GSEB Class 8 Maths Solutions Chapter 6 Square and Square Roots Exercise 6.4

Question 1. Find the square root of each of the following numbers by Division method?
(i) 2304
(ii) 4489
(iii) 3481
(iv) 529
(v) 3249
(vi) 1369
(vii) 5776
(viii) 7921
(ix) 576
(x) 1024
(xi) 3136
(xii) 900
Answer:
(i) \( \sqrt{2304} = 48 \)
(ii) \( \sqrt{4489} = 67 \)
(iii) \( \sqrt{3481} = 59 \)
(iv) \( \sqrt{529} = 23 \)
(v) \( \sqrt{3249} = 57 \)
(vi) \( \sqrt{1369} = 37 \)
(vii) \( \sqrt{5776} = 76 \)
(viii) \( \sqrt{7921} = 89 \)
(ix) \( \sqrt{576} = 24 \)
(x) \( \sqrt{1024} = 32 \)
(xi) \( \sqrt{3136} = 56 \)
(xii) \( \sqrt{900} = 30 \)
In simple words: To discover the square root using the division technique, we group digits from the right, then find the biggest digit whose square is less than or equal to the first pair. This process is repeated by bringing down the next pair and doubling the quotient to use as the new divisor, until no remainder is left.

Exam Tip: Remember to pair digits from the right for whole numbers and from the decimal point for decimal parts when applying the division method for square roots.

Question 2. Find the number of digits in the square root of each of the following numbers (without
1. 64
2. 144
3. 4489
4. 27225
5. 390625
Answer:
If 'n' represents the count of digits in the given number:
1. For 64, n = 2 [This is an even number]
The number of digits in its square root is \( \frac{n}{2} = \frac{2}{2} = 1 \).
2. For 144, n = 3 [This is an odd number]
The number of digits in its square root is \( \frac{n+1}{2} = \frac{3+1}{2} = \frac{4}{2} = 2 \).
3. For 4489, n = 4 [This is an even number]
The number of digits in its square root is \( \frac{n}{2} = \frac{4}{2} = 2 \).
4. For 27225, n = 5 [This is an odd number]
The number of digits in its square root is \( \frac{n+1}{2} = \frac{5+1}{2} = \frac{6}{2} = 3 \).
5. For 390625, n = 6 [This is an even number]
The number of digits in its square root is \( \frac{n}{2} = \frac{6}{2} = 3 \).
In simple words: To find how many digits a square root will have without actually calculating it, count the digits in the original number. If that count is even, divide it by two. If the count is odd, add one to it, then divide by two.

Exam Tip: Remember the simple rule for predicting the number of digits: \( n/2 \) for even \( n \) and \( (n+1)/2 \) for odd \( n \).

Question 3. Find the square root of the following decimal numbers?
1. 2.56
2. 7.29
3. 51.84
4. 42.25
5. 31.36
Answer:
1. For \( \sqrt{2.56} \): The number of decimal places is already even. We mark off the periods and discover the square root.
\( \sqrt{2.56} = 1.6 \)
2. For \( \sqrt{7.29} \): The number of decimal places is already even. Thus, we mark off the periods and discover the square root.
\( \sqrt{7.29} = 2.7 \)
3. For \( \sqrt{51.84} \): The decimal places are already even. We mark off the periods and discover the square root.
\( \sqrt{51.84} = 7.2 \)
4. For \( \sqrt{42.25} \): The decimal places are already even. We mark off the periods and discover the square root.
\( \sqrt{42.25} = 6.5 \)
5. For \( \sqrt{31.36} \): The decimal places are already even. We mark off the periods and discover the square root.
\( \sqrt{31.36} = 5.6 \)
In simple words: To find the square root of a decimal, treat it like a whole number but remember to put the decimal point in the answer at the right spot. The count of digits after the decimal in the answer is half the count of digits after the decimal in the original number.

Exam Tip: When finding square roots of decimals, ensure the number of decimal places in the original number is even; if not, add a zero to make it even before calculating.

Question 4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained?
1. 402
2. 1989
3. 3250
4. 825
5. 4000
Answer:
1. For 402: We get a remainder of 2. So, the smallest number to be removed from 402 is 2.
\( 402 - 2 = 400 \)
\( \implies \sqrt{400} = 20 \)
2. For 1989: We get a remainder of 53. So, the smallest number to be removed from 1989 is 53.
\( 1989 - 53 = 1936 \)
\( \implies \sqrt{1936} = 44 \)
3. For 3250: We get a remainder of 1. So, the smallest number to be removed from 3250 is 1.
\( 3250 - 1 = 3249 \)
\( \implies \sqrt{3249} = 57 \)
4. For 825: We get a remainder of 41. So, the required smallest number to be removed from 825 is 41.
\( 825 - 41 = 784 \)
\( \implies \sqrt{784} = 28 \)
5. For 4000: We get a remainder of 31. So, the required smallest number to be removed from 4000 is 31.
\( 4000 - 31 = 3969 \)
\( \implies \sqrt{3969} = 63 \)
In simple words: To find the smallest number to subtract to make a perfect square, perform the long division method for square roots. The remainder you get is the number to subtract. Once subtracted, find the square root of the new number.

Exam Tip: When using the long division method, the remainder directly tells you the amount to subtract to achieve the nearest perfect square below the original number.

Question 5. Find the least number which must be added to each of the following numbers so as to get a perfect perfect square. Also find the square root of the perfect square so obtained?
1. 525
2. 1750
3. 252
4. 1825
5. 6412
Answer:
1. For 525: We get a remainder of 41. This means \( 525 > 22^2 \). The next square number is \( 23^2 = 529 \). The number needed to be added is \( 529 - 525 = 4 \).
So, \( 525 + 4 = 529 \)
\( \implies \sqrt{529} = 23 \)
2. For 1750: We get a remainder of 69. This means \( 1750 > 41^2 \). The next square number is \( 42^2 = 1764 \). The number needed to be added is \( 1764 - 1750 = 14 \).
So, \( 1750 + 14 = 1764 \)
\( \implies \sqrt{1764} = 42 \)
3. For 252: We get a remainder of 27. This means \( 252 > 15^2 \). The next square number is \( 16^2 = 256 \). The number needed to be added is \( 256 - 252 = 4 \).
So, \( 252 + 4 = 256 \)
\( \implies \sqrt{256} = 16 \)
4. For 1825: We get a remainder of 61. This means \( 1825 > 42^2 \). The next square number is \( 43^2 = 1849 \). The number needed to be added is \( 1849 - 1825 = 24 \).
So, \( 1825 + 24 = 1849 \)
\( \implies \sqrt{1849} = 43 \)
5. For 6412: We get a remainder of 12. This means \( 6412 > 80^2 \). The next square number is \( 81^2 = 6561 \). The number needed to be added is \( 6561 - 6412 = 149 \).
So, \( 6412 + 149 = 6561 \)
\( \implies \sqrt{6561} = 81 \)
In simple words: To find the smallest number to add to get a perfect square, first find the square root using the division method. If there's a remainder, it means the number is not a perfect square. Find the next whole number after your square root result, square it, and then subtract your original number from this new perfect square to find what needs to be added.

Exam Tip: When determining the number to add, always look for the *next* perfect square larger than the given number, not the one immediately below it.

Question 6. Find the length of the side of a square whose area is 441 m²?
Answer: Let the side of the square be \( x \) metres.
Area of a square = side \( \times \) side
\( = x \times x = x^2 \) square metres.
We are given that the area is 441 m².
So, \( x^2 = 441 \)
\( \implies x = \sqrt{441} \)
\( \implies x = 21 \)
Therefore, the required side length is 21 m.
In simple words: If you know the area of a square, to find the length of one side, you just need to calculate the square root of that area. In this case, the square root of 441 gives you the side length.

Exam Tip: Always remember that the area of a square is the side length squared, so to find the side from the area, calculate the square root.

Question 7. In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm. find AC.
(b) If AC = 13 cm, BC = 5 cm, find AB.
Answer:
In a right triangle, the hypotenuse is the longest side, opposite the right angle.
According to the Pythagorean theorem, the square of the hypotenuse equals the sum of the squares of the other two sides.
(a) Given: \( \angle B = 90^\circ \), AB = 6 cm, BC = 8 cm. We need to find AC (the hypotenuse).
Using Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
\( \implies AC^2 = 6^2 + 8^2 \)
\( \implies AC^2 = 36 + 64 \)
\( \implies AC^2 = 100 \)
\( \implies AC = \sqrt{100} \)
\( \implies AC = 10 \) cm.
The length of AC is 10 cm.

(b) Given: \( \angle B = 90^\circ \), AC = 13 cm, BC = 5 cm. We need to find AB.
Using Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
\( \implies 13^2 = AB^2 + 5^2 \)
\( \implies 169 = AB^2 + 25 \)
\( \implies AB^2 = 169 - 25 \)
\( \implies AB^2 = 144 \)
\( \implies AB = \sqrt{144} \)
\( \implies AB = 12 \) cm.
The length of AB is 12 cm.

In simple words: For any right-angled triangle, the Pythagorean theorem helps you find a missing side if you know the other two. Just square the two known sides, add or subtract them as needed, and then take the square root of the result to get the third side.

Exam Tip: Always label the hypotenuse correctly (the side opposite the 90-degree angle) before applying the Pythagorean theorem to avoid errors.

Question 8. 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 same. Find the minimum number of plants he needs more for this?
Answer: The gardener wishes to arrange 1000 plants so that the number of rows and columns are equal. This means the total number of plants must be a perfect square.
We need to find the smallest perfect square greater than 1000.
Let's find the square root of 1000 by the division method:
The square root of 1000 is approximately 31.6.
\( 31^2 = 961 \)
\( 32^2 = 1024 \)
Since 1000 is not a perfect square, the next perfect square after 1000 is 1024 (which is \( 32^2 \)).
The number of plants required to be added is \( 1024 - 1000 = 24 \).
Thus, the gardener needs 24 more plants to make a perfect square arrangement.
In simple words: To arrange plants into an equal number of rows and columns, the total number of plants must be a perfect square. Since 1000 isn't a perfect square, find the next perfect square bigger than 1000 (which is 1024). Subtract 1000 from 1024 to find how many more plants are needed.

Exam Tip: When a question asks for the "minimum number to add" to make a perfect square, always find the next perfect square that is greater than the given number.

Question 9. 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: The children must stand in an arrangement where the number of rows matches the number of columns. This means the total number of children in the formation must be a perfect square.
We need to find the largest perfect square that is less than or equal to 500.
Let's find the square root of 500 by the division method:
The square root of 500 is approximately 22.3.
\( 22^2 = 484 \)
\( 23^2 = 529 \)
Since we only have 500 children, the largest possible perfect square formation we can make is with 484 children (which is \( 22^2 \)).
The number of children who would be left out is \( 500 - 484 = 16 \).
Therefore, 16 children would not be part of this arrangement.
In simple words: If children need to stand in equal rows and columns, the total in the formation must be a perfect square. Since there are 500 children, find the largest perfect square number that is 500 or less (which is 484). The children remaining after forming this square (500 minus 484) are the ones left out.

Exam Tip: When a question asks how many items will be "left out," it implies finding the largest perfect square *less than or equal to* the given total number of items.