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Joy of Mathematics Class 8 Solutions Chapter 12 Identities
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 12 Identities. Here students can easily find step by step solutions of all the problems for Identities. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 12
Identities Exercise 12 Solution :
Question no – (1)
Solution :
(a) x + 2y
= we know that (x + 2y)2 = x2 + 2y + y2
=> (x + 2y)2 = x2 + 2.x.2y + (2y)2
= x2 + 4xy + 4y2
Thus the required square x2 + 4xy + 4y2
(b) 3x + 4y
=> (3x + 4y)2 = (3x)2 + 2.3x.4y + (4y)2
= 9x2 + 24xy + 16y2
∴ Thus the required square 9x2 + 24xy + 16y2
(c) 1 2/3 x + 3/5 y
= 5/3 x + 3/5 y
=> (5/3 x + 3/5 y)2 = (5/3 x)2 + 2. 5/3 x. 3/5 y + (3/5 y)
= 25x2/9 + 2xy + 9y2/25
∴ Thus the required square 25x2/9 + 2xy + 9/25y2
(d) 2x/3y + 3y/2x
=> (2x/3y + 3y/2x)2 = (2x/3y)2 + 2.2x/3y.3y/2x + (3y/2x)2
= 4x2/9y2 + 2 + 9y2/4x2
∴ Thus the required square 4×2/9y2 + 2 + 9y2/4×2
(e) √2x + 1/√2y
=> (√2x + 1/√2y)2 = (√2x)2 + 2. √2.x.1/√2y + (1/√2y)2
= 2×2 + 2xy + 1/2y2
∴ Thus the required square 2x2 + 2xy + 1/2y2
(f) 2/x + 3/y
=> (2/x + 3/y)2 = (2/x)2 + 2.2/x.3/y + (3/y)2
= 4/x2 + 12/xy + 9/y2
∴ Thus the required square 4/x2 + 12/xy + 9/y2
Question no – (2)
Solution :
(a) (3a – 7)2 = (3a)2 – 2.3a.7 + (7)2
we know that (x – y)2 = x2 – 2xy + y2
= 9a2 – 42a + 49
(b) (2a – 3b)2 = (2a)2 – 2.2a + 3b + (3b)2
= 4a2 – 12ab + 9b2
(c) (2x – 1/x)2 = (2x)2 – 2.2x.1/x + (1/x)2
= 4x2 – 4 + 1/x2
(d) (x/2 – y/3)2 = (x/2)2 – 2.x/2.y/3 + (y/3)2
= x2/4 – xy/3 + y2/9
(e) (0.1x – 0.9)2 = (0.1x)2 – 2 × 0.1x × 0.9 + (0.9)2
= 0.01x2 – 0.18x + 0.81
(f) (√2p – 1/√2p)2 = (√2p)2 – 2. √2p.1/√2p + (1/√2p)2
= 2p2 – 2 + 1/2p2
Question no – (3)
Solution :
(a) (2/3a + 1/3b)2
= we know that (a + b)2 = a2 + 2ab + b2
= (2/3a)2 + 2.2/3a.1/3b + (1/3b)2
= 4/9a2 + 4/9ab + b2/9
(b) (3/4a + 4/5b)2
= (3/4a)2 + 2.3/4a.4/5.b + (4/5b)2 [∵ (a + b)2 = a2 +2ab + b2]
= 9/12a2 + 6/5ab + 16/5b2
(c) (a – 1/a)2
= a2 – 2.a.1/a + (1/a)2 [∵ (a – b)2 = a2 – 2ab + b2]
= a2 – 2 + 1/a2
(d) 2(a + 1/a)2
= 2{(a)2 + 2.a.1/a + (1/a)2} [∵ (a + b)2 = a2 + 2ab + b2]
= 2 (a2 + 2 + 1/a2)
= 2a2 + 4 + 2/a2
(e) (2a – 1/2a)2
= (2a)2 – 2.2a.1/2a + (1/2a)2 [∵ (a – b)2 = a2 – 2ab + b2]
= 4a2 – 2 + 1/4a2
(f) 4 (2/3a – 3/4b)2
= 4 {(2/3a)2 – 2.2/3a.3/4b + (3/4b)} [∵ (a – b)2 = a2 – 2ab + b2]
= 4 {4a2/9 – ab + 9/16 b2}
= 16a2/9 – 4ab + 36/16 b2
Question no – (4)
Solution :
(a) (x2 – 3/2) (x2 + 3/2)
= (x2)2 – (3/2)2 [∵ a2 – b2 = (a + b) (a – b)]
= x4 – 9/4
∴ Thus the required result x4 – 9/4
(b) (1 – 1/2x) (1 + 1/2x)
= {(1)2 – (1/2x)2} [∵ a2 – b2 = (a + b) (a – b)]
= 12 – 1/4x2
= 1- 1/4x2
∴ Thus the required result 1 – 1/4x2
(c) (x + 2/y) (x – 2/y)
= {(x)2 – (2/y)2} [∵ a2 – b2 = (a + b) (a – b)]
= x2 – 4/y2
∴ Thus the required result x2 – 4/y2
(d) (1.5/x + 3/y2) (1.5/x – 3/y2)
= {(1.5/x)2 – (3/y2)2} [∵ a2 – b2 = (a + b) (a – b)]
= 3/x2 – 9/y4
∴ Thus the required result 3/x2 – 9/y4
(e) (4x2 – 2y4) (4x2 + 2y4)
= {(4x2)2 – (2y4)2} [∵ a2 – b2 = (a + b) (a – b)]
= 16x4 – 4y8
∴ Thus the required result 16×4 – 4y8
(f) (2 3/5x2 + 5/2y2) (2 3/5 x2 – 5/2 y2)
= (13/5 x2 + 5/2 y2) (13/5 x2 – 5/2 y2)
= (13/5 x2)2 – (5/2 y2)2 [∵ a2 – b2 = (a +b) (a – b)]
= 169/25 x4 – 25/4 y4
∴ Thus the required result 169/25 x4 – 25/4 y4
Question no – (5)
Solution :
Here, A = x2 + 4, B = y2 – 4
(a) A2
= (x2 + 4)2
= (x2)2 + 2.x2.4 + (4)2 [∵ (a + b)2 = a2 + 2ab + b2]
= x4 + 8x2 + 16
∴ Thus the value of A2 = x4 + 8x2 + 16
(b) B2
= (y2 – 4)2
= (y2)2 – 2.y2.4 + (4)2 [∵ (a – b)2 = a2 – 2ab + b2]
= y4 – 8y2 + 16
∴ Thus the value of B2 = y4 – 8y2 + 16
(c) (A +B)2
= (x2 + 4 + y2 – 4)2 [Putting the value of A and B]
= (x2 + y2)2
= (x2)2 + 2x2y2 + (y2)2 [∵ (x + y)2 = x2 + 2xy + y2]
= x4 + 2x2y2 + y4
∴ Thus the value of (A + B)2 = x4 + 2x2y2 + y2
Question no – (6)
Solution :
(a) (106)2 = (100 + 6)2
= (100)2 + 2.100.6 + (6)2 {(a + b)2 = a2 + 2ab + b2}
= 10000 + 1200 + 36
= 11236
∴ (106)2 = 11236
(b) (999)2 = (1000 – 1)2
= (1000)2 – 2 × 1000.1 + (1)2
= 1000000 – 2000 + 1
= 998001
∴ (999)2 = 998001
(c) (103)2 = (100 + 3)2
= (100)2 + 2.100.3 + (3)2
= 10000 + 600 + 9 [∵ (a + b)2 = a2 + 2ab + b2]
= 1609
∴ (103)2 = 1609
(d) (298)2
= (300 – 2)2
= (300)2 – 2.300.2 + (2)2 [∵ (a – b)2 = a2 – 2ab + b2]
= 90000 – 1200 + 4
= 88804
∴ (298)2 = 88804
(e) (58)2
= (60 – 2)2
= (60)2 – 2.60.2 + (2)2
= 3600 – 240 + 4
= 3364
(f) (151)2
= (150 + 1)2
= (150)2 + 2.150.(1)2 + (1)2
= 22500 + 300 + 1
= 22801
∴ (151)2 = 22801
Question no – (7)
Solution :
(a) 157 × 143
= (150 + 7) (150 – 7)
= (150)2 – (7)2 [∵ (a + b) (a – b) = a2 – b2]
= 22500 – 49
= 22,451
∴ (157 × 143) = 22,451
(b) 96 × 104
= (100 – 4) (100 + 4) [∵ (a + b) (a – b) = a2 – b2]
= (100)2 – (4)2
= 10000 – 16
= 9984
∴ (96 × 104) = 9984
(c) 1005 × 995
= (1000 + 5) (1000 – 5)
= (1000)2 – (5)2 [∵ (a + b) (a – b) = a2 – b2]
= 1000000 – 25
= 999975
∴ (1005 × 995) = 999975
(d) (3.1)2
= (3 + 0.1)2
= (3)2 + 2 × 3 × (0.1) + (0.1)2 [∵ (a + b)2 = a2 + 2ab + b2]
= 9 + 0.6 + 0.01
= 9.61
∴ (3.1)2 = 9.61
(e) (3.1)2 – (2.9)2
= (3.1 + 2.9) (3.1 – 2.9) [∵ a2 – b2 = (a + b) (a – b)]
= 6 × 0.2
= 1.2
(f) 2.07 × 1.93
= (2 + 0.7) (2 – 0.7)
= (2)2 – (0.7)2 [∵ (a + b) (a – b) = a2 – b2]
= 4 – 0.49
= 3.51
∴ 2.07 × 1.93 = 3.51
Question no – (8)
Solution :
(a) 8x = 372 – 292
=> 8x = (37 + 29) (37 – 29) [∵ a2 – b2 = (a + b) (a – b)]
=> 8x = 66 × 8
=> x = 66 × 8/8
=> x = 6
∴ The value of x = 6
(b) 13x = 852 – 84
=> 13x = (85 + 84) (85 – 84)
=> 13x = (169 × 1)
=> x = 169 × 1/13
=> x = 13
∴ The value of x = 13
(c) xyz = (3y + z)2 – (3y – z)2
=> xyz = (3y)2 + 2.3y.z + z2 – {(3y)2 – 2.3y.z + z2}
=> xyz = 9y2 + 6yz + z2 – 9y2 + 6yz – z2
= xyz = 12yz
=> x = 12yz/yz
=> x = 12
∴ The value of x = 12
(d) ax = b [(2a + 1/b)2 – (2a – 1/b)2]
=> ax = b [(2a)2 + 2.2a.1/b + 1/b2 – (2a)2 + 2.2a.1/b – (1/b)2]
=> ax = b [4a2 + 4a/b + 1/b2 – 4a2 + 4a/b – 1/b2]
=> ax = b [8a/b]
=> ax = 8a
=> x = 8a/a = 8
∴ The value of x = 8
Question no – (9)
Solution :
(a) 6.25 × 6.25 – 1.75 × 1.75/4.5
= (6.25)2 – (1.75)2/4.5
= (6.25 + 1.75) (6.25 – 1.75)/4 [∵ a2 – b2 = (a + b) (a – b)]
= 8 × 4.5/4
= 9
(b) 3.7 × 3.7 – 1.3 × 1.3/5
= (3.7)2 – (1.3)2/5
= (3.7 + 1.3) (3.7 – 1.3)/5 [∵ a2 – b2 = (a + b) (a – b)]
= 5 × 2.4/5
= 2.4
(c) 629 × 629 – 371 × 371/1000
= (629)2 – (371)2/1000
= (629 + 371) (629 – 371)/1000 [∵ a2 – b2 = (a + b) (a – b)]
= 1000 × 258/1000
= 258
(d) 1.25 × 1.25 – 0.75 × 0.75/0.5
= (1.25)2 – (0.75)2/0.5
= (1.25 + 0.75) (1.25 – 0.75)/0.5 [∵ a2 – b2 = (a + b) (a – b)]
= 2 × 0.5/0.5
= 2
Question no – (10)
Solution :
(a) 16m2 – 4mn + 1/4n2
= (4m)2 – 2.5m.1/2m + (1/2n)2
= (4m – 1/2n)2 [∵ (a – b)2 = a2 – 2ab + b2]
(b) 81p2 + 6 + 1/9p2
= (9p)2 + 2.9p.1/3p + (1/3p)2
= (9p + 1/3p)2 [∵ (a + b)2 = a2 + 2ab + b2]
(c) 9b2 + 24bc + 16c2
= (3b)2 + 2.3b.4c + (4c)2
= (3b + 4c)2 [∵ (a + b)2 = a2 + 2b + b2]
(d) 25x2 – 70xy + 49y2
= (5x)2 – 2.5x.7y + (7y)2
= (5x – 7y)2 [∵ (a – b)2 = a2 – 2ab + b2]
Question no – (11)
Solution :
x + 1/x = 3
Then, x2 + 1/x2 = ?
=> Here, x + 1/x = 3
Both side square
(x + 1/x)2 = (3)2
=> x2 + 2.x.1/x + (1/x)2 = 9 [∵ (a + b)2 = a2 + 2ab + b2]
=> x2 + 1/x2 = 9 – 2
=> x2 + 1/x2 = 7
∴ Thus the value of x2 + 1/x2 = 7
Question no – (12)
Solution :
Here, x + 1/x = 2
=> (x + 1/x)2 = (2)2 [Both side square]
=> x2 + 2.x.1/x + 1/x2 = 4
=> x2 + 1/x2 = 4 – 2
=> x2 + 1/x2 = 2
∴ Thus the value of x2 + 1/x2 = 2
Now,
x2 + 1/x2 = 2
=> (x2 + 1/x2)2 = (2)2 (Again both side square)
=> x4 + 2.x2.1/x2 + 1/x4 = 4
=> x4 + 1/x4 = 4 – 2
=> x4 + 1/x4 = 2
∴ Thus the value of x4 + 1/x4 = 2
Question no – (13)
Solution :
Here,
(x – 1/x)2 = 22 [Both side square]
=> x2 – 2.x.1/x + (1/x)2 = 4
=> x2 + 1/x2 = 4 + 2
=> x2 + 1/x2 = 6
=> (x2 + 1/x2)2 = (6)2 (Again both side square)
=> x4 – 2.x2.1/x2 + 1/x4 = 36
=> x4 + 1/x4 = 36 – 2
=> x4 + 1/x4 = 34
∴ Thus the value of x4 + 1/x4 = 34
Question no – (14)
Solution :
Here,
3y – 1/3y = 3
(a) (3y – 1/3y)2 = (3)2 (Both side square)
=> (3y)2 – 2.3y.1/3y + (1/3y)2 = 9
=> 9y2 + 1/9y2 = 9 – 2
=> 9y2 + 1/9y2 = 7
(b) (9y2 + 1/9y2)2 = (7)2
=> (9y2)2 + 2.9y2.1/9y2 + (1/9y2)2 = 49
=> 81y4 + 2 + 1/81y4 = 49
=> 81y4 + 1/81y4 = 49 – 2
=> 81y4 + 1/81y4 = 47
∴ The value of 81y4 + 1/81y4 = 47
Question no – (15)
Solution :
x + 2y = 9
=> (x + 2y)2 = (9)2 (Both side square)
=> x2 + 2.x.2y + (2y)2 = 81
=> x2 + 4xy + 4y2 = 81
=> x2 + 4.7 + 4y2 = 81 [∵ xy = 7]
=> x2 + 28 + 4y2 = 81
=> x2 + 4y2 = 81 – 28
=> x2 + 4y2 = 53
∴ Thus the value of x2 + 4y2 = 53
Question no – (16)
Solution :
25x2 – 70xy + 49y2
= (5x)2 – 2.5x.7y + (7y)2
= (5x – 7y)2 [∵ (a – b)2 = a2 – 2ab + b2]
= (5 × 15 – 7 × 15)2 [∵ x = 15, y = 5]
= (75 – 35)2
= (40)2 = 1600
∴ The value of 25x2 – 70xy + 49y2 = 1600
Question no – (17)
Solution :
x – y = 16
=> (x – y)2 = (16)2 (Both side square)
=> x2 – 2xy + y2 = 256
=> x2 + y2 – 2xy = 256
=> 400 – 2xy = 256
=> 400 – 256 = 2xy
= 2xy = 144
=> xy = 144/2 = 72
∴ The value of xy = 72
Question no – (18)
Solution :
We know,
(x + y)2 = x2 + 2xy + y2
= x2 + y2 + 2xy
= 74 + 2 × 35 [∵ x2 + y2 = 74, xy = 35]
=> (x + y)2 = 74 + 70
=> (x + y)2 = 144
=> (x + y) = √144
=> (x + y) = √12 × 12
=> (x + y) = 12
∴ The value of x+ y = 12
Question no – (19)
Solution :
Let, the side of the square = a cm
∴ Area of the square = a2 cm2
According to the question,
a2 = 9x2 + 24xy + 16y2
=> a2 = (3x)2 + 2.3x.4y + (4y)2
=> a2 = (3x + 4y)2
=> a = 3x + 4y
∴ Thus the side of the square (3x + 4y)cm
Question no – (20)
Solution :
Let, the radius of the circle = r cm
∴ Area of the circle = πr2 cm2
According to the question,
πr2 = 9πx2 – 12πx + 4π
=> πr2 = π(9x2 – 12x + 4)
=> r2 = π/π (9x2 – 12x + 4)
=> r2 = (3x)2 – 2.3x.2 + (2)2
=> r2 = (3x – 2)2 [∵ (a – b)2 = a2 – 2ab + b2]
=> r = (3x – 2)
∴ Thus the radius of the circle
= 3x – 2
= 3 × 4 – 2 (x = 4 given)
= 12 – 2
= 10
∴ Therefore the radius of the circle 10cm.
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