ML Aggarwal CBSE Solutions Class 8 Math 3rd Chapter Squares and Square Roots Exercise 3.3
∴ 6 should be multiplied to get a perfect square
∴ 294 x 6 = 1764 = 6 x 6 x 7 x 7
∴ 1764 is perfect square root.
√1764 = √6x6x7x7 = 6 x 7
= 42
∴ square root is 42
∵ 3 is unpaired
∴ 3 should be divided
∴ The perfect sq root = 613747/3
= √20449 = 143
(7) If 64 student does not accommodated
Then, the num of row = the num of column
No of students arrange = 2000
∴ If 64 student could not accommodated,
then no of students = 2000 – 64
= 1936
∴ No of row = √1936 = No of columns
= 2 x 2 x 2 x 11 x 11
= 2 x 2 x 11
= 44
(8) According to question
No. of students = No of contributed rupee
Total rupees Rs. 2304
∴ No of students = √2304
= 2 x 2 x 2 x 2 x 12 x 12
= 2 x 2 x 12
= 48
(9) One number = x
another number = 15x
According to the question
15x X x = 7260
Or, 15x2 = 7260
Or, x2 = 7260/15
Or, x = √484
= √2x2x11x11
= 2 x 11
= 22
One number = 22
Another number = 22 x 15 = 330
(10) Common factor = (2+3+5) 10
1st positive number = 2/10 x 950
190
2nd positive number = 3/10 x 950
285
3rd positive number = 5/10 x 950
= 475
(11) 1st sq’s perimeter = 60m = 4L
∴ 1st sq’s one side = L1 = 60/4 = 15 m
∴ 1st sq’s area = 15 x 15 = 225 m2
2nd sq’s perimeter = 144 m = 4 L
∴ 2nd sq’s one side = L2 = 144/4 = 36
∴ 2nd sq’s area = 36 x 26 = 1296 m2
Sum of area of 1st & 2nd sq = (225 + 1296) m2
= 1521 m2
∴ According to question –
3rd sq’s area = 1521 m2
∴ Let, 3rd sq’s area = L32
∴ L32 = 1521
Or, L3 = √1521 = 39
∴ 3rd sq’s perimeter = 4L = 39 x 4
156 m
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