# RS Aggarwal Class 8 Math Third Chapter Squares and Square Roots Exercise 3D Solution

## EXERCISE 3D

Find the square root of each of the following numbers by using the method of prime fraction:

(1) 225

Solution: By prime factorization,

225 = 3 × 3 × 5 × 5

∴ √225 = (3 × 5) = 15.

(2) 441 = 3 × 3 × 7 × 7

∴ √441 = (3 × 7) = 21.

(3) 729 = 3 × 3 × 3 × 3 × 3 × 3

∴ √729 = (3 × 3 × 3) = 27.

(4) 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

∴ √1296 = (2 × 2 × 3 × 3) = 36.

(5) 2025 = 3 × 3 × 3 × 3 × 5 × 5

∴ √2025 = (3 × 3 × 3 × 5) = 45.

(6) 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ √4096 = (2 × 2 × 2 × 2 × 2 × 2) = 64.

(7) 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

∴ √7056 = (2 × 2 × 3 × 7) = 84.

(8) 8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

∴ √8100 = (2 × 3 × 3 × 5) = 90.

(9) 9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

∴ √9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96.

(10) 11025 = 3 × 3 × 5 × 5 × 7 × 7

∴ √11025 = (3 × 5 × 7) = 105.

(11) 15876 = 2 × 2 × 3 × 3 × 3 × 3 × 7 × 7

∴ √15876 = (2 × 3 × 3 × 7) = 126.

(12) 17424 = 2 × 2 × 2 × 2 × 3 × 3 × 11 ×11

∴ √17424 = (2 × 2 × 3 × 11) = 132.

(13) Find the smallest number by which 252 must be multiplied to get a perfect square. Also, find the square root the perfect square so obtained.

Solution: By prime factorization, we get

252 = 2 × 2 × 3 × 3 × 7

So, the given number should be multiplied by 7to make the product a perfect square.

New number = 252 × 7 = 1764

∴ 1764 = 2 × 2 × 3 × 3 × 7 × 7

√1764 = 2 × 3 × 7 = 42

(14) Find the smallest number by which 2925 must be divided to obtain a perfect square. Also find the square root of the perfect square so obtained.

Solution: By prime factorization, we get

2925 = 3 × 3 × 5 × 5 × 13

So, the given number should be divided by 13 to make the product a perfect square.

New number = 2925 ÷ 13 = 225

∴ 225 = 3 × 3 × 5 × 5

√225 = 3 × 5 = 15

(15) 1225 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Solution: Let the number of row be x.

∴ x2 = 1225

or, x = √1225

or, x = 5 × 7 = 35

Hence, the number of the rows is 35.

(16) The students of a class arranged a picnic. Each student contributed is Rs 1156, find the strength of the class.

Solution: Let the number of students be x.

∴ x2 = 1156

or, x = √1156

or, x = 2 × 17 = 34

(17) Find the least square number which is exactly divisible by each of the numbers 6, 9, 15 and 20.

Solution: The least number divisible by each one of 6, 9, 15 and 20 is their LCM.

Now, LCM of 6, 9, 15 and 20 = (2 × 3 × 5 × 3 × 2) = 180.

By prime factorization, we get

180 = 2 × 2 × 3 × 3 × 5

To make it perfect square it must be multiplied by 5.

Hence, required number = (180 × 5) = 900.

(18) Find the least square number which is exactly divisible by each of the numbers 8, 12, 15 and 20.

Solution: The least number divisible by each one of 8, 12, 15 and 20 is their LCM.

Now, LCM of 8, 12, 15, 20 = (2 × 2 × 3 × 5 × 2) = 120.

By prime factorization, we get

120 = 2 × 2 × 2 × 3 × 5

To make it a perfect square it must be multiplied by (2 × 3 × 5), i.e., 30.

Hence, required number = (120 × 30) = 3600.

Updated: May 30, 2022 — 2:01 pm

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