RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3

Other Exercises

Question 1.
Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication :
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96
Solution:
(i) (25)2
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 1
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 2

Question 2.
Find the squares of the following numbers using diagonal method :
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171
Solution:
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 3
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 4
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 5

Question 3.
Find the squares of the following numbers :
(i) 127
(ii) 503
(iii) 451
(iv) 862
(v) 265
Solution:
(i) (127)2 = (120 + 7)2
{(a + b)2 = a2 + lab + b2}
= (120)2 + 2 x 120 x 7 + (7)2
= 14400+ 1680 + 49 = 16129

(ii) (503)2 = (500 + 3)2
{(a + b)2 = a2 + lab + b1}
= (500)2 + 2 x 500 x 3 + (3)2
= 250000 + 3000 + 9 = 353009

(iii) (451)2 = (400 + 51)2
{(a + b)2 = a2 + lab + b2}
= (400)2 + 2 x 400 x 51 + (5l)2
= 160000 + 40800 + 2601 = 203401

(iv) (451)2 = (800 + 62)2
{(a + b)2 = a2 + lab + b2}
= (800)2 + 2 x 800 x 62 + (62)2
= 640000 + 99200 + 3844 = 743044

(v) (265)2
{(a + b)2 = a2 + 2ab + b2}
(200 + 65)2 = (200)2 + 2 x 200 x 65 + (65)2
= 40000 + 26000 + 4225 = 70225

Question 4.
Find the squares of the following numbers
(i) 425
(ii) 575
(iii) 405
(iv) 205
(v) 95
(vi) 745
(vii) 512
(viii) 995
Solution:
(i) (425)2
Here n = 42
∴ n (n + 1) = 42 (42 + 1) = 42 x 43 = 1806
∴ (425)2 = 180625

(ii) (575)2
Here n = 57
∴ n (n + 1) = 57 (57 + 1) = 57 x 58 = 3306
∴ (575)2 = 330625

(iii) (405)2
Here n = 40
∴ n (n + 1) = 40 (40 + 1) -40 x 41 = 1640
∴ (405)2 = 164025

(iv) (205)2
Here n = 20
∴ n (n + 1) = 20 (20 + 1) = 20 x 21 = 420
∴ (205)2 = 42025

(v) (95)2
Here n = 9
∴ n (n + 1) = 9 (9 + 1) = 9 x 10 = 90
∴ (95)2 = 9025

(vi) (745)2
Here n = 74
∴ n (n + 1) = 74 (74 + 1) = 74 x 75 = 5550
∴ (745)2 = 555025

(vii) (512)2
Here a = 1, b = 2
∴ (5ab)2 = (250 + ab) x 1000 + (ab)2
∴ (512)2 = (250 + 12) x 1000 + (12)2
= 262 x 1000 + 144
= 262000 + 144 = 262144

(viii) (995)2
Here n = 99
∴ n (n + 1) = 99 (99 + 1) = 99 x 100 = 9900
∴ (995)2 = 990025

Question 5.
Find the squares of the following numbers using the identity (a + b)1 = a2 + lab + b1
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605
Solution:
a + b)2 = a2 + lab + b2

(i) (405)2 = (400 + 5)2
= (400)2 + 2 x 400 x 5 + (5)2
= 160000 + 4000 + 25 = 164025

(ii) (510)2 = (500 + 10)2
= (500)2 + 2 x 500 x 10 x (10)2
= 250000 + 10000 + 100
= 260100

(iii) (1001)2 = (1000+1)2
= (1000)2 + 2 X 1000 x 1 + (1)
= 1000000 + 2000 + 1
=1002001

(iv) (209)2 = (200 + 9)2
= (200)2 + 2 x 200 x 9 x (9)2
= 40000 + 3600 +81
= 43681

(v) (605)2 = (600 + 5)2
= (600)2 + 2 x 600 x 5 +(5)2
= 360000 + 6000       25
=366025

Question 6.
Find the squares of the following numbers using the identity (a – b)2 = a2 – 2ab + b2 :
(i) 395
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599
Solution:
a – b)2 = a2 – lab + b2

(i) (395)2 = (400 – 5)2
= (400)2 – 2 x 400 x 5 + (5)2
= 160000-4000 + 25
= 160025-4000
= 156025

(ii) (995)2 = (1000 – 5)2
= (1000)2 – 2 x 1000 x 5 + (5)2
= 1000000- 10000 + 25
= 1000025- 10000
= 990025

(iii) (495)2 = (500 – 5)2
= (500)2 – 2 x 500 x 5 + (5)2
= 250000 – 5000 + 25
= 250025 – 5000
= 245025

(iv) (498)2 = (500 – 2)2
= (500)2 – 2 x 500 x 2 + (2)2
= 250000 – 2000 + 4
= 250004 – 2000
= 248004

(v) (99)2 = (100 – l)2
= (100)2 – 2 x 100 x 1 + (1)2
= 10000 – 200 + 1
= 10001 – 200
= 9801

(vi) (999)2 = (1000- l)2
= (1000)2 – 2 x 1000 x 1+ (1)2
= 1000000-2000+1
= 10000001-2000=998001

(vii) (599)2 = (600 – 1)2
= (600)2 -2 x 600 X 1+ (1)2
= 360000 -1200+1
= 360001 – 1200 = 358801

Question 7.
Find the squares of the following numbers by visual method :
(i) 52
(ii) 95
(iii) 505
(iv) 702
(v) 99
Solution:
(a + b)2 = a2 – ab + ab + b2
(i) (52)2 = (50 + 2)2
= 2500 + 100 + 100 + 4
= 2704
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 6
(ii) (95)2 = (90 + 5)2
= 8100 + 450 + 450 + 25
= 9025
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 7
(iii) (505)2 = (500 + 5)2
= 250000 + 2500 + 2500 + 25
= 255025
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 8
(iv) (702)2 = (700 + 2)2
= 490000 + 1400+ 1400 + 4
= 492804
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 9
(v) (99)2 = (90 + 9)2
= 8100 + 810 + 810 + 81
= 9801
RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3 10

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