Squares and Cube Roots Class 8 RS Aggarwal Exe-3A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3. We provide step by step Solutions of council prescribe publication / textbook. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Squares and Squares Roots Cube and Cube Roots
Class 8 RS Aggarwal Exe-3A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3
| Board | ICSE |
| Publications | Goyal Brothers Prakshan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Chapter-3 | Squares and Squares Roots Cube and Cube Roots |
| Exe-3A | Method of Finding Squire Roots |
| Edition | 2024-2025 |
Method of Finding Squire Roots
Class 8 RS Aggarwal Exe-3A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3
Page- 42
Exercise- 3A
Squares and Squares Roots Cube and Cube Roots Class 8 RS Aggarwal Exe-3A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-3
Que-1: Find the square of each of the following numbers :
(i) 14 (ii) 137 (iii) 4/17 (iv) 2*(3/4) (v) 0.01 (vi) 1.2 (vii) 0.17 (viii) 4.6
Solution- (i) 14² = 196
(ii) 137² = 18769
(iii) (4/17)² = 16/289
(iv) [2*(3/4)]² = (11/4)² = 121/16 = 7*(9/16)
(v) 0.01² = 0.0001
(vi) 1.2² = 1.44
(vii) 0.17² = 0.0289
(viii) 4.6² = 21.16
Que-2: Using prime factorization method, find which of the following are perfect squares numbers :
(i) 252 (ii) 324 (iii) 676 (iv) 1225 (iv) 2916 (vi) 3528 (vii) 5625 (viii) 9075
Solution- (i) Prime factorization of 252 = 2^2×3^2×7
the exponent of 7 is not even, so 252 is not a perfect square.
(ii) Prime factorization of 324 = 2^2×3^4
In this case, both exponents are even, so 324 is a perfect square.
(iii) Prime factorization of 676 = 2^2×13^2
Both exponents are even, so 676 is a perfect square.
(iv) Prime factorization of 1225 = 5^2×7^2
Both exponents are even, so 1225 is a perfect square.
(v) Prime factorization of 2916 = 2^2×3^4×17
The exponents of 2 and 3 are even, but the exponent of 17 is not even, so 2916 is not a perfect square.
(vi) Prime factorization of 3528 : 2^3×3^2×7^2
Each prime factor occurs an even number of times (3, 2, 2), so 3528 is a perfect square.
(vii) Prime factorization of 5625 : 3^4×5^2
Each prime factor occurs an even number of times (4, 2), so 5625 is a perfect square.
(viii) Prime factorization of 9075 : 3^2×5^2×11
The power of 11 is odd, so 9075 is not a perfect square
Que-3: Show that each of the following numbers is a perfect square. Also, find the square root of each :
(i) 441 (ii) 784 (iii) 1225 (iv) 3969 (v) 2601 (vi) 5929 (vii) 7056 (viii) 8281
Solution- (i) √441 = 21
(ii) √784 = 28
(iii) √1225 = 35
(iv) √3969 = 63
(v) √2601 = 51
(vi) √5929 = 77
(vii) √7056 = 84
(viii) √8281 = 81
Que-4: Find the smallest number by which each of the following numbers must be multiplied to get a perfect square number. Also, find the square root of the resulting number :
(i) 588 (ii) 2592 (iii) 3332 (iv) 3380
Solution- (i) 588 = 2 × 2 × 7 × 7 × 3
Common factors = 2 × 7
Uncommon factor = 3
Now only 3 which is an uncommon factor that is not in pair. So the least number by which 588 be multiplied to get a perfect number is 3.
The square root of resulting number is
2 x 7 x 3 = 42.
(ii) let’s factorise it
factors of 2592: 2×2×2×2×2×9×9
=(2×2)(2×2)(9×9)(2)
so the smallest no by which 2592 should be multiplied to get a perfect square 2.
so the square no is 5184
let’s find its square root :
√5184 = 72.
(iii) 3332 = 2×2×7×7×17
= (2)^2×(7)^2×17
In order to get a perfect square, the given number should be multiplied by 17
The square root of the new number =(2)×(7)×17=238.
(iv) let’s factorise it
factors : 2×2×5×13×13
= (2×2)(13×13)(5)
so the smallest no by which 3380 should be multiplied to get a perfect square is 5
resulted no : 3380×5 = 16900
square root of 16900 = 130
Que-5: By which least number should be given number be divided to obtain a perfect square number? Also, find the resulting perfect square number and its square root :
(i) 1728 (ii) 4500 (iii) 7776 (iv) 8820
Solution- (i) The least number is 3.
1728 ÷ 3 = 576
Square root of 576 is 24.
(ii) Resolve 4500 into prime factors, we get
4500 = 2×2×3×3×5×5×5 = (2^2×3^2×5^2×5)
Now to get perfect square we have to divide the above equation by 5 Ans.
Then we get, 4500/5 = 2×2×3×3×5×5
New number =(4×9×25)
=(2^2×3^2×5^2) = 900 Ans.
Taking squares as common from the above equation we get
∴ New number = (2×3×5)^2
=(30)^2
Hence, the new number is square of 30 Ans.
(iii) The least number is 6.
7776 ÷ 6 = 1296
Square root of 1296 is 36.
(iv) The least number is 5.
8820 ÷ 5 = 1764
Square root of 1764 is 42.
Que-6: The students of a class arranged a picnic. Each student contributed as many rupees as the number of students in the class. If the total contribution is Rs 1156, find the strength of a class.
Solution-Let no. of students in the class = x
Then contribution of each student = x
∴x × x = 1156
⇒ x^2 = 1156
∴ x = √1156 = √(2×2)(17×17)
= 2 × 17 = 34 Ans.
Que-7: In an auditorium, the number of rows is equal to the number of chairs in each row. If the capacity of the auditorium is 2025, find the number of chairs in each row.
Solution- Let the number of chairs in each row be x.
Then, the number of rows = x.
Total number of chairs in the auditorium
= (x × x) = x²
But, the capacity of the auditorium = 2025 (given).
Therefore, x^2 = 2025
2025 = 5×5×3×3×3×3
= 5^2×3^4
⇒ √(2025)=√(5^2×3^4)
= 5×3×3 = 45
∴x = 5×3×3 = 45 Ans.
Que-8: Find the least square number which is exactly divisible by each of the numbers 8,9,10 and 15.
Solution- Now, LCM of 8, 9 and 10 = 2 × 2 × 2 × 3 × 3 × 5 = 360
We observe that 2 and 5 do not occur in pairs.
So, 360 is not a perfect square.
Now, 360 must be multiplied by 2 × 5 to get a perfect square
= 360 × 5 × 2
= 360 × 10
= 3600
— : End of Squares and Cube Roots Class 8 RS Aggarwal Exe-3A Goyal Brothers Brothers Prakashan ICSE Foundation Maths Solutions :–
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