Class 8 RD Sharma Solutions - Chapter 3 Squares and Square Roots - Exercise 3.3 | Set 2

Exercise 3.3 Set 2 of RD Sharma's Class 8 Mathematics textbook focuses on squares and square roots. This section builds upon students' understanding of perfect squares and introduces methods for finding square roots of numbers that are not perfect squares. Students will learn to estimate square roots, use long division method for square root calculation, and solve related word problems.

Question 4. Find the squares of the following numbers:

(i) 425

Solution:

425 = 420 + 5 = (42 x 10) + 5

Here, n = 42

=> n(n + 1) = 42(42 + 1) = 42 x 43 = 1806

=> 425² = 180625

(ii) 575

Solution:

575 = 570 + 5 = (57 x 10) + 5

Here, n = 57

=> n(n + 1) = 57(57 + 1) = 57 x 58 = 3306

=> 575² = 330625

(iii) 405

Solution:

405 = 400 + 5 = (40 x 10) + 5

Here, n = 40

=> n(n + 1) = 40(40 + 1) = 1640

=> 405² = 164025

(iv) 205

Solution:

205 = 200 + 5 = (20 x 10) + 5

Here, n = 20

=> n(n + 1) = 20(20 + 1) = 420

=> 205² = 42025

(v) 95

Solution:

95 = 90 + 5 = (9 x 10) + 5

Here, n = 9

=> n(n + 1) = 9(9 + 1) = 90

=> 95² = 9025

(vi) 745

Solution:

745 = 740 + 5 = (74 x 10) + 5

Here, n = 74

=> n(n + 1) = 74(74 + 1) = 5550

=> 745² = 555025

(vii) 512

Solution:

512² = (250 + 12)1000 + (12)² 

= (262)1000 + 144

= 262000 + 144

= 262144

Hence, 512² = 262144

(viii) 995

Solution:

995 = 990 + 5 = (99 x 10) + 5

Here, n = 99

=>n(n + 1) = 99(99 + 1) = 9900

995² = 990025

Question 5. Find the square of the following numbers using the formula: (a + b)² = a² + 2ab + b²

(i) 405 

Solution:

405 = 400 + 5 

Here, a = 400, b = 5

Using the identity, (a + b)² = a² + 2ab + b²

405² = 400² + (2 x 400 x 5) + 5²

= 160000 + 4000 + 25

= 164025

Thus, 405² = 164025

(ii) 510

Solution:

510 = 500 + 10

Here, a = 500, b = 10 

Using the identity, (a + b)² = a² + 2ab + b²

510² = 500² + (2 x 500 x 10) + 10²

= 250000 + 10000 + 100

= 260100

Thus, 510² = 260100

(iii) 1001

Solution:

1001 = 1000 + 1

Here, a = 1000, b = 1

Using the identity, (a + b)² = a² + 2ab + b²

1001² = 1000² + (2 x 1000 x 1) + 1²

= 1000000 + 2000 + 1

= 1002001

Thus, 1001² = 1002001 

(iv) 209

Solution:

209 = 200 + 9

Here, a = 200, b = 9

Using the identity, (a + b)² = a² + 2ab + b²

209² = 200² + (2 x 200 x 9) + 9²

= 40000 + 3600 + 81

= 43681

Thus, 209² = 43681

(v) 605

Solution:

605 = 600 + 5

Here, a = 600, b = 5

Using the identity, (a + b)² = a² + 2ab + b²

605² = 600² + (2 x 600 x 5) + 5²

= 360000 + 6000 + 25

= 366025

Thus, 605² = 366025

Question 6. Find the square of the following numbers using the formula: (a - b)² = a² - 2ab + b²

(i) 395

Solution:

395 = 400 - 5

Here, a = 400, b = 5

Using the identity, (a - b)² = a² - 2ab + b²

395² = (400 - 5)² = 400² - (2 x 400 x 5) + 5²

= 160000 - 4000 + 25

= 156025

Thus, 395² = 156025

(ii) 995

Solution:

995 = 1000 - 5

Here, a = 1000, b = 5

Using the identity, (a - b)² = a² - 2ab + b²

995² = (1000 - 5)² = 1000² - (2 x 1000 x 5) + 5²

= 1000000 - 10000 + 25

= 990025

Thus, 995² = 990025

(iii) 495

Solution:

495 = 500 - 5

Here, a = 500, b = 5

Using the identity, (a - b)² = a² - 2ab + b²

495² = (500 - 5)² = 500² - (2 x 500 x 5) + 5²

= 250000 - 5000 + 25

= 245025

Thus, 495² = 245025

(iv) 498

Solution:

498 = 500 - 2

Here, a = 500, b = 2

Using the identity, (a - b)² = a² - 2ab + b²

498² = (500 - 2)² = 500² - (2 x 500 x 2) + 2²

= 250000 - 2000 + 4

= 248004

Thus, 498² = 248004

(v) 99

Solution:

99 = 100 - 1

Here, a = 100, b = 1

99² = (100 - 1)² = 100² - (2 x 100 x 1) + 1²

= 10000 - 200 + 1

= 9801

Using the identity, (a - b)² = a² - 2ab + b²

Thus, 99² = 9801

(vi) 999

Solution:

999 = 1000 - 1

Here, a = 1000, b = 1

Using the identity, (a - b)² = a² - 2ab + b²

999² = (1000 - 1)² = 1000² - (2 x 1000 x 1) + 1²

= 1000000 - 2000 + 1

= 998001 

Thus, 999² = 998001

(vii) 599

Solution:

599 = 600 - 1

Here, a = 600, b = 1

Using the identity, (a - b)² = a² - 2ab + b²

599² = (600 - 1)² = 600² - (2 x 600 x 1) + 1²

= 360000 - 1200 + 1

= 358801

Thus, 599² = 358801

Question 7. Find the squares of the following numbers by the visual method:

(i) 52

Solution:

52 = 50 + 2

Here, a = 50, b = 2

👁 Image

Thus, the area of square of side 52 = 2500 + 100 + 100 + 4

= 2704 

Thus, 52² = 2704

(ii) 95

Solution:

95 = 90 + 5

Here, a = 90, b = 5

👁 Image

Thus, the area of square of side of 95 = 8100 + 450 + 450 + 25

= 9025

Thus, 95² = 9025

(iii) 505

Solution:

505 = 500 + 5

Here, a = 500, b = 5

👁 Image

Thus, the area of square of side 505 = 250000 + 2500 + 2500 + 25

= 255025

Thus, 505² = 255025

(iv) 702

Solution:

702 = 700 + 2

Here, a = 700, b = 2

👁 Image

Thus, the area of square of side 702 = 490000 + 1400 + 1400 + 4

= 492804

Thus, 702² = 492804

(v) 99

Solution:

99 = 90 + 9

Here, a = 90, b = 9.

👁 Image

Thus, the area of square of side of 99 = 8100 + 810 + 810 + 81

= 9801 

Thus, 99² = 9801

Summary

This exercise set reinforces students' skills in working with squares and square roots. It covers various aspects including finding square roots using long division, estimating square roots, identifying perfect squares, and applying these concepts to solve real-world problems. The questions range from straightforward calculations to more complex word problems, helping students develop a comprehensive understanding of the topic.