NCERT Solutions for Class 9 Ganita Manjari Chapter 4 Exploring Algebraic Identities बीजीय सर्वसमिकाओं का अन्वेषण
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Table of Contents
Exercise Set 4.1
Question 1:
Using the identity (a + b)2 = a2 + 2ab + b2, expand the following:
सर्वसमिका (a + b)2 = a2 + 2ab + b2 का उपयोग करके निम्नलिखित का विस्तार कीजिए।
(i) (7x + 4y)2 (ii) \(\left(\frac{7}{5}x + \frac{3}{2}y\right)^2\) (iii) (2.5p + 1.5q)2
(iv) \(\left(\frac{3}{4}s + 8t\right)^2\) (v) \(\left(x + \frac{1}{2y}\right)^2\) (iv) \(\left(\frac{1}{x} + \frac{1}{y}\right)^2\)
Solution
(i) (7x + 4y)2
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) (7x + 4y)2 = (7x)2 + 2×(7x)×(4y) + (4y)2
\(\quad\quad\quad\quad\;\) = 49x2 + 56xy + 16y2
(ii) \(\left(\frac{7}{5}x + \frac{3}{2}y\right)^2\)
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) \(\left(\frac{7}{5}x + \frac{3}{2}y\right)^2\) = \(\left(\frac{7}{5}x\right)^2\) + 2×\(\left(\frac{7}{5}x\right)\)×\(\left(\frac{3}{2}y\right)\) + \(\left(\frac{3}{2}y\right)^2\)
\(\quad\quad\quad\quad\quad\quad\) = \(\frac{49}{25}x^2\) + \(\frac{21}{5}xy\) + \(\frac{9}{4}y^2\)
(iii) (2.5p + 1.5q)2
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) (2.5p + 1.5q)2 = (2.5p)2 + 2×(2.5x)×(1.5y) + (1.5y)2
\(\quad\quad\quad\quad\;\) = 6.25p2 + 7.5pq + 2.25q2
(iv) \(\left(\frac{3}{4}s + 8t\right)^2\)
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) \(\left(\frac{3}{4}s + 8t\right)^2\) = \(\left(\frac{3}{4}s\right)^2\) + 2×\(\left(\frac{3}{4}s\right)\)× 8t + \((8t)^2\)
\(\quad\) \(\left(\frac{3}{4}s + 8t\right)^2\) = \(\left(\frac{3}{4}s\right)^2\) + 2×\(\left(\frac{3}{4}s\right)\)× 8t + \((8t)^2\)
\(\quad\quad\quad\quad\quad\quad\) = \(\frac{4}{16}s^2\) + 12st + \(64t^2\)
(v) \(\left(x + \frac{1}{2y}\right)^2\)
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) \(\left(x + \frac{1}{2y}\right)^2\) = \((x)^2\) + 2×(x)× \(\frac{1}{2y}\)+ \(\left(\frac{1}{2y}\right)^2\)
\(\quad\quad\quad\quad\quad\) = \(x^2\) + \(\frac{x}{y}\)+ \(\frac{1}{4y^2}\)
(iv) \(\left(\frac{1}{x} + \frac{1}{y}\right)^2\)
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) \(\left(\frac{1}{x} + \frac{1}{y}\right)^2\) = \(\left(\frac{1}{x}\right)^2\) + 2×\(\left(\frac{1}{x}\right)\)×\(\left(\frac{1}{y}\right)\) + \(\left(\frac{1}{y}\right)^2\)
\(\quad\quad\quad\quad\quad\quad\) = \(\frac{1}{x^2}\) + \(\frac{2}{xy}\) + \(\frac{1}{y^2}\)
Question 2:
Using the same identity, find the values of the following:
उपर्युक्त सर्वसमिका का उपयोग करते हुए निम्नलिखित का मान ज्ञात कीजिए ।
(i) (64)2 (ii) (105)2 (iii) (205)2
Soution
(i) (64)2 = (60 + 4)2
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) (60 + 4)2 = (60)2 + 2×(60)×(4) + (4)2
\(\quad\quad\quad\quad\;\) = 3600 + 480 + 16
\(\quad\quad\quad\quad\;\) = 4096
(ii) (105)2 = (100 + 5)2
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) (100 + 5)2 = (100)2 + 2×(100)×(5) + (5)2
\(\quad\quad\quad\quad\;\) = 10000 + 1000 + 25
\(\quad\quad\quad\quad\;\) = 11025
(iii) (205)2 = (200 + 5)2
\(\quad\) [(a + b)2 = a2 + 2ab + b2]
\(\quad\) (200 + 5)2 = (200)2 + 2×(200)×(5) + (5)2
\(\quad\quad\quad\quad\;\) = 40000 + 2000 + 25
\(\quad\quad\quad\quad\;\) = 42025
NCERT Solutions for Class 9 Ganita Manjari Chapter 4 Exploring Algebraic Identities बीजीय सर्वसमिकाओं का अन्वेषण
Exercise Set 4.2
Question 1:
Factor completely:
पूर्णतः गुणनखंडन कीजिए:
(i) 9x2 + 24xy + 16y2
(ii) 4s2 + 20st + 25t2
(iii) 49x2 + 28xy + 4y2
(iv) 64p2 + \(\frac{32}{3}\)pq + \(\frac{4}{9}\)q2
(v) 3a2 + 4ab + \(\frac{4}{3}\)b2
(vi) \(\frac{9}{5}\)s2 + 6sv + 5v2
Solution
(i) 9x2 + 24xy + 16y2
\(\quad\) = \((3x)^2\) + 2 × (3x) × (4y) + \((4y)^2\)
\(\quad\) [a2 + 2ab + b2 = (a + b)2]
\(\quad\) = (3x + 4y)2
\(\quad\) = (3x + 4y)(3x + 4y)
(ii) 4s2 + 20st + 25t2
\(\quad\) = \((2s)^2\) + 2 × (2s) × (5t) + \((5t)^2\)
\(\quad\) [a2 + 2ab + b2 = (a + b)2]
\(\quad\) = (2s + 5t)2
\(\quad\) = (2s + 5t)(2s + 5t)
(iii) 49x2 + 28xy + 4y2
\(\quad\) = \((7x)^2\) + 2 × (7x) × (2y) + \((2y)^2\)
\(\quad\) [a2 + 2ab + b2 = (a + b)2]
\(\quad\) = (7x + 2y)2
\(\quad\) = (7x + 2y)(7x + 2y)
(iv) 64p2 + \(\frac{32}{3}\)pq + \(\frac{4}{9}\)q2
\(\quad\) = \((8p)^2\) + 2 × (3x) × (4y) + \(\left(\frac{2}{3}q\right)^2\)
\(\quad\) [a2 + 2ab + b2 = (a + b)2]
\(\quad\) = (8p + \(\frac{2}{3}q\))2
(v) 3a2 + 4ab + \(\frac{4}{3}\)b2
\(\quad\) = \((\sqrt{3}a)^2\) + 2 × \(\sqrt{3} a\) × \(\left(\frac{2}{\sqrt{3}}b\right)\) + \(\left(\frac{2}{\sqrt{3}}b\right)^2\)
\(\quad\) [a2 + 2ab + b2 = (a + b)2]
\(\quad\) = (3x + 4y)2
(vi) \(\frac{9}{5}\)s2 + 6sv + 5v2
\(\quad\) = \((3x)^2\) + 2 × (3x) × (4y) + \((4y)^2\)
\(\quad\) [a2 + 2ab + b2 = (a + b)2]
\(\quad\) = (3x + 4y)2
Question 2:
Find the values of the following using the identity (a – b)2 = a2 – 2ab + b2.
सर्वसमिका (a – b)2 = a2 – 2ab + b2 का उपयोग करके निम्नलिखित के मान ज्ञात कीजिए।
(i) (79)2 (ii) (193)2 (iii) (299)2
Solution
(i) (79)2 = (80 – 1)2
\(\quad\) [(a – b)2 = a2 – 2ab + b2]
\(\quad\) (80 – 1)2 = (80)2 – 2×(80)×(1) + (1)2
\(\quad\quad\quad\quad\;\) = 6400 – 160 + 1
\(\quad\quad\quad\quad\;\) = 6401 – 160
\(\quad\quad\quad\quad\;\) = 6241
(ii) (193)2 = (200 – 7)2
\(\quad\) [(a – b)2 = a2 – 2ab + b2]
\(\quad\) (200 – 7)2 = (200)2 – 2×(200)×(7) + (7)2
\(\quad\quad\quad\quad\;\) = 40000 – 2800 + 49
\(\quad\quad\quad\quad\;\) = 40049 – 2800
\(\quad\quad\quad\quad\;\) = 37249
(iii) (299)2 = (300 – 1)2
\(\quad\) [(a – b)2 = a2 – 2ab + b2]
\(\quad\) (300 – 1)2 = (300)2 – 2×(300)×(1) + (1)2
\(\quad\quad\quad\quad\;\) = 90000 – 600 + 1
\(\quad\quad\quad\quad\;\) = 90001 – 600
\(\quad\quad\quad\quad\;\) = 89401
NCERT Solutions for Class 9 Ganita Manjari Chapter 4 Exploring Algebraic Identities बीजीय सर्वसमिकाओं का अन्वेषण
Exercise Set 4.3
Question 1:
Find the following squares using one of the above identities. Determine which of these identities will make these calculations
easier
(i) 1172 (ii) 782 (iii) 1982
(iv) 2142 (v) 11042 (vi) 11202
Answer
(i) 1172
\(\quad\) = (100 + 17)2
\(\quad\) [(A + B)2 = A2 + 2AB + B2]
\(\quad\) (100 + 17)2 = 1002 + 2(100)(17) + 172
\(\quad\quad\quad\quad\quad\quad\) = 10000 + 3400 + 289
\(\quad\quad\quad\quad\quad\quad\) = 13689
(ii) 782
\(\quad\) = (80 – 2)2
\(\quad\) [(A – B)2 = A2 – 2AB + B2]
\(\quad\) (100 – 2)2 = 802 – 2(80)(2) + 22
\(\quad\quad\quad\quad\quad\quad\) = 6400 – 320 + 4
\(\quad\quad\quad\quad\quad\quad\) = 6404 – 320 = 6084
(iii) 1982
\(\quad\) = (200 – 2)2
\(\quad\) [(A – B)2 = A2 – 2AB + B2]
\(\quad\) (200 – 2)2 = 2002 – 2(200)(2) + 22
\(\quad\quad\quad\quad\quad\quad\) = 40000 – 800 + 4
\(\quad\quad\quad\quad\quad\quad\) = 40004 – 800 = 39204
(iv) 2142
\(\quad\) = (200 + 10 + 4)2
\(\quad\) [(A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2CA]
\(\quad\) (200 + 10 + 4)2 = 2002 + 102 + 42 + 2(200)(10) + 2(10)(4) + 2(4)(200)
\(\quad\quad\quad\quad\quad\quad\) = 40000 + 100 + 16 + 4000 + 80 + 1600
\(\quad\quad\quad\quad\quad\quad\) = 45796
(v) 11042
\(\quad\) = (1000 + 100 + 4)2
\(\quad\) [(A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2CA]
\(\quad\) (1000 + 100 + 4)2 = 10002 + 1002 + 42 + 2(1000)(100) + 2(100)(4) + 2(4)(1000)
\(\quad\quad\quad\quad\quad\quad\) = 1000000 + 10000 + 16 + 200000 + 800 + 8000
\(\quad\quad\quad\quad\quad\quad\) = 1218816
(vi) 11202
\(\quad\) = (1000 + 100 + 20)2
\(\quad\) [(A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2CA]
\(\quad\) (1000 + 100 + 20)2 = 10002 + 1002 + 202 + 2(1000)(100) + 2(100)(20) + 2(20)(1000)
\(\quad\quad\quad\quad\quad\quad\) = 1000000 + 10000 + 400 + 200000 + 4000 + 40000
\(\quad\quad\quad\quad\quad\quad\) = 1254400
Question 2:
Factor using suitable identities:
(i) 16y2 – 24y + 9 (ii) \(\frac{9}{4}\)s2 + 6st + 4t2
(iii) \(\frac{m^2}{9}\) + \(\frac{mk}{3}\) + \(\frac{k^2}{4}\) + 3nk + 2nm + 9n2 (iv) \(\frac{p^2}{16}\) – 2 + \(\frac{16}{p^2}\)
(v) 9a2 + 4b2 + c2 – 12ab + 6ac – 4bc
Question 3:
Expand the following using the identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (p + 3q + 7r)2 (ii) (3x – 2y + 4z)2
Question 4:
Is this an identity?
(a + b – c)2 + (a – b + c)2 + (a – b – c)2 = 2a2 + 2b2 + 2c2.
NCERT Solutions for Class 9 Ganita Manjari Chapter 4 Exploring Algebraic Identities बीजीय सर्वसमिकाओं का अन्वेषण
Exercise Set 4.4
Question 1:
Fill in the blanks to complete the following identities:
(i) s2 – 11s + 24 = (_______) (_______)
(ii) (______) (x + 1) = (3x2 – 4x – 7)
(iii) 10x2 – 11x – 6 = (2x – __) (__ + 2)
(iv) 6x2 +7x + 2 = (_____)(_____)
Question 2:
Select and use the identity that will help you to find the following products without multiplying directly:
(i) (41)2 (ii) (27)2 (iii) (23 x 17)
(iv) (135)2 (v) (97)2 (vi) (18 x 29)
(vii) (34 x 43) (viii) (205)2
Question 3:
Factor the following:
(i) 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc (ii) 16s2 + 25t2 – 40st
(iii) r2 – r – 42 (iv) 49g2 + 14gh + h2
(v) 64u2 + 121v2 + 4w2 – 176uv – 32uw + 44vw
NCERT Solutions for Class 9 Ganita Manjari Chapter 4 Exploring Algebraic Identities बीजीय सर्वसमिकाओं का अन्वेषण
Exercise Set 4.5
Question 1:
Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
NCERT Solutions for Class 9 Ganita Manjari Chapter 4 Exploring Algebraic Identities बीजीय सर्वसमिकाओं का अन्वेषण
End-of-Chapter Exercises
Question 1:
Use suitable identities to find the following products:
Question 2:
Find the values using suitable identities:
Question 3:
Factor the following algebraic expressions:
Question 4:
Simplify the following:
Question 5:
Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
Question 6:
Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
Question 7:
The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Question 8:
If a number plus its reciprocal equals \(\frac{10}{3}\) , find the number.
Question 9:
A rectangular pool has area 2x2 + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Question 10:
If both x – 2 and x – \(\frac{1}{2}\) are factors of px2 + 5x + r, show that p = r.
Question 11:
If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 –3abc = – 25.
Question 12:
By factoring the expression, check that n3 – n is always divisible by 6 for all natural numbers n. Give reasons.
Question 13:
Find the value of
(i) x3 + y3 – 12xy + 64, when x + y = – 4
(ii) x3 – 8y3 – 36xy – 216, when x = 2y + 6
NCERT Solutions for Class 9 Ganita Manjari Chapter 4 Exploring Algebraic Identities बीजीय सर्वसमिकाओं का अन्वेषण
