CBSE Class 8 Mathematics Squares and Square Roots Practice Paper Worksheet Set 11 with Answers (Old NCERT Syllabus)

This Class 8 Maths Squares and Square Roots practice paper is a 40-mark test of 17 questions in five sections (MCQs with an Assertion–Reason item, short answers, long answers and a case study), with step-by-step answers for every question. It follows the earlier (old) NCERT Class 8 Maths syllabus, so it suits state-board students and anyone revising this chapter. The questions test square roots by the long division method, the least number to add or subtract to get a perfect square, Pythagorean triplets and the smallest square divisible by given numbers.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 7. Find the value of \( \sqrt{248 + \sqrt{52 + \sqrt{144}}} \)
Answer:
Work from the innermost root outwards.
√144 = 12, so the expression = \( \sqrt{248 + \sqrt{52 + 12}} = \sqrt{248 + \sqrt{64}} \).
√64 = 8, so it becomes \( \sqrt{248 + 8} = \sqrt{256} \).
16 × 16 = 256, so √256 = 16.
Value = 16.
In simple words: Open the roots one at a time from the inside: √144 = 12, √64 = 8 and √256 = 16.

Teacher's Note:
1. Always simplify the innermost square root first.
2. Check: 16² = 256 = 248 + 8.

Question 8. Find the side of a square whose area is equal to the area of a rectangle with sides 6.4 m and 2.5 m.
Answer:
Area of the rectangle = length × breadth = 6.4 × 2.5 = 16 m².
Let the side of the square be a m. Then a² = 16.
a = √16 = 4.
Side of the square = 4 m.
In simple words: The rectangle has area 16 m², and a square with area 16 m² has side 4 m.

Teacher's Note:
1. 6.4 × 2.5 is easy as 64 × 25 = 1600, then place the decimal: 16.00.
2. Equal areas, not equal perimeters, is the condition here.

Question 9. What is the least number should be subtracted from 1385 to get a perfect square? Also find the square root of the perfect square.
Answer:
Find √1385 by the long division method. Pair the digits from the right: 13 | 85.
Largest square ≤ 13 is 3² = 9. Write 3 in the quotient; 13 − 9 = 4.
Bring down 85 to get 485. Double the quotient: 6. Find the digit d with 6d × d ≤ 485: 67 × 7 = 469.
485 − 469 = 16. The quotient is 37 and the remainder is 16.
So 37² = 1369 is the largest perfect square below 1385, and 1385 − 1369 = 16.
Least number to be subtracted = 16; the perfect square is 1369 and √1369 = 37.
In simple words: 1385 is 16 more than 37² = 1369, so take away 16 to get a perfect square whose root is 37.

Teacher's Note:
1. When you have to subtract, the remainder of the long division is the answer.
2. Check: 37 × 37 = 1369 and 1369 + 16 = 1385.

Section C: Short Answer Questions (3 Marks Each)

Question 10. Rahul walks 12 m north from his house and turns west to walk 35 m to reach his friend's house. While returning, he walks diagonally from his friend's house to reach back to his house. What distance did he walk while returning?
Answer:
Let P be Rahul's house, Q the point where he turns and R his friend's house.

PQ = 12 m (north) and QR = 35 m (west). North and west are at right angles, so ∠PQR = 90°.
By Pythagoras' theorem in right triangle PQR: PR² = PQ² + QR² = 12² + 35².
PR² = 144 + 1225 = 1369.
PR = √1369 = 37 (since 37 × 37 = 1369).
Distance walked while returning = 37 m.
In simple words: His walk makes a right triangle with legs 12 m and 35 m, so the diagonal path back is √(144 + 1225) = 37 m.

Teacher's Note:
1. North-then-west is a right-angle turn, so Pythagoras' theorem applies.
2. 12, 35, 37 is a Pythagorean triplet; it is worth remembering.

Question 11. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?
Answer:
Rows = columns, so the number of children who stand must be a perfect square.
Find √500 by long division. Pairs: 5 | 00.
Largest square ≤ 5 is 2² = 4; 5 − 4 = 1. Bring down 00 to get 100.
Double the quotient: 4. 42 × 2 = 84 ≤ 100; 100 − 84 = 16.
Quotient 22, remainder 16, so 22² = 484 is the largest square below 500.
Children left out = 500 − 484 = 16 (they form 22 rows of 22).
In simple words: The biggest square arrangement possible is 22 × 22 = 484, so 500 − 484 = 16 children are left out.

Teacher's Note:
1. "Rows equal to columns" always means a perfect square number.
2. Left out means subtract; needed more means add to reach the next square.

Question 12. Write Pythagorean triplet of 18.
Answer:
For any natural number m > 1, the numbers 2m, m² − 1 and m² + 1 form a Pythagorean triplet.
Take 2m = 18, so m = 9.
m² − 1 = 81 − 1 = 80 and m² + 1 = 81 + 1 = 82.
Check: 18² + 80² = 324 + 6400 = 6724 = 82².
The Pythagorean triplet is 18, 80, 82.
In simple words: Put 2m = 18, so m = 9; then m² − 1 = 80 and m² + 1 = 82.

Teacher's Note:
1. For an even number, use it as 2m; this always gives whole numbers.
2. Always check by squaring: the two smaller squares must add to the largest square.

Question 13. Find the least number that must be added to 1500 so as to get a perfect square. Also find the square root of the perfect square.
Answer:
Find √1500 by long division. Pairs: 15 | 00.
Largest square ≤ 15 is 3² = 9; 15 − 9 = 6. Bring down 00 to get 600.
Double the quotient: 6. 68 × 8 = 544 ≤ 600 (69 × 9 = 621 is too big); 600 − 544 = 56.
So 38² = 1444 < 1500 < 39² = 1521.
Number to be added = 39² − 1500 = 1521 − 1500 = 21.
Least number to be added = 21; the perfect square is 1521 and √1521 = 39.
In simple words: 1500 lies between 38² and 39²; adding 21 takes it up to 39² = 1521.

Teacher's Note:
1. When you must add, go to the next square: (quotient + 1)².
2. Do not add the remainder 56; that is a common mistake.

Section D: Long Answer Questions (4 Marks Each)

Question 14. Find the smallest square number which is exactly divisible by 4, 9 and 10. Also find the square root of the square number.
Answer:
The smallest number divisible by 4, 9 and 10 is their LCM.
4 = 2 × 2, 9 = 3 × 3, 10 = 2 × 5, so LCM = 2 × 2 × 3 × 3 × 5 = 180.
In 180 = (2 × 2) × (3 × 3) × 5, the factors 2 and 3 are in pairs but 5 is not paired.
Multiply by 5 to complete the pair: 180 × 5 = 900.
900 = (2 × 2) × (3 × 3) × (5 × 5), so √900 = 2 × 3 × 5 = 30.
Smallest square number = 900; its square root is 30.
In simple words: The LCM 180 needs one more 5 to become a perfect square, giving 900 = 30².

Teacher's Note:
1. Start with the LCM, then multiply by the unpaired prime factors.
2. 900 ÷ 4 = 225, 900 ÷ 9 = 100 and 900 ÷ 10 = 90, so it really is divisible by all three.

Question 15. If the area of a square picture is 20.25 cm², find the length of its side.
Answer:
Let the side be x cm. Then x² = 20.25, so x = √20.25.
Long division: pair the digits from the decimal point: 20 . 25.
Largest square ≤ 20 is 4² = 16; 20 − 16 = 4. Put the decimal point in the quotient: 4.
Bring down 25 to get 425. Double the quotient: 8. 85 × 5 = 425; remainder 0.
So √20.25 = 4.5. Check: 4.5 × 4.5 = 20.25.
Length of the side = 4.5 cm.
In simple words: The side is the square root of the area: √20.25 = 4.5 cm.

Teacher's Note:
1. For decimals, pair the digits on both sides of the decimal point.
2. Another way: 20.25 = 2025 ÷ 100, so √20.25 = 45 ÷ 10 = 4.5.

Question 16. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain the same. Find the minimum number of plants he needs more for this. Also find the number of rows and the number of columns.
Answer:
Rows = columns, so the number of plants must be a perfect square.
Find √1000 by long division. Pairs: 10 | 00.
Largest square ≤ 10 is 3² = 9; 10 − 9 = 1. Bring down 00 to get 100.
Double the quotient: 6. 61 × 1 = 61 ≤ 100; 100 − 61 = 39.
So 31² = 961 < 1000 < 32² = 1024.
Plants needed more = 1024 − 1000 = 24.
Number of rows = number of columns = √1024 = 32.
In simple words: 1000 is not a square; the next square is 32² = 1024, so he needs 24 more plants and plants 32 rows of 32.

Teacher's Note:
1. "Needs more" means go up to the next perfect square.
2. Check: 32 × 32 = 1024 = 1000 + 24.

Section E: Case Study Based Question (4 Marks)

Question 17. During dance practice in school 6570 students of different schools are arranged in rows such that the number of students in each row is equal to the number of rows. In doing so, the instructor finds out that few children are left out.
(a) How many students were left out in arrangement? (1)
(b) What is the number of students forming a square? (1)
(c) Find the number of children in each row of the square. (2)
Answer:
(a) Find √6570 by long division. Pairs: 65 | 70.
Largest square ≤ 65 is 8² = 64; 65 − 64 = 1. Bring down 70 to get 170.
Double the quotient: 16. 161 × 1 = 161 ≤ 170; 170 − 161 = 9.
The remainder is 9, so 9 students were left out.
(b) Students forming the square = 6570 − 9 = 6561.
(c) Let each row have x students; then there are x rows, so x × x = 6561.
x = √6561 = 81 (the quotient of the long division). Check: 81 × 81 = 6561.
Number of children in each row = 81.
In simple words: The largest square up to 6570 is 81² = 6561, so 9 children are left out and the rest stand 81 in a row.

Teacher's Note:
1. Rows equal to students in each row means a square arrangement.
2. The quotient of the long division gives the side of the square; the remainder gives the children left out.