Bihar Board Class 7 Hindi Book Solutions किसलय Kislay Bhag 2

Bihar Board Class 7 Hindi Book Solutions किसलय Kislay Bhag 2

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Bihar Board Class 8 English Book Solutions Radiance Part 3

Bihar Board Class 8 English Book Solutions Radiance Part 3

Bihar Board Class 8 English Book Solutions Radiance Part 3 helps you revise the syllabus in a smart way. BSEB Bihar Board Class 8th Radiance English Text Book Solutions are as per the latest syllabus. You can acquire knowledge on the English Subject by referring to Bihar Board Solutions for Class 8 English. Check out Chapterwise Radiance English Book Class 8 Solutions Bihar Board through the quick links available. Download the Bihar Board Class 8 English Solutions PDF free of cost.

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Bihar Board Class 8 Hindi Book Solutions किसलय Kislay Bhag 3

Bihar Board Class 8 Hindi Book Solutions किसलय Kislay Bhag 3

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Bihar Board Class 8th Social Science Solutions सामाजिक विज्ञान

Bihar Board Class 8th Social Science Solutions सामाजिक विज्ञान

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Bihar Board Class 8 Sanskrit Book Solutions अमृता Amrita Bhag 3

Bihar Board Class 8 Sanskrit Book Solutions अमृता Amrita Bhag 3

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Bihar Board Class 8th Maths Solutions गणित

Bihar Board Class 8th Maths Solutions गणित

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Bihar Board Class 8th Science Solutions विज्ञान

Bihar Board Class 8th Science Solutions विज्ञान

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Bihar Board Class 7th Science Solutions विज्ञान

Bihar Board Class 7th Science Solutions विज्ञान

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Bihar Board Class 7 English Book Solutions Radiance Part 2

Bihar Board Class 7 English Book Solutions Radiance Part 2

Bihar Board Class 7 English Book Solutions Radiance Part 2 is a very important preparation resource to learn the subject clearly and improve the subject knowledge. Students should consider BSEB Bihar Board Class 7th Radiance English Text Book Solutions Pdf while preparation and understand the concepts covered in the board prescribed syllabus. It will help students to clear their doubts and perform well in the final Bihar Class 7 board examinations. Radiance English Book Class 7 Solutions Bihar Board are written by our subject expertise after ample research so you all can rely on them while preparing for English board exams & score maximum marks in all exams.

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Bihar Board Class 7 Sanskrit Book Solutions अमृता Amrita Bhag 2

Bihar Board Class 7 Sanskrit Book Solutions अमृता Amrita Bhag 2

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Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Text Book Questions and Answers.

BSEB Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

Bihar Board Class 10 Maths निर्देशांक ज्यामिति Ex 7.4

प्रश्न 1.
बिन्दुओं A (2, -2) और B(3, 7) को जोड़ने वाले रेखाखण्ड को रेखा 2x + y – 4 = 0 जिस अनुपात में विभाजित करती है, उसे ज्ञात कीजिए।
हल
दिया है, बिन्दु A = (2, -2) तथा B = (3, 7)
यहाँ x1 = 2, y1 = -2, x2 = 3, y2 = 7
माना दिए हुए बिन्दुओं से बना रेखाखण्ड रेखा 2x + y – 4 = 0 को m1 : m2 के अनुपात में विभाजित करता है जबकि प्रतिच्छेद बिन्दु (x, y) है।
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q1
बिन्दु (x, 3) रेखा 2x + y – 4 = 0 पर स्थित होगा;
अतः इसके निर्देशांक रेखा 2x + y – 4 = 0 को सन्तुष्ट करेंगे।
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q1.1
अत: अभीष्ट अनुपात = 2 : 9

Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

प्रश्न 2.
x और y में एक सम्बन्ध ज्ञात कीजिए यदि बिन्दु (x, y), (1, 2) और (7, 0) संरेखी है।
हल
माना बिन्दु A = (x, y), B = (1, 2) तथा C = (7, 0)
यहाँ, x1 = x, y1 = y, x2 = 1, y2 = 2, x3 = 7, y3 = 0
∆ का क्षेत्रफल = \(\frac{1}{2}\) [{x1y2 + x2y3 + x3y1} – {y1x2 + y2x3 + y3x1}]
= \(\frac{1}{2}\) [{x × 2 + 1 × 0 + 7 × y} – {y × 1 + 2 × 7 + 0 × x}]
= \(\frac{1}{2}\) [{2x + 0 + 7y} – {y + 14 + 0}]
= \(\frac{1}{2}\) [2x + 7y – y – 14]
= \(\frac{1}{2}\) [2x + 6y – 14]
= \(\frac{2}{2}\) (x + 3y – 7)
= x + 3y – 7
परन्तु यदि बिन्दु A, B, C संरेख हों तो ΔABC का क्षेत्रफल शून्य होना चाहिए।
x + 3y – 7 = 0
अतः x और में सम्बन्ध : x + 3y – 7 = 0

Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

प्रश्न 3.
बिन्दुओं (6, -6), (3, -7) और (3, 3) से होकर जाने वाले वृत्त का केन्द्र ज्ञात कीजिए।
हल
माना A(6, -6), B(3, -7) तथा C(3, 3) बिन्दु एक वृत्त की परिधि पर हैं और वृत्त का केन्द्र O(h, k) है।
तब, OA, OB तथा OC वृत्त की त्रिज्याएँ होंगी।
अतः OA = OB = OC
⇒ OA2 = OB2 = OC2
OA2 = [ केन्द्र O(h, k) और बिन्दु A (6, -6) के बीच की दूरी]2
⇒ OA2 = (h – 6)2 + (k + 6)2
⇒ OA2 = h2 – 12h + 36 + k2 + 12k + 36
⇒ OA2 = h2 + k2 – 12h + 12k + 72 ……..(1)
OB2 = [केन्द्र O (h, k) और बिन्दु B (3, -7) के बीच की दूरी]2
⇒ OB2 = (h – 3)2 + (k + 7)2
⇒ OB2 = h2 – 6h + 9 + k2 + 14k + 49
⇒ OB2 = h2 + k2 – 6h + 14k + 58 ………(2)
OC2 = [केन्द्र O(h, k) और बिन्दु C(3, 3) की दूरी]2
⇒ OC2 = (h – 3)2 + (k – 3)2
⇒ OC2 = h2 – 6h + 9 + k2 – 6k + 9
⇒ OC2 = h2 + k2 – 6h – 6k + 18 ………(3)
समीकरण (2) में से समीकरण (3) को घटाने पर,
20k + 40 = OB2 – OC2 = 0
⇒ k = -2
समीकरण (1) में से समीकरण (2) को घटाने पर,
-6h – 2k + 14 = OA2 – OB2 = 0
⇒ 6h + 2k = 14
⇒ 6h + (2 × -2) = 14 (∵ k = -2)
⇒ 6h – 4 = 14
⇒ 6h = 14 + 4 = 18
⇒ h = 3
अत: वृत्त का केन्द्र = (3, -2)

Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

प्रश्न 4.
किसी वर्ग के दो सम्मुख शीर्ष (-1, 2) और (3, 2) हैं। वर्ग के अन्य दोनों शीर्ष ज्ञात कीजिए।
हल
दिया है, वर्ग के दो सम्मुख शीर्ष (-1, 2) व (3, 2) हैं।
वर्ग के एक विकर्ण का मध्य-बिन्दु = \(\left(\frac{-1+3}{2}, \frac{2+2}{2}\right)\) = (1, 2)
वर्ग के विकर्ण की लम्बाई = \(\sqrt{(-1-3)^{2}+(2-2)^{2}}\)
= \(\sqrt{(-4)^{2}+0}\)
= √16
= 4 मात्रक
तब, विकर्णों के प्रतिच्छेद बिन्दु E(मध्य बिन्दु) से प्रत्येक शीर्ष विकर्ण × \(\frac {1}{2}\) = 2 मात्रक दूरी पर होगा।
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q4
चित्र से स्पष्ट है कि शेष दोनों बिन्दु विकर्ण BD पर होंगे जो AC पर लम्ब होगा। तब प्रत्येक बिन्दु का भुज +1 होगा। माना कोटि y है।
तब, बिन्दु (1, 2) की बिन्दु (+1, y) से दूरी = 2 मात्रक
\(\sqrt{(1-1)^{2}+(y-2)^{2}}=2\)
⇒ \(\sqrt{0+(y-2)^{2}}=2\)
⇒ ±(y – 2) = 2
⇒ y – 2 = ±2
⇒ y = ±2 + 2
⇒ y = 0 या 4
अत: वर्ग के शेष दोनों शीर्ष (1, 0) व (1, 4) हैं।

Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

प्रश्न 5.
कृष्णानगर के एक सेकेण्डरी स्कूल के कक्षा X के विद्यार्थियों को उनके बागवानी क्रियाकलाप के लिए एक आयताकार भूखण्ड दिया गया है। गुलमोहर की पौध (sapling) को परस्पर 1 मीटर की दूरी पर इस भूखण्ड की परिसीमा (boundary) पर लगाया जाता है। इस भूखण्ड के अन्दर एक त्रिभुजाकार घास लगा हुआ लॉन (lawn) है, जैसा कि आकृति में दर्शाया गया है। विद्यार्थियों को भूखण्ड के शेष भाग में फूलों के पौधे के बीज बोने हैं।
(i) A को मूलबिन्दु मानते हुए, त्रिभुज के शीर्षों के निर्देशांक ज्ञात कीजिए।
(ii) यदि मूलबिन्दु C हो तो ∆PQR के शीर्षों के निर्देशांक क्या होंगे?
साथ ही उपर्युक्त दोनों स्थितियों में, त्रिभुजों के क्षेत्रफल ज्ञात कीजिए। आप क्या देखते हैं?
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q5
हल
बिन्दुओं P, Q व R से सम्मुख अक्षों पर लम्ब खींचे गए हैं। (चित्र देखिए)
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q5.1
(i) यदि A मूलबिन्दु हो तो
बिन्दु P = (4, 6),Q = (3, 2) तथा R = (6, 5)
यहाँ x1 = 4, y1 = 6, x2 = 3, y2 = 2, x3 = 6, y3 = 5
∆PQR का क्षेत्रफल = \(\frac{1}{2}\) [{x1y2 + x2y3 + x3y1} – {y1x2 + y2x3 + y3x1}]
= \(\frac{1}{2}\) {{4 × 2 + 3 × 5 + 6 × 6} – {6 × 3 + 2 × 6 + 5 × 4}]
= \(\frac{1}{2}\) [(8 + 15 + 36) – (18 + 12 + 20)]
= \(\frac{1}{2}\) [59 – 50]
= \(\frac{1}{2}\) × 9
= \(\frac{9}{2}\) वर्ग मात्रक

(ii) जब C मूलबिन्दु हो तो
बिन्दु P = (-12, -2), Q = (-13, -6) तथा R = (-10, -3)
यहाँ x1 = -12, y1 = -2, x2 = -13, y2 = -6, x3 = -10, y3 = -3
∆PQR का क्षेत्रफल = \(\frac{1}{2}\) [{x1y2 + x2y3 + x3y1} – {y1x2 + y2x3 + y3x1}]
= \(\frac{1}{2}\) [{(-12 × -6) + (-13 × -3) + (-10 × -2)} – {(-2 × -13) + (-6 × -10) + (-3 × -12)}]
= \(\frac{1}{2}\) (72 + 39 + 20) – (26 + 60 + 36)]
= \(\frac{1}{2}\) [(131) – (122)]
= \(\frac{1}{2}\) × 9
= \(\frac{9}{2}\) वर्ग मात्रक
अत: दोनों ही स्थितियों में त्रिभुज का क्षेत्रफल समान है।

Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

प्रश्न 6.
एक त्रिभुज ABC के शीर्ष A (4, 6), B(1, 5) और C (7, 2) हैं। भुजाओं AB और AC को क्रमशः D और E पर प्रतिच्छेद करते हुए एक रेखा इस प्रकार खींची गई है कि \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) है। ΔADE का क्षेत्रफल परिकलित कीजिए और इसकी तुलना ΔABC के क्षेत्रफल से कीजिए।
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q6
हल
दिया है, ΔABC के शीर्ष A (4, 6), B(1, 5) और C (7, 2) हैं।
\(\frac{A D}{A B}=\frac{1}{4}\)
⇒ AB = 4AD
⇒ AD + DB = 4AD
⇒ DB = 3AD
⇒ \(\frac{A D}{D B}=\frac{1}{3}\)
माना D के निर्देशांक यदि (x, y) हों तो
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q6.1
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q6.2
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q6.3
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q6.4

Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

प्रश्न 7.
मान लीजिए A(4, 2), B(6, 5) और C (1, 4)एक त्रिभुज ABC के शीर्ष हैं।
(i) A से होकर जाने वाली माध्यिका BC से D पर मिलती है। बिन्दु D के निर्देशांक ज्ञात कीजिए।
(ii) AD पर स्थित ऐसे बिन्दु P के निर्देशांक ज्ञात कीजिए कि AP : PD = 2 : 1 हो।
(iii) माध्यिकाओं BE और CF पर ऐसे बिन्दुओं Q और R के निर्देशांक ज्ञात कीजिए कि BQ : QE = 2 : 1 हो और CR : RF = 2 : 1 हो।
(iv) आप क्या देखते हैं?
[नोट – वह बिन्दु जो तीनों माध्यिकाओं में सार्वनिष्ठ हो, उस त्रिभुज का केन्द्रक (centroid) कहलाता है और यह प्रत्येक माध्यिका को 2 : 1 के अनुपात में विभाजित करता है।]
(v) यदि A(x1, y1), B(x2, y2) और C(x3, y3) त्रिभुज ABC के शीर्ष हैं तो इस त्रिभुज के केन्द्रक के निर्देशांक ज्ञात कीजिए।
हल
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q7
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q7.1
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q7.2 (1)
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q7.3 (1)
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q7.4

Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4

प्रश्न 8.
बिन्दुओं A(-1, -1), B(-1, 4), C(5, 4) और D(5, -1) से एक आयत ABCD बनता है। P, Q, R और S क्रमश: भुजाओं AB, BC, CD और DA के मध्य-बिन्दु हैं। क्या चतुर्भुज PQRS एक वर्ग है? क्या यह एक आयत है? क्या यह एक समचतुर्भुज है? सकारण उत्तर दीजिए।
हल
दिए हुए बिन्दु A = (-1, -1), B = (-1, 4), C = (5, 4) और D = (5, -1)
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q8
Bihar Board Class 10 Maths Solutions Chapter 7 निर्देशांक ज्यामिति Ex 7.4 Q8.1
∵ चतुर्भुज PQRS में, PQ = QR = RS = SP और विकर्ण PR ≠ विकर्ण QS
अत: चतुर्भुज PQRS एक समचतुर्भुज है।