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

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

EXERCISE 3E

Evaluate:

(1) √576

(2) √1444

(3) √4489

(4) √6241

(5) √7056

(6) √9025

(7) √11449

(8) √14161

(9) √10404

(10) √17956

(11) √19600

(12) √92416

(13) Find the least number which must be subtracted from 2509 to make it a perfect square.

Solution: Let us try to find the square root of 2509.

This shows that (50)2 is less than 2509 by 9.

Required perfect square number = (2509 – 9) = 2500.

And, √2500 = 50.

(14) Find the least number which must be subtracted from 7581 to obtain a perfect square. Find this perfect square ant its square root.

Solution: Let us try to find the square root of 7581.

This shows that (87)2 is less than 7581 by 12.

Required perfect square number = (7581 – 12) = 7569.

And, √7569 = 87.

(15) Find the least number which must be added to 6203 to obtain a perfect square. Find this perfect square and its square root.

Solution: We try to find the square root of 6203.

We observe here that (78)2 < 6203 < (79)2.

The required number to be added = (79)2 – 6203 = (6241 – 6203) = 38.

Clearly, the required perfect square = 6241 and √6241 = 79.

(16) Find the least number which must be added to 8400 to obtain a perfect square. Find this perfect square and its square root.

Solution: We try to find the square root of 8400.

We observe here that (91)2 < 8400 < (92)2.

The required number to be added = (92)2 – 8400 = (8464 – 8400) = 64.

Clearly, the required perfect square = 8464 and √8464 = 92.

(17) Find the least number of four digits which is a perfect square. Also find the square root of the number so obtained.

Solution: the least number of four digits = 1000, which is not a perfect square.

Now, we must find the least number which when added to 1000 gives a perfect square. This perfect square is required number.

Now, we find out the square root of 1000.

Clearly, (31)2 < 1000 < (32)2.

∴ The least number to be added = (32)2 – 1000 = (1024 – 1000) = 24.

Hence, the required number = (1000 + 24) = 1024.

Also, √1024 = 32.

(18) Find the greatest number of five digits which is perfect square. Also find the square root of the number so obtained.

Solution: the least number of four digits = 99999, which is not a perfect square.

Now, we must find the least number which when added to 99999 gives a perfect square. This perfect square is required number.

Now, we find out the square root of 99999.

Clearly, (316)2 < 99999 < (317)2.

∴ The least number to be added = (317)2 – 99999 = (100489 – 99999) = 490.

Hence, the required number = (99999 + 490) = 100489.

Also, √100489 = 317.

(19) The area of a square field is 60025 m2. A man cycles along its boundary at 18 km/h. In how much time will he return to the starting point?


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