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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