – 5n – 14 = 0
⇒ n² – 7n + 2n – 14 = 0
⇒ n(n – 7) + 2 (n – 7) = 0
⇒ (n – 7) (n + 2) = 0
⇒ n = 7 या n = -2
n एक धन पूर्णांक होना चाहिए।
n = 7
तब, a = 6 – 2n = 6 – (2 × 7) = 6 – 14 = -8
a = -8 तथा n = 7

(ix) दिया है, a = 3, n = 8 और S_n = 192
S_n = \(\frac{n}{2}\) [2a + (n – 1) d] से,
⇒ \(\frac{n}{2}\) [2a + (n – 1)d] = 192 [∵ S = 192, दिया है]
⇒ \(\frac{8}{2}\) [(2 × 3) + (8 – 1) d] = 192
⇒ 4[6 + 7d] = 192
⇒ 24 + 28d = 192
⇒ 28d = 192 – 24 = 168
⇒ d = 6
अत: d = 6

(x) दिया है, अन्तिम पद, l = 28, S = 144 और कुल पद = 9
सूत्र, S = \(\frac{n}{2}\) [a + l] से,
⇒ 144 = \(\frac{9}{2}\) [a + 28]
⇒ 288 = 9[a + 28]
⇒ 288 = 9a + 252
⇒ 9a = 288 – 252
⇒ 9a = 36
⇒ a = 4
अतः a = 4

Class 10 Maths Bihar Board प्रश्न 4.
636 योग प्राप्त करने के लिए A.P.: 9, 17, 25,….. के कितने पद लेने चाहिए?
हल
दी गई A.P. : 9, 17, 25, ……..
यहाँ a = 9 तथा d = 17 – 9 = 8
माना पदों की संख्या n है। .
S_n = 636 (दिया है)
⇒ \(\frac{n}{2}\) [2a + (n – 1)d] = 636
⇒ \(\frac{n}{2}\) [2 × 9 + (n – 1)8] = 636
⇒ \(\frac{n}{2}\) [18 + 8n – 8] = 636
⇒ \(\frac{n}{2}\) [8n + 10] = 636
⇒ n(4n + 5) = 636
⇒ 4n² + 5n – 636 = 0
⇒ 4n² + 53n – 48n – 636 = 0
⇒ n(4n + 53) – 12(4n + 53) = 0
⇒ (4n + 53) (n – 12) = 0
⇒ n – 12 = 0 या 4n + 53 = 0
⇒ n = 12 या \(-\frac{53}{4}\)
परन्तु n एक धन पूर्णांक होना चाहिए।
n = 12
अत: 12 पद लेने चाहिए।

Bihar Board Class 10th Math Solution प्रश्न 5.
किसी A.P. का प्रथम पद 5, अन्तिम पद 45 और योग 400 है। पदों की संख्या और सार्वान्तर ज्ञात कीजिए।
हल
दिया है, प्रथम पद (a) = 5, अन्तिम पद (l) = 45 योग (S) = 400
माना पदों की संख्या n है।
सूत्र, S = \(\frac{n}{2}\) (a + l) से,
400 = \(\frac{n}{2}\) [5 + 45]
400 = \(\frac{n}{2}\) × 50
25n = 400
n = 16
अन्तिम पद (l) = 45 परन्तु 16 वाँ पद भी अन्तिम पद है।
a₁₆ = 45
a + (16 – 1)d = 45
5 + 15d = 45
15d = 45 – 5 = 40
d = \(\frac{40}{15}=\frac{8}{3}\)
अतः पदों की संख्या n = 16 तथा सार्वान्तर = \(\frac{8}{3}\)

Bihar Board Class 10th Math Solution In Hindi प्रश्न 6.
किसी A.P. के प्रथम और अन्तिम पद क्रमशः 17 और 350 हैं। यदि सार्वान्तर 9 है तो इसमें कितने पद हैं और इनका योग क्या है?
हल
दिया है, प्रथम पद (a) = 17 अन्तिम पद (l) = 350 तथा सार्वान्तर (d) = 9
माना दी गई A.P. में पदों की संख्या n हैं।
तब, अन्तिम पद, l = n वाँ पद
l = a + (n – 1)d
350 = 17 + (n – 1)9
350 – 17 = 9n – 9
350 – 17 + 9 = 9n
9n = 342
n = 38
तब, 38 पदों का योग, S38 = \(\frac{n}{2}\) (a + l)
= \(\frac{38}{2}\) (17 + 350)
= 19 × 367
= 6973
अतः पदों की संख्या = 38 तथा पदों का योग = 6973

Bihar Board Class 10th Math Book प्रश्न 7.
उस A.P. के प्रथम 22 पदों का योग ज्ञात कीजिए, जिसमें d = 7 है और 22 वाँ पद 149 है।
हल
दिया है, d = 7 तथा n = 22
22 वाँ पद = 149
a₂₂ = a + (22 – 1)d = 149
a + 21 × 7 = 149
a + 147 = 149
a = 2
तब, प्रथम 22 पदों का योग, S₂₂ = \(\frac{n}{2}\) (a + a₂₂)
= \(\frac{22}{2}\) (2 + 149)
= 11 × 151
= 1661
अत: दी गई A.P. के प्रथम 22 पदों का योग = 1661

Bihar Board Class 10 Maths Solution प्रश्न 8.
उस A.P. के प्रथम 51 पदों का योग ज्ञात कीजिए, जिसके दूसरे और तीसरे पद क्रमश: 14 और 18 हैं।
हल
दिया है, A.P. का दूसरा पद (a₂) = 14
तीसरा पद (a₃) = 18
सार्वान्तर (d) = a₃ – a₂ = 18 – 14 = 4
अब पुनः दूसरा पद = 14
a + d = 14
a + 4 = 14 [∵ d = 4]
a = 14 – 4
a = 10
तब, सूत्र S_n = \(\frac{n}{2}\) [2a + (n – 1)d] से,
51 पदों का योग, S₅₁ = \(\frac{51}{2}\) [2 × 10 + (51 – 1) 4] [∵ n = 51]
= \(\frac{51}{2}\) [20 + (50 × 4)]
= \(\frac{51}{2}\) [20 + 200]
= \(\frac{51}{2}\) × 220
= 51 x 110
= 5610
अत: दी गई A.P. के प्रथम 51 पदों का योग 5610 है।

Class 10 Bihar Board Math Solution प्रश्न 9.
यदि किसी A.P. के प्रथम 7 पदों का योग 49 है और प्रथम 17 पदों का योग 289 है, तो इसके प्रथम n पदों का योग ज्ञात कीजिए।
हल
माना A.P. का पहला पद a तथा सार्वान्तर d है।
दिया है, प्रथम 7 पदों का योग (S₇) = 49
\(\frac{7}{2}\) [2a + (7 – 1) d] = 49
\(\frac{7}{2}\) [2a + 6d] = 49
7(a +3d) = 49
a + 3d = 7 ……..(1)
इसी प्रकार, प्रथम 17 पदों का योग = 289
\(\frac{17}{2}\) [2a + (17 – 1) d] = 289
\(\frac{17}{2}\) [2a + 16d] = 289
\(\frac{17}{2}\) × 2[a + 8d] = 289
a + 8d = 17 …….(2)
समीकरण (2) में से समीकरण (1) को घटाने पर,
a + 8d – (a + 3d) = 17 – 7
5d = 10
d = 2
समीकरण (1) में d का मान रखने पर,
a + 3 × 2 = 7
a + 6 = 7
a = 1
a = 1, तथा d = 2
तब, प्रथम n पदों का योग, S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{n}{2}\) [2 × 1 + (n – 1)2]
= \(\frac{n}{2}\) [2 + (n – 1)2]
= \(\frac{n}{2}\) [2 + 2n – 2]
= \(\frac{n}{2}\) (2n)
= n²
अत: प्रथम n पदों का योग = n²

Bihar Board Math Solution Class 10 प्रश्न 10.
दर्शाइए कि a₁, a₂,….., a_n,…..से एक A.P. बनती है, यदि a_n नीचे दिए अनुसार परिभाषित है :
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
साथ ही, प्रत्येक स्थिति में, प्रथम 15 पदों का योग ज्ञात कीजिए।
हल
(i) दिया है, किसी अनुक्रम का n वाँ पद (a_n) = 3 + 4n
n = 1 रखने पर, पहला पद (a₁) = 3 + 4(1) = 7
n = 2 रखने पर, दूसरा पद (a₂) = 3 + 4(2) = 11
n = 3 रखने पर, तीसरा पद (a₃) = 3 + 4(3) = 15
अत: अभीष्ट अनुक्रम = 7, 11, 15, ……,(3 + 4n) है।
सार्वान्तर = दूसरा पद (a₂) – पहला पद (a₁) = 11 – 7 = 4
अथवा तीसरा पद (a₃) – दूसरा पद (a₂) = 15 – 11 = 4
सार्वान्तर नियत है; अत: अनुक्रम एक A.P. है।
तब, प्रथम 15 पदों का योगफल,

अत: अनुक्रम = 7, 11, 15, …… , (3 + 4n) A.P. है तथा योगफल = 525

(ii) दिया है, अनुक्रम का n वा पद (a_n) = 9 – 5n
n = 1 रखने पर, पहला पद (a₁) = 9 – 5(1) = 4
n = 2 रखने पर, दूसरा पद (a₂) = 9 – 5(2) = -1
n = 3 रखने पर, तीसरा पद (a₃) = 9 – 5(3) = -6
अत: अनुक्रम 4, -1, -6,….., (9 – 5n) है।
पदों का सार्वान्तर (d) = दूसरा पद (a₂) – पहला पद (a₁) = -1 – (4) = -5
अथवा तीसरा पद (a₃) – दूसरा पद (a₂) = -6 – (-1) = -5
चूँकि सार्वान्तर नियत है; अत: अनुक्रम एक A.P. है।
तब, प्रथम 15 पदों का योगफल,

अत: अनुक्रम = 4, -1, -6,……,(9 – 5n) A.P. है तथा योगफल = -465

Bihar Board Class 10 Math Solution प्रश्न 11.
यदि किसी A.P. के प्रथम n पदों का योग 4n – n² है, तो इसका प्रथम पद (अर्थात S₁) क्या है? प्रथम दो पदों का योग क्या है? दूसरा पद क्या है? इसी प्रकार, तीसरे, 10 वें और nवें पद ज्ञात कीजिए।
हल
दिया है, A.P. के प्रथम n पदों का योगफल, S_n = 4n – n²
n = 1 रखने पर, S₁ = (4 × 1) – (1)² = 3
प्रथम पद (a₁) = 3
n = 2 रखने पर,
S2 = (4 × 2) – (2)² = 8 – 4 = 4
प्रथम दो पदों का योगफल, S₂ = 4
प्रथम पद (a₁) = 3
दूसरा पद (a₂) = S₂ – S₁ = 4 – 3 = 1
n = 3 रखने पर,
S₃ = 4n – n²
= (4 × 3) – (3)²
= 12 – 9
= 3
तीसरा पद (a₃) = S₃ – S₂ = 3 – 4 = -1
n = 9 रखने पर, S₉ = 4n – n² = 4 × 9 – 9² = 36 – 81 = -45
n = 10 रखने पर, S₁₀ = 4n – n² = 4 × 10 – 10² = 40 – 100 = -60
10 वाँ पद (a₁₀) = S₁₀ – S₉ = -60 – (-45) = -60 + 45 = -15
S_n = 4n – n² और S_n-1 = 4(n – 1) – (n – 1)² [n के स्थान पर (n – 1) रखने पर]
= (n – 1) [4 – (n – 1)]
= (n – 1)[4 – n + 1]
= (n – 1) (5 – n)
= 5n – n² – 5 + n
= 6n – n² – 5
n वाँ पद (a_n) = S_n – S_n-1
= (4n – n²) – (6n – n² – 5)
= 4n – n² – 6n + n² + 5
= 5 – 2n
अत: S₁ = 3, प्रथम दो पदों का योग, S₂ = 4, दूसरा पद, a₂ = 1, तीसरा पद,(a₃) = -1,
10 वाँ पद, a₁₀ = -15 तथा n वाँ पद, a_n = 5 – 2n

Bihar Board Solution Class 10 Math प्रश्न 12.
ऐसे प्रथम 40 धन पूर्णांकों का योग ज्ञात कीजिए जो 6 से विभाज्य हो।
हल
6 से विभाज्य धन पूर्णांक क्रमशः
6, 12, 18, 24, 30, …….., 40 पदों तक
पहला पद (a) = 6, सार्वान्तर (d) = 12 – 6 = 6, तथा n = 40
प्रथम n पदों का योगफल, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
प्रथम 40 पदों का योगफल, S₄₀ = \(\frac{40}{2}\) [(2 × 6) + (40 – 1) 6]
= 20 [12 + 39 × 6]
= 20 [12 + 234]
= 20 × 246
= 4920
अत: 6 से विभाज्य प्रथम 40 धन पूर्णांकों का योग = 4920

Bihar Board Class 10 Math Solution In Hindi प्रश्न 13.
8 के प्रथम 15 गुणजों का योग ज्ञात कीजिए।
हल
8 के प्रथम 15 गुणज क्रमश:
8, 16, 24, 32, ………., 15 पदों तक
S = 8 + 16 + 24 + 32 +…….+ 15 × 8
= 8[1 + 2 + 3 + 4 +……+ 15]
= 8[\(\frac{15}{2}\) (1 + 15] [∵ S_n = \(\frac{n}{2}\) [a + l]]
= 8[\(\frac{15}{2}\) × 16]
= 8 × 120
= 960
अत: 8 के प्रथम 15 गुणजों का योगफल = 960

Class 10th Math Solution In Hindi Bihar Board प्रश्न 14.
0 और 50 के बीच की विषम संख्याओं का योग ज्ञात कीजिए।
हल
0 और 50 के बीच की विषम संख्याएँ क्रमश:
1, 3, 5, 7, ……….., 49
यहाँ a = 1, d = 3 – 1 = 2, तथा a_n = 49
a_n = 49
a + (n – 1)d = 49
1 + (n – 1)2 = 49
(n – 1) 2 = 48
(n – 1) = 24
n = 25
A.P.: 1, 3, 5, 7, ………. का 25 पदों तक योगफल

अतः शून्य और 50 के बीच की विषम संख्याओं का योगफल = 625

Bihar Board Class 10 Math Book Solution प्रश्न 15.
निर्माण कार्य से सम्बन्धित किसी ठेके में, एक निश्चित तिथि के बाद कार्य को विलम्ब से पूरा करने के लिए, जुर्माना लगाने का प्रावधान इस प्रकार है : पहले दिन के लिए ₹ 200, दूसरे दिन के लिए ₹ 250, तीसरे दिन के लिए ₹ 300 इत्यादि, अर्थात् प्रत्येक उत्तरोत्तर दिन का जुर्माना अपने से ठीक पहले दिन के जुर्माने से ₹ 50 अधिक है। एक ठेकेदार को जुर्माने के रूप में कितनी राशि अदा करनी
पड़ेगी, यदि वह इस कार्य में 30 दिन का विलम्ब कर देता है?
हल
यहाँ, पहले दिन के विलम्ब के लिए अर्थदण्ड = ₹ 200
दूसरे दिन के विलम्ब के लिए अर्थदण्ड = ₹ 250
तीसरे दिन के विलम्ब के लिए अर्थदण्ड = ₹ 300
………………………..
………………………..
a = 200, d = 250 – 200 = 50, तथा n = 30 दिन
30 दिन के विलम्ब के बाद अर्थदण्ड का योगफल,
S₃₀ = \(\frac{30}{2}\) [(2 × 200) + (30 – 1) × 50]
[∵ सूत्र, S_n = \(\frac{n}{2}\) [2a + (n – 1)d] से]
= 15[400 + 29 × 50]
= 15[400 + 1450]
= 15 × 1850
= 27750
अत: ठेकेदार को जुर्माने के रूप में ₹ 27750 देने होंगे।

Class 10th Math Bihar Board प्रश्न 16.
किसी स्कूल के विद्यार्थियों को उनके समग्र शैक्षिक प्रदर्शन के लिए 7 नकद पुरस्कार देने के लिए ₹ 700 की राशि रखी गई है। यदि प्रत्येक पुरस्कार अपने से ठीक पहले पुरस्कार से ₹ 20 कम है, तो प्रत्येक पुरस्कार का मान ज्ञात कीजिए।
हल
माना पहला पुरस्कार ₹ a है।
दूसरा पुरस्कार (a₂) = (a – 20)
तीसरा पुरस्कार (a₃) = ₹ (a – 20 – 20) = ₹ (a – 40)
चौथा पुरस्कार (a₄) = ₹ (a – 40 – 20) = ₹ (a – 60)
पाँचवाँ पुरस्कार (a₅) = ₹ (a – 60 – 20) = ₹ (a – 80)
छठा पुरस्कार (a₆) = ₹ (a – 80 – 20) = ₹ (a – 100)
सातवा पुरस्कार (a₇) = ₹ (a – 100 – 20) = ₹ (a – 120)
कुल पुरस्कारों की धनराशि = a + a₂ + a₃ + a₄ + a₅ + a₆ + a₇
= a + (a – 20) + (a – 40) + (a – 60) + (a – 80) + (a – 100) + (a – 120)
= 7a – 420
प्रश्नानुसार, यह धनराशि ₹ 700 है।
7a – 420 = 700
7a = 700 + 420
7a = 1120
a = 160
पहला पुरस्कार = ₹ 160, शेष पुरस्कार क्रम से ₹ 20 – 20 कम है।
अतः पुरस्कार ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40 हैं।

Math Class 10 Bihar Board प्रश्न 17.
एक स्कूल के विद्यार्थियों ने वायु प्रदूषण कम करने के लिए स्कूल के अन्दर और बाहर पेड़ लगाने के बारे में सोचा। यह निर्णय लिया गया कि प्रत्येक कक्षा का प्रत्येक अनुभाग अपनी कक्षा की संख्या के बराबर पेड़ लगाएगा। उदाहरणार्थ, कक्षा I का एक अनुभाग 1पेड़ लगाएगा, कक्षा II का एक अनुभाग 2 पेड़ लगाएगा, कक्षा III का एक अनुभाग 3 पेड़ लगाएगा, इत्यादि और ऐसा कक्षा XII तक के लिए चलता रहेगा। प्रत्येक कक्षा के तीन अनुभाग हैं। इस स्कूल के विद्यार्थियों द्वारा लगाए गए कुल पेड़ों की संख्या कितनी होगी?
हल
प्रत्येक कक्षा में तीन अनुभाग हैं।
कक्षा I द्वारा लगाए गए कुल पेड़ = 3 × 1 = 3
कक्षा II द्वारा लगाए गए कुल पेड़ = 3 × 2 = 6
कक्षा III द्वारा लगाए गए कुल पेड़ = 3 × 3 = 9
कक्षा IV द्वारा लगाए गए कुल पेड़ = 3 × 4 = 12
………………………..
………………………..
तब, अनुक्रम A.P. : 3, 6, 9, 12, ………. बनता है।
a = 3, तथा d = 6 – 3 = 3
तब, कक्षा XII तक के कुल विद्यार्थियों द्वारा लगाए गए पेड़ों का योगफल
सूत्र, S_n = \(\frac{n}{2}\) [2a + (n – 1)d] से,
S₁₂ = \(\frac{12}{2}\) [(2 × 3) + (12 – 1) × 3]
= 6[6 + 33]
= 6 × 39
= 234
अत: स्कूल के विद्यार्थियों द्वारा लगाए कुल पेड़ = 234

Bihar Board 10th Class Maths Book Solution In Hindi प्रश्न 18.
केन्द्र A से प्रारम्भ करते हुए, बारी-बारी से केन्द्रों A और B को लेते हुए, त्रिज्याओं 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ….. वाले उत्तरोत्तर अर्द्धवृत्तों को खींचकर एक सर्पिल (spiral) बनाया गया है, जैसा कि आकृति में दर्शाया गया है। तेरह क्रमागत अर्द्धवृत्तों से बने इस सर्पिल की कुल लम्बाई क्या है?(π = \(\frac{22}{7}\)) लीजिए। [संकेत : क्रमशः केन्द्रों A, B, A, B… वाले अर्धवृत्तों की लम्बाइयाँ l1, l2, l3, l4 हैं।

हल
पहले अर्द्धवृत्त की त्रिज्या, r₁ = 0.5 cm
दूसरे अर्द्धवृत्त की त्रिज्या, r₂ = 1.0 cm
तीसरे अर्द्धवृत्त की त्रिज्या, r₃ = 1.5 cm
चौथे अर्द्धवृत्त की त्रिज्या, r₄ = 2.0 cm
……………………………….
……………………………….
13 वें अर्द्धवृत्त की त्रिज्या, r₁₃ = ?
r₁ = a = 0.5 cm, d = 1.0 – 0.5 = 0.5 cm तथा n = 13
r₁₃ = a + (n – 1) d = 0.5 + (13 – 1) × 0.5
= 0.5 + 12 × 0.5
= 0.5 + 6.0
= 6.5
अर्द्धवृत्तों की वृत्तीय परिधियाँ :
πr₁, πr₂, πr₃, ………., πr₁₃
13 क्रमागत अर्द्धवृत्तों से बने सर्पिल की लम्बाई

अत: सर्पिल की लम्बाई = 143 cm

Bihar Board 10th Math Book Solution प्रश्न 19.
200 लट्ठों (logs) को ढेरी के रूप में इस प्रकार रखा जाता है : सबसे नीचे वाली पंक्ति में 20 लढे, उससे अगली पंक्ति में 19 लटे, उससे अगली पंक्ति में 18 लट्टे, इत्यादि जैसा कि चित्र में प्रदर्शित है। ये 200 लटे कितनी पंक्तियों में रखे हुए हैं तथा सबसे ऊपरी पंक्ति में कितने लढे हैं?

हल
दिया है, सबसे निचली पंक्ति में 20 लटे हैं।
अर्थात् नीचे से प्रारम्भ कर प्रथम पंक्ति में = 20 लढे
दूसरी पंक्ति में = 19 लढे
तीसरी पंक्ति में = 18 लढे
चौथी पंक्ति में = 17 लढे ……… इत्यादि
तब, एक A.P. बनती है : 20, 19, 18, 17, …..
a = 20, तथा d = 19 – 20 = -1
माना पंक्तियों की संख्या n हैं।

यदि n = 25, तो a_n = a + (n – 1)d
= 20 + (25 – 1) × (-1)
= 20 – 24
= -4
अत: n = 25 स्वीकार्य नहीं है।
तब, n = 16 से,
a_n = a + (n – 1) d
= 20 + (16 – 1) × -1
= 20 + (15 × (-1))
= 20 – 15
= 5
अत: कुल पंक्तियाँ = 16 और सबसे ऊपर की पंक्ति में लट्ठों की संख्या = 5

प्रश्न 20.
एक आलू दौड़ (potato race) में, प्रारम्भिक स्थान पर एक बाल्टी रखी हुई है, जो पहले आलू से 5 मीटर की दूरी पर है तथा अन्य आलुओं को एक सीधी रेखा में परस्पर 3 m की दूरियों पर रखा गया है। इस रेखा पर 10 आलू रखे गए हैं। जैसा कि चित्र में दिखाया गया है।

प्रत्येक प्रतियोगी बाल्टी से चलना प्रारम्भ करती है, निकटतम आलू को उठाती है, उसे लेकर वापस आकर दौड़कर बाल्टी में डालती है, दूसरा आलू उठाने के लिए वापस दौड़ती है, उसे उठाकर वापस बाल्टी में डालती है और वह ऐसा तब तक करती रहती है, जब तक सभी आलू बाल्टी में न आ जाएँ। इसमें प्रतियोगी को कुल कितनी दूरी दौड़नी पड़ेगी?
[संकेत : पहले और दूसरे आलुओं को उठाकर बाल्टी में डालने तक दौड़ी गई दूरी = 2 × 5 + 2 × (5 + 3) है।]
हल
पहले आलू की बाल्टी से दूरी = 5 m
दूसरे आलू की बाल्टी से दूरी = (5 + 3) = 8 m
तीसरे आलू की बाल्टी से दूरी = (8 + 3) = 11 m
चौथे आलू की बाल्टी से दूरी = (11 + 3) = 14 m
इस प्रकार बाल्टी से आलुओं की दूरी A.P. में है जिसका
पहला पद (a) = 5 m तथा सार्वान्तर (d) = 3 m
एक बार बाल्टी से चलकर आलू को उठाना होता है और उसे फिर वापस बाल्टी में डालना पड़ता है।
आलू बाल्टी में डालने के लिए चली दूरियाँ :
= 2 × 5 m, 2 × 8 m, 2 × 11 m, 2 × 14 m, …….
= 10 m, 16 m, 22 m, 28 m, …………
यहाँ a = 10, d = 16 – 10 = 6, तथा n = 10
n आलुओं को उठाकर बाल्टी में डालने के लिए चली दूरी = \(\frac{n}{2}\) [2a + (n – 1)d]
10 आलुओं की रेस में चली दूरी = \(\frac{10}{2}\) [2 × 10 + (10 – 1) × 6]
= 5[20 + (9 × 6)]
= 5[20 + 54]
= 5[74]
= 370 m
अतः प्रतियोगी द्वारा चली दूरी = 370 m

Get Updated Bihar Board Class 9th English Book Solutions in PDF Format and download them free of cost. Bihar Board Class 9 English Book Solutions Prose Chapter 6 The Shehnai of Bismillah Khan Questions and Answers provided are as per the latest exam pattern and syllabus. Access the topics of Panorama English Book Class 9 Solutions Chapter 6 The Shehnai of Bismillah Khan through the direct links available depending on the need. Clear all your queries on the Class 9 English Subject by using the Bihar Board Solutions for Chapter 6 The Shehnai of Bismillah Khan existing.

Panorama English Book Class 9 Solutions Chapter 6 The Shehnai of Bismillah Khan

If you are eager to know about the Bihar Board Solutions of Class 9 English Chapter 6 The Shehnai of Bismillah Khan Questions and Answers you will find all of them here. You can identify the knowledge gap using these Bihar Board Class 9 English Solutions PDF and plan accordingly. Don’t worry about the accuracy as they are given after extensive research by people having subject knowledge alongside from the latest English Textbooks.

Bihar Board Class 9 English The Shehnai of Bismillah Khan Text Book Questions and Answers

A. Work in small groups and discuss the following:

The Shehnai Of Bismillah Khan Class 9 Answers Bihar Board Question 1.
Have you ever heard a shehnai being played in marriage ceremonies or festivals?
Answer:
Yes, I have heard a shehnai being played in marriage ceremonies and festivals on so many occasions.

The Shehnai Of Bismillah Khan Question Answer Bihar Board Question 2.
How do you like this instrument?
Answer:
I like this instrument very much. It has magical power to charm every one.

Question Answer Of The Shehnai Of Bismillah Khan Bihar Board Question 3.
Discuss any pipe instrument which is played in your local-ity in marriage ceremonies or festivals.
Answer:
Clarinent is a pipe instrument which is played in my locality in marriage and festivals. This is a part of the band party. The man who plays on the clarinet is the master of the musical party, he heads the party.

The Shehnai Of Bismillah Khan Class 9 Solutions Bihar Board Question 4.
Do you know that it was Bismillah Khan, the great Shehnai maestro, who made this instrument a reality? Can you name some leading players of other popular musical instruments?
Answer:
I know that it was Bismillah Khan the great Shehnai maestro. Who made this instrument a reality. Pandit Ravishanker plays Sifar, Guddai Maharaj Tabla, Amjad Ali Khan SaTod. They are leading players of these popular instruments.

B.1.1. Write ‘T’ for true and ‘F’ for false statements:

1. Bismillah Khan belongs to a family of musicians from Uttar Pradesh.
2. Bismillah Khan’s ancestors were also great shehnai players.
3. The flowing water of the Ganga gave inspiration to Bismillah to create ragas.
4. He learnt shehnai from his parent and grandfather.

Answer:

1. — F
2. — T
3. — T
4. — F

B.1.2. Complete the sentences on the basis of the unit you have just studied.

1. The pungi is a __________ instrument.
2. The pungi became the generic name for ________ noise makers.
3. The instrument which is so different from the pungi is called __________
4. _______ holes were made on the body of a pipe.
5. _________ was the Shehnai nawaj of Bhojpuri king’s court.
6. __________ was Bismillah’s grandfather.
7. Bismillah accompanied _________ to the Vishnu temple of Benaras.
8. Bismillah played at the temple of ________ and at the banks of ________ as a young apprentice.

Answer:

1. musical
2. reeded
3. Shehnai
4. Seven
5. Rasool Bux Khan
6. Rasool bux Khan
7. Ali Bux
8. Balaji and Mangla Maiya, the Ganga.

B.1.3. Answer the following questions very briefly:

Shehnai Of Bismillah Khan Class 9 Solutions Bihar Board Question 1.
Who banned the playing of the pungi?
Answer:
Emperor Aurangzeb banned the playing of the pungi.

Shehnai Of Bismillah Khan Class 9 Question Answers Bihar Board Question 2.
What generic name did the pungi come to acquire?
Answer:
The pungi became the generic name for reeded the pungi.

The Shehnai Of Bismillah Khan Class 9 Questions And Answers Bihar Board Question 3.
Who revived the pungi?
Answer:
A barber of a family of professional musician revived the pungi.

Question Answer Of Chapter The Shehnai Of Bismillah Khan Bihar Board Question 4.
Where was the Shehnai played for the first time?
Answer:
The Shehnai was played for the first time in the Shah’s chamber.

Class 9 The Shehnai Of Bismillah Khan Question Answer Bihar Board Question 5.
Who played the instrument for the first time so different from the pungi?
Answer:
A barber of a family of professional musicians played the instrument for the first time so different from the pungi.

The Shehnai Of Bismillah Khan Class 9 Question Answer Bihar Board Question 6.
What is naubat called?
Answer:
The naubat is traditional ensemble of nine instruments.

The Shehnai Of Bismillah Khan Class 9 Bihar Board Question 7.
Who brought the instrument shehnai on the classical stage?
Answer:
Ustad Bismillah Khan brought the instrument Shehnai on the classical stage.

Class 9 English Chapter The Shehnai Of Bismillah Khan Question Answer Bihar Board Question 8.
Which sport did Bismillah Khan play in his childhood?
Answer:
In his childhood Bismillah Khan played gilli-danda.

Question Answer Of Shehnai Of Bismillah Khan Bihar Board Question 9.
Where did he play the sports?
Answer:
He played gilli-danda near a pond in the ancient estate of Dumraon in Bihar.

The Shehnai Of Bismillah Khan Question Answer Class 9 Bihar Board Question 10.
Where did he go to sing the Bhojpuri “Chaita”?
Answer:
He went to Bihariji temple to sing the Bhdjpuri “Chaita”.

Class 9 The Sound Of Music Solutions Bihar Board Question 11.
What is the highest civilian award in India?
Answer:
‘The Bharat Ratna’ is the highest civilian award in In-dia.

The Shehnai Of Bismillah Khan Question Answers Bihar Board Question 12.
Who was Bismillah’s father?
Answer:
Paigamber Bux was Bismillah’s father.

Bismillah Khan Class 9 Question Answer Bihar Board Question 13.
Who was Bismillah’s mau-nial uncle?
Answer:
Ali Buxwis Bismillah’s maternal uncle.

B.2.1. Complete the sentences on the basis of the unit you have just Studied:

1. At the age of fourteen Bismillah accompanied his uncle to __________
2. In 1938 came Bismillah’s first break in _________ in _________
3. He sang __________ on 15th August 1947.
4. His first trip abroad was to __________
5. Film director Vijay Bhatt named his film as __________ after being impressed by the shehnai.
6. National awards like the ________ and the Padma Vibhushan were conferred on him.
7. Bismillah Khan was also referred to as __________
8. An auditorium in Tehran named after him is called __________

Answer:

1. The Allahabad music conference
2. All India Radio, Lucknow
3. Ragg Kafi
4. Afganistan
5. Gunj Uthi Shehnai
6. Padma Shri, the Padma Bhushan
7. Khansaab
8. Tahar Musiquee Ustad Bismillah Khan.

B.2.2. Write ‘T’ for true and ‘F’ for false statements:

1. An auditorium in Tehran was named after Bismillah Khan.
2. Bismillah Khan was fondly called ‘Khansaab’
3. Khan Saab was a shehnai player of international repute even then no National Awards were conferred on him.

Answer:

1. — T
2. — T
3. — F

B.2.3. Answer the following questions very briefly:

1. When was Bismillah Khan awarded India’s highest civilian award?
2. Of which two cities was he most found of?

Answer:

1. In 2001 Bismillah Khan was awarded India’s highest civilian award
2. Benaras and Dumraon.

C. Long Answer Type Questions

Question 1.
Which emperor banned the playing of the pungi? Do you think that is against the right to expression?
Answer:
Emperor Aurangzeb banned the playing of the pungi. I think it is against the right to expression. But it must be thought that it was time of kingship, not ‘democracy.’ Those days a king was free to do that mind it playing Pungi was banned in the palace only not for public places.

Question 2.
Ustad Khan refused the celluloid world after two films. Was it a loss tathe cinema world or gain to Hindustani Music? Discuss.
Answer:
Ustad Bismillah Khan refused the celluloid world only after two films. It was not much loss to the cinema world because it was a great gain to Indian Hindustani music. Classical music has not much importance to the general public, who go to see a popular cinema. Ustad rejected films because it was an artificial world and it was much too glamorous. He preferred music to money. So it is clear that according to Ustad outside cinema. Natural Hindustani music exists.

Question 3.
Capture in your own words the feeling of the Ustad, when he received Bharat Ratna.
Answer:
When Ustad Bismillah Khan received the Bharat Ratna he became very happy. The covered award was resting on his chest. His eyes were glinting with more happiness. He told that “All I would like to say is. Teach your children music this is Hindustani’s rishest tradition even the West is now coming to Meant our music.”

Question 4.
‘Only in India it is possible that a devout Muslim like Khan Saheb can very naturally play the shehnai every morning at Kashi Vishwanath temple.’ What light does this statement throw on India’s cultural heritage? Discuss.
Answer:
India is a land of composite culture. There is religious tolerance in our society. Temple is no place for a muslim. But a muslim like Bismillah Khan used to play shehnai in Hindu Temple. Not only Bismillah but his uncle used to play in different Hindu temples. It makes the composite culture of India tellingly clear. Their playing in temple and getting the love and admiration of all in the bargain is a wonderful example of our rich cultural heritage that denies any sort of discrimination on the basis of religion. It embraces all.

Question 5.
Describe the incident of Ustad Khan visiting Pakistan.
Answer:
After partition of India Ustad Khan did not want to go to Pakistan. He could not leave Benaras and the Ganga, He went to Pakistan only once. He crossed the boarder justto say that he had been to Pakisten. He was there for only about an hour. He said ‘Namaskar’ to the Pakistanis and ‘Salam Alai-kum’ to the Indians. This was an exchange of language. He had a good laugh at that incident.

Question 6.
How did shehnai get its name? Describe in your own words the process how the Pungi became the shehnai.
Answer:
The musical instrument was named shehnai because it was bom in the chamber of shah or emperor. The nai or the barber who perfected it also deserved credit. So Shah and nai put together become Shehnai. Shehnai has its origin from the pungi which had an unpleasant sound. A barber decided to improve the tonal quality of this instrument. He chose a pipe.This hollow stem was longer and broader than the Pungi. He drilled seven holes in it. It now ‘produced a musical sound. The nai played it in the chamber of Emperor Aurangzeb. It became popular in course of time.

Comprehension Based Questions with Answers

1. Emperor Aurangzeb banned the playing of a musical instrument called pungi in the royal residence, for it had a shrill unpleasant sound. The pungi became the generic name for reeded noisemakers. Few had thought that it would one day be revived. A barber of a family of professional musicians, who had access to the royal palace, decided to improve the tonal quality of the pungi. He chose a pipe with a natural hollow stem that was longer and broader than the pungi, and made seven holes on the body of the pipe. When he played on it, closing and opening some of these holes, soft and melodious sounds were produced. He played the instrument before royalty and every one yvas impressed. The instrument so different, from the pungi had to be given a new name. As the story goes, since it was first played in the Shah’s chambers and was played by a nai (barber), the instrument was named the ‘shehnai’.

Questions:

1. Name the lesson and author.
2. Why did emperor Aurangzeb bap the playing of pungi?
3. Who decided to improve it?
4. What did he do to bring pungi back?
5. How did shehnai get its name?
6. Which word in the passage means the following, ruler.

Answers:

1. The name of lesson is The Shehnai of Bismillah Khan and it is adapted.
2. Emperor Aurangzeb banned the playing of pungi because of its shrill unpleasant sound.
3. A barber decided to improve it.
4. He improved its quality of sound.
5. Since the new instrument was played in the shah’s chambers and was played by a nai, it was named the ‘Shehnai’.
6. Emperor.

2. The sound of shehnai began to be considered auspieious. And for this reason it is still played in temples and is an indispensable component of any North Indian wedding. In the past, the shehnai was part of the naubat or traditional ensemble of nine instruments found at royal courts. Till recently it was used only in temples and weddings. The credit for bringing this instrument on to the classical stage goes to Ustad Bismillah Khan.
As a five-year old boy, Bismillah Khan played gilli-danda near a-pond in the ancient estate of Dumraon in Bihar. He would regularly go to the nearby Bihariji temple to sing the Bhojpuri ‘Chaita’, at the end of which he would earn a big laddu weighing 1.25 kg. a prize given by the local Maharaja. This happened 80 years ago, and the little boy has travelled far to earn the highest civilian award in India – the Bharat Ratna.
Born on 21 March 1916, Bismillah belongs to a well-known family of musicians from Bihar. His grandfather, Rasool Bux Khan, was the shehnai-nawaz of the Bhojpur king’s court. His father, Paigambar Bux, and other paternal ancestors were also shehnai players.

Questions:

1. For what reason is it played in temples and weddings?
2. Who was the credit man and for what?
3. How did Bismillah earn a big laddu?
4. When and where was Bismillah bom?
5. Which word in the passage mean ‘group’.

Answers:

1. Shehnai is played in the temples on the occasion of wedding in North India because the sound of shehnai is considered auspicious.
2. Bismillah was the credit man, who brought shehnai to the classical stage.
3. He would regularly go to the near by Bihariji temple to sing the Bhojpuri ‘Chaita’, at the end of which he would earn a big Laddu weighing 1.25 kg. a prize given by the local Maharaja.
4. Bismillah Khan was bom on 21 March 1916, in a well known family of musicians from Bihar.
5. Ensemble.

3. The young boy took to music early in life. At the age of three when his mother took him to his maternal uncle’s house in Benaras (now Varanasi), Bismillah was fascinated watching his uncles practise the shehnai. Soon Bismillah started accompanying his uncle, Ali Bux, to the Vishnu temple of Benaras where Bux was employed to play the shehnai. Ali Bux would play the shehnai and Bismillah would sit captivated for hours. Slowly, he started getting lesson in playing the instrument and would sit practising throughout the day. For years to come the temple of Balaji and Mangala Maiya and the banks of the Ganga became the young apprentice’s favourite haunts, where he could practise in solitude. The flowing waters of the Ganga inspired him to improvise and invent ragas that were earlier considered to be beyond the range of the shehnai.

Questions:

1. Who was Bismillah’s first Gum in Shehnai Vadan?
2. Where did he practice Shehnai?
3. What did he invent?
4. What was his motivating force?
5. Which word in the passage means, a person who is learning a trade?

Answers:

1. Bismillah’s maternal uncle Ali Bux was his first Guru.
2. He practised in the temple of Balaji, Mangala Maiya and on the bank of the Ganga.
3. He invented ragas that were earlier considered to be beyond the range of Shehnai.
4. The flowing water of the Ganga was his motivating force.
5. Apprentice.

4. At the age of 14, Bismillah accompanied his uncle to the Allahabad Music Conference. At the end of hlTTccital, Ustad Faiyaz Khan patted the young boy’s back and said, “Work hard and you shall make it.” With the opening of the All India Radio in Lucknow in 1938 came Bismillah’s big break. He sdoh became an often heard shehnai player on the radio. When India gained independence on 15 August 1947, Bismillah Khan became the first Indian to greet the nation with his shehnai. He poured his heart out into Ragg Kafi from the Red Fort to an audience which included Pandit Jawaharlal Nehru, who later gave his famous Tryst with Destiny’ speech.

Questions:

1. Who patted the back of young Bismillah and what did he say?
2. When did he get big break?
3. On what occasion did he pour his heart?
4. Among the audiance, who was a big personality there?

Answers:

1. Ustad Faiyaz Khan patted the back of Bismillah and he said, “work hard and you shall make it”.
2. Bismillah got a big break in 1938 when. AH India Radio was opened at Lucknow. He became an often heard Shehnai player, on the radio.
3. It was the occasion qf 15th August 1947 Bismillah Khan was the first Indian to greet the nation with his Shehnai from the Red fort. He had played Ragg Kafi in which he poured his heart.
4. Among audience a big personality was Pandit Jawahar Lai Nehru.

5. Bismillah Khan has given many memorable performances both in India and abroad. His first trip abroad was to Afganistan where King Zahir Shah was so taken in by the maestro that he gifted him priceless Persian carpets and other souvenirs. The King of Afganistan was not the only one to be fascinated with Bismillah’s music. Film director Vijay Bhatt was so impressed after hearing Bishmillah play at a festival that he named a film after the instrument called Gunj Uthi Shehnai. The film was a hit, and one of Bismillah Khan’s composition, “Dil ka khilona hai toot gaya turned out to be nationwide chartbuster! Despite this huge success in the celluloid world, Bismillah Khan’s success in film music was limited to two: Vijay Bhatt’s Gunj Uthi Shehnai and Vikram Srinivas’s Kannada venture, Sanadhi Apanna. “I just can’t come to terms with the artificiality and glamour of the film world,” he says with emphasis.

Questions:

1. Where was his first trip?
2. Who gave him priceless gift and what was that?
3. Which two film^did he work in and with what result?
4. Why did he not continue working in the films?
5. What does ‘come to term’ imply?
6. Which quality of his character and ideology is revealed in the above passage?
7. Which words in the passage mean the following (a) of film (b) risk in under taking.

Answers:

1. His first trip was to Afganistan.
2. The king Zahir Shah gave him priceless gifts those were Persian carpets.
3. He worked in ‘Gung UthiShehnai’ and ‘Sanadhi Apanna’. He was quite successful.
4. He did not like the glamour and the artificial atmosphere ‘ over there.
5. It means that he could not agree to or feel at home with the film industry.
6. He does not give preference to money. Music is more important to him. He is true artist who knows that art can not be developed in artificial atmosphere.
7. (a) celluloid (b) venture.

6. Awards and recognition came thick and fast. Bismillah Khan became the first Indian to be invited to perform at the prestigious Lincoln Centre Hall in the United States of America. He also took part in the World Exposition in Montreal, in the Cannes Art Festival and in the Osaka Trade Fair. So well known did he become internationally that an auditorium in Teheran was named after him — Tahar Mosiquee Ustad Bismillah Khan. National awards like the Padmashri, the Padma Bhushan and the Padma Vibushan were conferred on him. In 2001, Ustad Bismillah Khan was awarded India’s highest civilian award, the Bharat Ratna. With the coveted award resting on his chest and his eyes glinting with’rare happiness, he said, “All I would like to say is : Teach your children music, this is Hindustan’s richest tradition; even the West is now coming to leam our music.”

Questions:

1. What does ‘thick and fast’ imply in the first sentence?
2. Mention two places in foreign countries that he visited?
3. How can you say that people of Tehran has a great regard for him?
4. Name the National Awards that he got.
5. What did he say when he got ‘Bharat Ratna’?
6. Which word in the passage mean ‘a place where audience assemble’.

Answers:

1. It means that big awards and great recognition came in quick succession one after the. other.
2. They are Montreal in the Cannes and Osaka.
3. They have named an Auditorium in Tehran after his name.
4. He got the following National awards The Padmashri, The Padma Bhushan, the Padma Vibhushan qnd the Bharat Ratna.
5. He said, All I would like to say is : Teach your children music, this is Hindustan’s richest tradition even the West is now coming to leam our music.
6. Auditorium.

7. In spite of having travelled all over the world Khansaab, as he is fondly called, is exceedingly fond of Benaras and Dumraon and they remain for him the most wonderful towns of the world. A student of his once wanted him to head a shehnai school in the U.S.A., and the student promised to recreate the atmosphere of Benaras by replicating the temples there. But Khansaab asked him if he would be able to transport River Ganga as well. Later he is remembered to have said, “That is why whenever I am in a foreign country, I think of only Benaras and the holy Ganga. And while in Benaras, I miss the unique mattha of Dumraon. Shekhar Gupta: When partition happened, didn’t you and your family think of moving to Pakistan?
Bismillah Khan: God forbid! Me, leave Benaras? Never! I went to Pakistan once I crossed the border just to say I have been to Pakistan. I was there for about an hour. I said nature to the Pakistanis and salaam Walaikum to the Indians! I had a good laugh.
Ustad Bismillah Khan’s life is a perfect example of the rich, cultural heritage of India, one that effortlessly accepts that a devout Muslim like him can very naturally play the shehnai every morning at the Kashi Vishwanath temple.

Questions:

1. What did one of his students offer him, once?
2. How did he think he could bring in the atmosphere of Benaras in the U.S.A.
3. What could the student not able to bring in the U.S.A. to create the same atmosphere?
4. Which two towns in India are Ustad’s favourite? Why?
5. Bismillah Khan is an example of a glorious tradition oi India, what is that tradition?
6. Which words in the passage mean the following.
   (a) exact copying
   (b) make again

Answers:

1. One of his students offered him to become the head of a Shehnai- school in the U.S.A.
2. He thought he could bring in the same atmosphere by building in the U.S.A. the exact copies of temples of Benaras.
3. The student could not bring the Ganga there.
4. The favourite towns in India are-Benaras and Dunraon. Benaras is the town where he practised and invented many ragas because an expert and earned the name of ‘ Shehnai maestro. His memories and associations to these towns made them his favourite.
5. Ustad Bismillah Khan’s life is a perfect example of the rich cultural heritage of India which accepts a devout Muslim like him naturally playing every morning in the Kashi Vishwanath temple.
6. (a) Photo state
   (b) Repeat.

Hope you liked the article on Bihar Board Solutions of Class 9 English Chapter 6 The Shehnai of Bismillah Khan Questions and Answers and share it among your friends to make them aware. Keep in touch to avail latest information regarding different state board solutions instantly.

Bihar Board Class 8 Maths Solutions Chapter 11 सीधा और प्रतिलोम समानुपात Text Book Questions and Answers.

BSEB Bihar Board Class 8 Maths Solutions Chapter 11 सीधा और प्रतिलोम समानुपात

Bihar Board Class 8 Maths सीधा और प्रतिलोम समानुपात Ex 11.1

8th Class Math Bihar Board प्रश्न 1.
निम्नलिखित तालिका में x तथा y समानुपाती (अनुक्रमानुपाती) हैं। या नहीं? ज्ञात कीजिए।
उत्तर

Class 8th Bihar Board Math प्रश्न 2.
टाइपिंग की परीक्षा पास करने के लिए कम से कम 30 शब्द प्रति मिनट टाइप करने होते हैं। एक परीक्षार्थी को पास होने के लिए आधे घंटे में कम से कम कितने शब्द टाइप करने होंगे?
उत्तर
प्रति मिनट कम से कम शब्द = 30 शब्द
आधे घंटे में = 30 min
कम से कम शब्द टाइप करने होंगे = 30 × 30 = 900 शब्द

Sidha Aur Pratilom Samanupat Bihar Board प्रश्न 3.
मुकुंद के पास एक सड़क का मानचित्र है जिसके पैमाने में 1 सेमी. की दूरी 15 किमी. निरूपित करती है। गाँधी नगर से जाकिर हुसैन सर्कल तक जाने वाली सड़क यदि 75 किमी. है तो मानचित्र में उसे कितने सेमी. से निरूपित किया गया होगा?
उत्तर
1 cm = 15 km
x cm = 75 km
75 × 1 = 15 × x
x = \(\frac{75}{15}\)
x = 5 cm

Bihar Board 8th Class Math Solution प्रश्न 4.
यदि 25 मीटर कपड़े का मूल्य 337.50 रुपये हो तो,
(i) उसी प्रकार के 60 मीटर कपड़े का मूल्य क्या होगा?
(ii) 1620 रु. में इस तरह का कितनी लम्बाई का कपड़ा खरीदा जा सकता है?
उत्तर
(i) 25 m = 337.50
1 m = 337.50 = 25
1 m = 13.5
60m = 13.5 × 60 = 810
(ii) 1 m = 13.5
x m = 1620
1620 = 13.5 × x
x = \(\frac{1620}{13.5}\)
x = 120 m

बिहार बोर्ड क्लास 8 मैथ सलूशन प्रश्न 5.
मकान के एक मॉडल में उसकी ऊँचाई 5 सेमी. व क्रमशः लम्बाई व चौड़ाई 12 सेमी. व 8 सेमी. है। अब यदि वास्तविक परिस्थिति में उसकी ऊँचाई 25 फुट हो तो मॉडल में काम लिया गया पैमाना बताइए तथा वास्तविक लम्बाई व चौड़ाई ज्ञात कीजिए।
उत्तर
वास्तविक ऊँचाई = 25 फुट
मॉडल में उसकी ऊँचाई = 5 फुट
पैमाना = \(\frac{25}{5}\) = 5 cm
वास्तविक ऊँ = 5 × 5 = 25 फुट
ल० = 12 × 5 = 60 फुट
चौ० = 8 × 5 = 40 फुटः

बिहार बोर्ड क्लास 8 मैथ प्रश्न 6.
मान लीजिए 2 किग्रा. दाल में 7 × 10⁵ क्रिस्टल हैं। तब दी गई दालों की मात्रा में कितने क्रिस्टल होंगे?
(i) 8 किग्रा.
(ii) 5 किग्रा.
उत्तर
2 kg के दाल में क्रिस्टल = 7 × 10⁵
1 kg के दाल में क्रिस्टल = \(\frac{7 \times 10^{5}}{2}\) = 3.5 × 10⁵
(i) 8 kg के दाल में क्रिस्टल = 8 × 3.5 × 10⁵ =28 × 10⁵
(ii) 5 kg के दाल में क्रिस्टल = 5 × 3.5 × 10⁵ = 17.5 × 10⁵

Bihar Board Class 8 Math Solution प्रश्न 7.
एक मानचित्र का पैमाना 1 : 25,000000 दिया है। दो नगरों की मानचित्र में दूरी 3 सेमी. है तो वास्तविकता में उनके बीच कितनी दूरी होगी?
उत्तर
मानचित्र में पैमाना
1 cm = 250000 m
3 cm = 3 × 25000000 = 75000000

Math Class 8 Bihar Board प्रश्न 8.
यदि एक स्कूटर 3 लीटर पेट्रोल में 96 किमी. चलता है, तो 320 किमी. चलने के लिए इसे कितने पेट्रोल की आवश्यकता होगी?
उत्तर
3 लीटर = 96 km
x लीटर = 320 km
3 × 320 = x × 96
x = \(\frac{3 \times 320}{96}\)
x = 10 liter

Bihar Board Class 8 Maths सीधा और प्रतिलोम समानुपात Ex 11.2

Class 8 Math Bihar Board प्रश्न 1.
यदि x और y व्युत्क्रमानुपाती विचरण में हों, तो आवश्यकतानुसार रिक्त स्थानों की पूर्ति कीजिए-

उत्तर

Class 8 Maths Bihar Board प्रश्न 2.
निम्नांकित विचरण सारणी में रिक्त स्थानों की पूर्ति कीजिए-

उत्तर

Class 8 Math Book Bihar Board प्रश्न 3.
10 मजदूर किसी काम को 2 दिन में करते हैं। उसी काम को 2 मजदूर कितने दिनों में करेंगे?
उत्तर
10 मजदूर = 2 दिन
2 मजदूर = x दिन
\(\frac {10}{2}\) = \(\frac {2}{x}\)
x = \(\frac {10}{2}\) × 2 = 10 दिन

Bihar Board Math Class 8 प्रश्न 4.
45 मजदूर एक काम को 27 दिनों में पूरा करते हैं, तो कितने मजदूर उसी काम को 15 दिनों में पूरा करेंगे?
उत्तर
45 मजदूर = 27 दिन
x मजदूर = 15 दिन
\(\frac {45}{x}\) = \(\frac {27}{15}\)
x = \(\frac {27}{15}\) × 45
x = 27 × 3
x = 81 मजदूर

Bihar Board Class 8 Math प्रश्न 5.
एक बस 30 किमी./घण्टा की चाल से 6 घण्टे में एक निश्चित दूरी तय करती है । उसी दूरी को वह बस किस चाल से केवल 4 घण्टे में तय कर लेगी?
उत्तर
30 km = 6 hours = x km/h
30 km = 4 hour
x = \(\frac{30 \times 6}{4}\) = 45 km/h

Bihar Board Class 8 Math Book Pdf प्रश्न 6.
40 घोड़े एक क्विंटल चने को 7 दिनों में खाते हैं। कितने घोड़े उतने ही चने को 28 दिनों में खायेंगे?
उत्तर
40 घोड़े = 7 दिन
x घोड़े = 28 दिन
x = \(\frac{28}{7 \times 40}\)
x = 10 दिन

Bihar Board Class 8 Math Solution In Hindi प्रश्न 7.
एक छात्रावास में 300 छात्रों के लिए 15 दिनों की राशन सामग्री है। यदि अवकाश के कारण 200 छात्र बाहर चले जाएँ तो वह सामग्री कितने दिनों तक चलेगी?
उत्तर
300 छात्र = 15 दिन
100 छात्र = x दिन
x = \(\frac{15 \times 300}{100}\) = 45 दिन

Bihar Board Class 8 Ka Math Solution प्रश्न 8.
एक छावनी में 700 सैनिकों के लिए 25 दिनों की पर्याप्त खाद्य सामग्री है। किन्तु कुछ और सैनिकों के आ जाने के कारण वह खाद्य सामग्री केवल 20 दिनों में समाप्त हो जाती है। बताइए कि बाद में छावनी में और कितने सैनिक आए?
उत्तर
700 सैनिक = 25 दिन
x सैनिक = 20 दिन
x = \(\frac{700 \times 25}{20}\) = 875 सैनिक
बाद में छावनी में (700 – 875) = 175 सैनिक आ गए।

Class 8 Math Solution Bihar Board प्रश्न 9.
एक व्यक्ति प्रतिदिन किसी पुस्तक के 8 पृष्ठों को पढ़कर उसे 15 दिनों में पूरा पढ़ लेता है। यदि वह प्रतिदिन 12 पृष्ठ पढ़े तो पूरी पुस्तक को वह कितने दिनों में पढ़ लेगा?
उत्तर
8 पृष्ठ = 75 दिन
12 पृष्ठ = x दिन
x = \(\frac{8}{12}\) × 15 = 10 दिन

Bihar Board Class 8 Math Book प्रश्न 10.
एक सैनिक शिविर में 105 सैनिकों के लिए 21 दिनों की रसद सामग्री है। यदि शिविर में 42 सैनिक और शामिल हो जाएँ, तो रसद सामग्री कितने दिनों में समाप्त हो जायेगी?
उत्तर
105 सैनिक = 21
147 सैनिक = x
x = 105 × \(\frac{21}{147}\)
x = 7 दिन

प्रश्न 11.
निम्नलिखित में से कौन-कौन सी व्युत्क्रमानुपात में विचरण करती हैं?

1. खरीदी गई पुस्तकों की संख्या और प्रत्येक पुस्तक की कीमत।
2. बस द्वारा तय की गई दूरी और खपत पेट्रोल की कीमत।
3. साइकिल द्वारा किसी निश्चित दूरी को पार करने में लगा समय, और उसकी चाल।
4. एक पुल बनाने में लगाए गए मजदूरों की संख्या और पुल बनने में लगने वाला समय।
5. छात्रों की संख्या और प्रतिछात्र वितरित मिठाई का वजन। (यदि 40 किग्रा. मिठाई बाँटनी है)।
6. मजदूरी और कार्य के घण्टे।
7. वस्तुओं की संख्या और उनका कुल मूल्य।

उत्तर

1. नहीं
2. नहीं
3. हाँ
4. हाँ
5. हाँ
6. हाँ
7. हाँ।

प्रश्न 12.
26 जनवरी को एक विद्यालय के 800 छात्रों में 100 ग्राम प्रति छात्र के हिसाब से मिठाई बाँटी गई। उतनी ही मिठाई यदि 1000 छात्रों में बराबर-बराबर बाँटी जाए, तो प्रत्येक छात्र को कितने ग्राम मिठाई मिलेगी?
उत्तर
800 छात्र = 100 ग्राम
1000 छात्र = x ग्राम
x = \(\frac{800 \times 100}{1000}\) = 80 ग्राम

प्रश्न 13.
जब एक नल एक घंटे में 640 लीटर पानी भरता है तो एक पानी टंकी को भरने में 10 घण्टे का समय लगता है। यदि उसी टंकी को दूसरे नल से 8 घण्टे में भरा गया हो, तो दूसरे नल में प्रतिघंटा कितना पानी भरा?
उत्तर
640 ली० = 10 घंटा
x ली० = 8 घंटा
x = \(\frac{640 \times 10}{8}\) = 800 लीटर