CBSE Class 8 Mathematics Squares and Square Roots Practice Paper Assignment Set 15 with Answers (Old NCERT Syllabus)

This Class 8 Maths Squares and Square Roots practice paper has 20 questions for 40 marks, set out in Sections A to E with 10 MCQs (two of them Assertion–Reason), short and long answer questions with internal choices, and two case studies; every answer is worked out step by step. It follows the earlier (old) NCERT Class 8 Maths syllabus and is useful for state-board students and for revising these topics. It checks square roots of decimals, squares of decimals, writing squares as sums of odd numbers, the long division method and real-life problems on square fields and square arrangements.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. 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 12. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer:
The sum of the first n odd natural numbers is n².
(i) 49 = 7², so it is the sum of the first 7 odd numbers:
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13.
(ii) 121 = 11², so it is the sum of the first 11 odd numbers:
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.
In simple words: A square n² is the sum of the first n odd numbers, so 49 uses 1 to 13 and 121 uses 1 to 21.

Teacher's Note:
1. The last odd number in the list is 2n − 1: 13 for n = 7 and 21 for n = 11.
2. Add the list once to check that it really gives 49 or 121.

Question 13. 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 14. What is the least number should be subtracted from 1385 to get a perfect square? Also find the square root of the perfect square.
OR
Write Pythagorean triplet of 18.
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.
OR
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: 1385 − 16 = 1369 = 37²; in the OR question, 2m = 18 gives m = 9 and the triplet 18, 80, 82.

Teacher's Note:
1. When you have to subtract, the remainder of the long division is the answer.
2. For an even number, use it as 2m; this always gives whole numbers.

Section C: Short Answer Questions (3 Marks Each)

Question 15. 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 16. 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 17. If the area of a square picture is 20.25 cm², find the length of its side.
OR
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:
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.
OR
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: √20.25 = 4.5 cm; in the OR question 1500 + 21 = 1521 = 39².

Teacher's Note:
1. For decimals, pair the digits on both sides of the decimal point.
2. When you must add, go to the next square: (quotient + 1)².

Section D: Long Answer Question (5 Marks)

Question 18. Find the smallest square number which is exactly divisible by 4, 9 and 10. Also find the square root of the square number.
OR
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:
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.
OR
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: LCM 180 × 5 = 900 = 30²; in the OR question 1000 + 24 = 1024 = 32².

Teacher's Note:
1. Start with the LCM, then multiply by the unpaired prime factors.
2. "Needs more" means go up to the next perfect square.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. A farmer owns a perfectly square field and plans to install a strong fence along its entire boundary to protect crops from animals. The total length of fencing available is 480 meters. In addition, he dreams of planting a large tree exactly at the center of the field to provide shade and beauty. To carry out his plan, he needs to calculate the field’s side length and area accurately.
(a) Find the length of each side of the field. (1)
(b) What is the area of the field in square meters? (1)
(c) If the farmer decides to divide the field into 4 equal square parts by drawing two lines parallel to the sides, what will be the area of each part? (1)
(d) If the square field is divided into 4 equal square plots, find the side of each smaller square plot. (1)
Answer:
(a) The fence goes round the boundary, so perimeter = 4 × side = 480 m.
Side = 480 ÷ 4 = 120 m.
(b) Area = side² = 120² = 120 × 120 = 14400 m².
(c) The two lines through the centre, parallel to the sides, cut the field into 4 equal squares.
Area of each part = 14400 ÷ 4 = 3600 m².

(d) Each small square has side = 120 ÷ 2 = 60 m (check: √3600 = 60).
Side of each smaller plot = 60 m.
In simple words: Fence 480 m means side 120 m and area 14400 m²; four equal squares each have area 3600 m² and side 60 m.

Teacher's Note:
1. Perimeter of a square = 4 × side, so side = perimeter ÷ 4.
2. Halving the side divides the area by 4, not by 2.

Question 20. 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.