Chapter 3 - Squares and Square Roots (Ex-3.3) - Class 8 Math RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

Step 5. Write the underlined digits at the bottom of each column to obtain the square of the given number.
In this case, we have:
25² = 625
Using multiplication:

This matches with the result obtained by the column method. 

(ii) Here, a = 3, b = 7
Step 1. Make 3 columns and write the values of a², 2 x a x b, and b² in these columns. 

QUESTION 3: FIND THE SQUARES OF THE FOLLOWING NUMBERS:
(I) 127
(II) 503
(III) 451
(IV) 862
(V) 265
ANSWER 3: We will use visual method as it is the most efficient method to solve this problem.
(i) We have:
127 = 120 + 7
Hence, let us draw a square having side 127 units. Let us split it into 120 units and 7 units. 

Hence, the square of 127 is 16129. 

(ii) We have:
503 = 500 + 3
Hence, let us draw a square having side 503 units. Let us split it into 500 units and 3 units. 

Hence, the square of 503 is 253009.

(iii) We have:
451 = 450 + 1
Hence, let us draw a square having side 451 units. Let us split it into 450 units and 1 units. 

Hence, the square of 451 is 203401. 

(iv) We have:
862 = 860 + 2
Hence, let us draw a square having side 862 units. Let us split it into 860 units and 2 units. 

Hence, the square of 862 is 743044. 

(v) We have:
265 = 260 + 5
Hence, let us draw a square having side 265 units. Let us split it into 260 units and 5 units. 

Hence, the square of 265 is 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
Answer 4: Notice that all numbers except the one in question (vii) has 5 as their respective unit digits. We know that the square of a number with the form n5 is a number ending with 25 and has the number n(n + 1) before 25. Notice that all numbers except the one in question (vii) has 5 as their respective unit digits. We know that the square of a number with the form n5 is a number ending with 25 and has the number n(n + 1) before 25. 

(i) Here, n = 42
∴ n(n + 1) = (42)(43) = 1806
∴ 425² = 180625
(ii) Here, n = 57
∴ n(n + 1) = (57)(58) = 3306
∴ 575² = 330625
(iii) Here n = 40
∴ n(n + 1) = (40)(41) = 1640
∴ 405² = 164025
(iv) Here n = 20
∴∴ n(n + 1) = (20)(21) = 420
∴ 205² =  42025
(v) Here n = 9
∴ n(n + 1) = (9)(10) = 90
∴ 95² = 9025
(vi) Here n = 74
∴ n(n + 1) = (74)(75) = 5550
∴ 745² = 555025
(vii) We know:
The square of a three-digit number of the form 5ab = (250 + ab)1000 + (ab)²
∴ 512² = (250+12)1000 + (12)² = 262000 + 144 = 262144
(viii) Here, n = 99
∴ n(n + 1) = (99)(100) = 9900
∴ 995² = 990025 

Question 5: Find the squares of the following numbers using the identity (a + b)² = a² + 2ab + b²:
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605
Answer 5: (i) On decomposing:
405 = 400 + 5
Here, a = 400 and b = 5
Using the identity (a + b)² = a² + 2ab + b²:
405² = (400 + 5)² = 400² + 2(400)(5) + 5² = 160000 + 4000 + 25 = 164025
(ii) On decomposing:
510 = 500 + 10
Here, a = 500 and b = 10
Using the identity (a + b)² = a² + 2ab + b²:
510² = (500 + 10)² = 500² + 2(500)(10) + 10² = 250000 + 10000 + 100 = 260100
(iii) On decomposing:
1001 = 1000 + 1
Here, a = 1000 and b = 1
Using the identity (a + b)² = a² + 2ab + b²:
1001² = (1000 + 1)² = 1000² + 2(1000)(1) + 1² = 1000000 + 2000 + 1 = 1002001
(iv) On decomposing:
209 = 200 + 9
Here, a = 200 and b = 9
Using the identity (a + b)² = a² + 2ab + b²:
209² = (200 + 9)² = 200² + 2(200)(9) + 9² = 40000 + 3600 + 81 = 43681
(v) On decomposing:
605 = 600 + 5
Here, a = 600 and b = 5
Using the identity (a + b)² = a² + 2ab + b²:
605² = (600 + 5)² = 600² + 2(600)(5) + 5² = 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
Answer 6: (i) Decomposing: 395 = 400 − 5
Here, a = 400 and b = 5
Using the identity (a − b)² = a² − 2ab + b²:
395² = (400 − 5)² = 400² − 2(400)(5) + 5² = 160000 − 4000 + 25 = 156025
(ii) Decomposing: 995 = 1000 − 5
Here, a = 1000 and b = 5
Using the identity (a − b)² = a² − 2ab + b²:
995² = (1000 − 5)² = 1000² − 2(1000)(5) + 5² = 1000000 − 10000 + 25 = 990025
(iii) Decomposing: 495 = 500 − 5
Here, a = 500 and b = 5
Using the identity (a − b)² = a² − 2ab + b²:
495² = (500 − 5)² = 500² − 2(500)(5) + 5² = 250000 − 5000 + 25 = 245025
(iv) Decomposing: 498 = 500 − 2
Here, a = 500 and b = 2
Using the identity (a − b)² = a² − 2ab + b²:  
498² = (500 − 2)² = 500² − 2(500)(2) + 2² = 250000 − 2000 + 4 = 248004
(v) Decomposing: 99 = 100 − 1
Here, a = 100 and b = 1
Using the identity (a − b)² = a² − 2ab + b²:
99² = (100 − 1)² = 100² − 2(100)(1) + 1² = 10000 − 200 + 1 = 9801
(vi) Decomposing: 999 = 1000 - 1
Here, a = 1000 and b = 1
Using the identity (a − b)² = a² − 2ab + b²:
999² = (1000 − 1)² = 1000² − 2(1000)(1) + 1² = 1000000 − 2000 + 1 = 998001
(vii) Decomposing: 599 = 600 − 1
Here, a = 600 and b = 1
Using the identity (a − b)² = a² − 2ab + b²:
599² = (600 − 1)² = 600² − 2(600)(1) + 1² = 360000 − 1200 + 1 = 358801 

Question 7: Find the squares of the following numbers by visual method:
(i) 52
(ii) 95
(iii) 505
(iv) 702
(v) 99
Answer 7: (i) We have:
52 = 50 + 2
Let us draw a square having side 52 units. Let us split it into 50 units and 2 units. 

The sum of the areas of these four parts is the square of 52. Thus, the square of 52 is 2704.
(ii) We have:
95 = 90 + 5
Let us draw a square having side 95 units. Let us split it into 90 units and 5 units. 

The sum of the areas of these four parts is the square of 95. Thus, the square of 95 is 9025.
(iii) We have:
505 = 500 + 5
Let us draw a square having side 505 units. Let us split it into 500 units and 5 units. 

The sum of the areas of these four parts is the square of 505. Thus, the square of 505 is 255025. 

(iv) We have:
702 = 700 + 2
Let us draw a square having side 702 units. Let us split it into 700 units and 2 units. 

The sum of the areas of these four parts is the square of 702. Thus, the square of 702 is 492804. 

(v) We have:
99 = 90 + 9
Let us draw a square having side 99 units. Let us split it into 90 units and 9 units.

The sum of the areas of these four parts is the square of 99. Thus, the square of 99 is 9801.