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

### Chapter 3 Squares and Square Roots – Exercise 3.3 | Set 1

### 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

=> 425

^{2}= 180625

### (ii) 575

**Solution:**

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

Here, n = 57

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

=> 575

^{2}= 330625

### (iii) 405

**Solution:**

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

Here, n = 40

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

=> 405

^{2}= 164025

### (iv) 205

**Solution:**

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

Here, n = 20

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

=> 205

^{2}= 42025

### (v) 95

**Solution:**

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

Here, n = 9

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

=> 95

^{2}= 9025

### (vi) 745

**Solution:**

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

Here, n = 74

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

=> 745

^{2}= 555025

### (vii) 512

**Solution:**

512Â² = (250 + 12)1000 + (12)Â²

= (262)1000 + 144

= 262000 + 144

= 262144

Hence, 512

^{2}= 262144

### (viii) 995

**Solution:**

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

Here, n = 99

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

995

^{2}= 990025

### Question 5. Find the square of the following numbers using the formula: (a + b)^{2} = a^{2 }+ 2ab + b^{2}

### (i) 405

**Solution:**

405 = 400 + 5

Here, a = 400, b = 5

Using the identity, (a + b)

^{2}= a^{2}+ 2ab + b^{2}405

^{2}= 400^{2}+ (2 x 400 x 5) + 5^{2}= 160000 + 4000 + 25

= 164025

Thus, 405

^{2}= 164025

### (ii) 510

**Solution:**

510 = 500 + 10

Here, a = 500, b = 10

Using the identity, (a + b)

^{2}= a^{2}+ 2ab + b^{2}510

^{2}= 500^{2}+ (2 x 500 x 10) + 10^{2}= 250000 + 10000 + 100

= 260100

Thus, 510

^{2}= 260100

### (iii) 1001

**Solution:**

1001 = 1000 + 1

Here, a = 1000, b = 1

Using the identity, (a + b)

^{2}= a^{2}+ 2ab + b^{2}1001

^{2}= 1000^{2}+ (2 x 1000 x 1) + 1^{2}= 1000000 + 2000 + 1

= 1002001

Thus, 1001

^{2}= 1002001

### (iv) 209

**Solution:**

209 = 200 + 9

Here, a = 200, b = 9

Using the identity, (a + b)

^{2}= a^{2}+ 2ab + b^{2}209

^{2}= 200^{2}+ (2 x 200 x 9) + 9^{2}= 40000 + 3600 + 81

= 43681

Thus, 209

^{2}= 43681

### (v) 605

**Solution:**

605 = 600 + 5

Here, a = 600, b = 5

Using the identity, (a + b)

^{2}= a^{2}+ 2ab + b^{2}605

^{2}= 600^{2}+ (2 x 600 x 5) + 5^{2}= 360000 + 6000 + 25

= 366025

Thus, 605

^{2}= 366025

### Question 6. Find the square of the following numbers using the formula: (a – b)Â² = aÂ² – 2ab + bÂ²

### (i) 395

**Solution:**

395 = 400 – 5

Here, a = 400, b = 5

Using the identity, (a – b)

^{2}= a^{2}– 2ab + b^{2}395

^{2}= (400 – 5)^{2}= 400^{2}– (2 x 400 x 5) + 5^{2}= 160000 – 4000 + 25

= 156025

Thus, 395

^{2}= 156025

### (ii) 995

**Solution:**

995 = 1000 – 5

Here, a = 1000, b = 5

Using the identity, (a – b)

^{2}= a^{2}– 2ab + b^{2}995

^{2}= (1000 – 5)^{2}= 1000^{2}– (2 x 1000 x 5) + 5^{2}= 1000000 – 10000 + 25

= 990025

Thus, 995

^{2}= 990025

### (iii) 495

**Solution:**

495 = 500 – 5

Here, a = 500, b = 5

Using the identity, (a – b)

^{2}= a^{2}– 2ab + b^{2}495

^{2}= (500 – 5)^{2}= 500^{2}– (2 x 500 x 5) + 5^{2}= 250000 – 5000 + 25

= 245025

Thus, 495

^{2}= 245025

### (iv) 498

**Solution:**

498 = 500 – 2

Here, a = 500, b = 2

Using the identity, (a – b)

^{2}= a^{2}– 2ab + b^{2}498

^{2}= (500 – 2)^{2}= 500^{2}– (2 x 500 x 2) + 2^{2}= 250000 – 2000 + 4

= 248004

Thus, 498

^{2}= 248004

### (v) 99

**Solution:**

99 = 100 – 1

Here, a = 100, b = 1

99

^{2}= (100 – 1)^{2}= 100^{2}– (2 x 100 x 1) + 1^{2}= 10000 – 200 + 1

= 9801

Using the identity, (a – b)

^{2}= a^{2}– 2ab + b^{2}Thus, 99

^{2}= 9801

### (vi) 999

**Solution:**

999 = 1000 – 1

Here, a = 1000, b = 1

Using the identity, (a – b)

^{2}= a^{2}– 2ab + b^{2}999

^{2}= (1000 – 1)^{2}= 1000^{2}– (2 x 1000 x 1) + 1^{2}= 1000000 – 2000 + 1

= 998001

Thus, 999

^{2}= 998001

### (vii) 599

**Solution:**

599 = 600 – 1

Here, a = 600, b = 1

Using the identity, (a – b)

^{2}= a^{2}– 2ab + b^{2}599

^{2}= (600 – 1)^{2}= 600^{2}– (2 x 600 x 1) + 1^{2}= 360000 – 1200 + 1

= 358801

Thus, 599

^{2}= 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

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

= 2704

Thus, 52

^{2}= 2704

### (ii) 95

**Solution:**

95 = 90 + 5

Here, a = 90, b = 5

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

= 9025

Thus, 95

^{2}= 9025

### (iii) 505

**Solution:**

505 = 500 + 5

Here, a = 500, b = 5

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

= 255025

Thus, 505

^{2}= 255025

### (iv) 702

**Solution:**

702 = 700 + 2

Here, a = 700, b = 2

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

= 492804

Thus, 702

^{2}= 492804

### (v) 99

**Solution:**

99 = 90 + 9

Here, a = 90, b = 9.

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

= 9801

Thus, 99

^{2}= 9801

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