{"id":601887,"date":"2022-05-13T07:42:32","date_gmt":"2022-05-13T07:42:32","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=601887"},"modified":"2022-05-14T04:30:21","modified_gmt":"2022-05-14T04:30:21","slug":"ml-aggarwal-solutions-for-class-8-maths-chapter-3-squares-and-square-roots","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-8-maths-chapter-3-squares-and-square-roots\/","title":{"rendered":"ML Aggarwal Solutions for Class 8 Maths Chapter 3- Squares and Square Roots"},"content":{"rendered":"\n

Class 8: Maths Chapter 3 solutions. Complete Class 8 Maths Chapter 3 Notes.<\/p>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 3- Squares and Square Roots<\/h2>\n\n\n\n

ML Aggarwal 8th Maths Chapter 3, Class 8 Maths Chapter 3 solutions<\/p>\n\n\n\n

Exercise 3.1<\/h4>\n\n\n\n

1. Which of the following natural numbers are perfect squares? Give reasons in support of your answer.<\/strong><\/p>\n\n\n\n

(i) 729<\/strong><\/p>\n\n\n\n

(ii) 5488<\/strong><\/p>\n\n\n\n

(iii) 1024<\/strong><\/p>\n\n\n\n

(iv) 243<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 729<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

729 = 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

729 is the product of pairs of equal prime factors<\/p>\n\n\n\n

Therefore, 729 is a perfect square.<\/p>\n\n\n\n

(ii) 5488<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

5488 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 7 \u00d7 7 \u00d7 7<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same prime factors, one factor 7 is left unpaired.<\/p>\n\n\n\n

Therefore, 5488 is not a perfect square.<\/p>\n\n\n\n

(iii) 1024<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

1024 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same prime factors, there is no factor left.<\/p>\n\n\n\n

Therefore, 1024 is a perfect square.<\/p>\n\n\n\n

(iv) 243<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

243 = 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same prime factors, factor 3 is left unpaired.<\/p>\n\n\n\n

Therefore, 243 is not a perfect square.<\/p>\n\n\n\n

2. Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.<\/strong><\/p>\n\n\n\n

(i) 1296<\/strong><\/p>\n\n\n\n

(ii) 1764<\/strong><\/p>\n\n\n\n

(iii) 3025<\/strong><\/p>\n\n\n\n

(iv) 3969<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 1296<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

1296 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same prime factors, no factor is left.<\/p>\n\n\n\n

Therefore, 1296 is a perfect square of 2 \u00d7 2 \u00d7 3 \u00d7 3 = 36.<\/p>\n\n\n\n

(ii) 1764<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

1764 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 7 \u00d7 7<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same factors, no factor is left.<\/p>\n\n\n\n

Therefore, 1764 is a perfect square of 2 \u00d7 3 \u00d7 7 = 42.<\/p>\n\n\n\n

(iii) 3025<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3025 = 5 \u00d7 5 \u00d7 11 \u00d7 11<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same prime factors, no factor is left.<\/p>\n\n\n\n

Therefore, 3025 is a perfect square of 5 \u00d7 11 = 55.<\/p>\n\n\n\n

(iv) 3969<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3969 = 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 7 \u00d7 7<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same prime factors, no factor is left.<\/p>\n\n\n\n

Therefore, 3969 is a perfect square of 3 \u00d7 3 \u00d7 7 = 63.<\/p>\n\n\n\n

3. Find the smallest natural number by which 1008 should be multiplied to make it a perfect square.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

1008 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 7<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same kind of prime factors, one factor 7 is left.<\/p>\n\n\n\n

Now multiplying 1008 by 7<\/p>\n\n\n\n

We get a perfect square<\/p>\n\n\n\n

Therefore, the required smallest number is 7.<\/p>\n\n\n\n

4. Find the smallest natural number by which 5808 should be divided to make it a perfect square. Also, find the number whose square is the resulting number.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

5808 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 11 \u00d7 11<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

After pairing the same kind of prime factors, factor 3 is left.<\/p>\n\n\n\n

Now dividing the number by 3, we get a perfect square.<\/p>\n\n\n\n

Therefore, the square root of the resulting number is 2 \u00d7 2 \u00d7 11 = 44.<\/p>\n\n\n\n

Exercise 3.2<\/h4>\n\n\n\n

1. Write five numbers which you can decide by looking at their one\u2019s digit that they are not square numbers. Solution:<\/strong><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

A number which ends with the digits 2, 3, 7 or 8 at its unit places is not a perfect square.<\/p>\n\n\n\n

Example \u2013 111, 372, 563, 978, 1282 are not square numbers.<\/p>\n\n\n\n

2. What will be the unit digit of the squares of the following numbers?<\/strong><\/p>\n\n\n\n

(i) 951<\/strong><\/p>\n\n\n\n

(ii) 502<\/strong><\/p>\n\n\n\n

(iii) 329<\/strong><\/p>\n\n\n\n

(iv) 643<\/strong><\/p>\n\n\n\n

(v) 5124<\/strong><\/p>\n\n\n\n

(vi) 7625<\/strong><\/p>\n\n\n\n

(vii) 68327<\/strong><\/p>\n\n\n\n

(viii) 95628<\/strong><\/p>\n\n\n\n

(ix) 99880<\/strong><\/p>\n\n\n\n

(x) 12796<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 951<\/p>\n\n\n\n

The unit digit of the square is 1.<\/p>\n\n\n\n

(ii) 502<\/p>\n\n\n\n

The unit digit of the square is 4.<\/p>\n\n\n\n

(iii) 329<\/p>\n\n\n\n

The unit digit of the square is 1.<\/p>\n\n\n\n

(iv) 643<\/p>\n\n\n\n

The unit digit of the square is 9.<\/p>\n\n\n\n

(v) 5124<\/p>\n\n\n\n

The unit digit of the square is 6.<\/p>\n\n\n\n

(vi) 7625<\/p>\n\n\n\n

The unit digit of the square is 5.<\/p>\n\n\n\n

(vii) 68327<\/p>\n\n\n\n

The unit digit of the square is 9.<\/p>\n\n\n\n

(viii) 95628<\/p>\n\n\n\n

The unit digit of the square is 4.<\/p>\n\n\n\n

(ix) 99880<\/p>\n\n\n\n

The unit digit of the square is 0.<\/p>\n\n\n\n

(x) 12796<\/p>\n\n\n\n

The unit digit of the square is 6.<\/p>\n\n\n\n

3. The following numbers are obviously not perfect. Give reason.<\/strong><\/p>\n\n\n\n

(i) 567<\/strong><\/p>\n\n\n\n

(ii) 2453<\/strong><\/p>\n\n\n\n

(iii) 5298<\/strong><\/p>\n\n\n\n

(iv) 46292<\/strong><\/p>\n\n\n\n

(v) 74000<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In the given numbers<\/p>\n\n\n\n

If the square of a number does not have 2, 3, 7, 8 or 0 as its unit digit, the squares 567, 2453, 5208, 46292 and 74000 cannot be the perfect squares as they have 7, 2, 8, 2 digits at the unit place.<\/p>\n\n\n\n

4. The square of which of the following numbers would be an odd number or an even number? Why?<\/strong><\/p>\n\n\n\n

(i) 573<\/strong><\/p>\n\n\n\n

(ii) 4096<\/strong><\/p>\n\n\n\n

(iii) 8267<\/strong><\/p>\n\n\n\n

(iv) 37916<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

The square of an odd number is odd and a square of an even number is even.<\/p>\n\n\n\n

So 573 and 8262 are odd numbers and their squares will be an odd number.<\/p>\n\n\n\n

4096 and 37916 are even numbers and their square will also be even number.<\/p>\n\n\n\n

5. How many natural numbers lie between square of the following numbers?<\/strong><\/p>\n\n\n\n

(i) 12 and 13<\/strong><\/p>\n\n\n\n

(ii) 90 and 91<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that<\/p>\n\n\n\n

No. of natural numbers between the squares of 12 and 13 = (132<\/sup> \u2013 122<\/sup>) \u2013 1<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (13 + 12 \u2013 1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 25 \u2013 1<\/p>\n\n\n\n

= 24<\/p>\n\n\n\n

(ii) We know that<\/p>\n\n\n\n

No. of natural numbers between the squares of 90 and 91 = (912<\/sup> \u2013 902<\/sup>) \u2013 1<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (91 + 90 \u2013 1)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 181 \u2013 1<\/p>\n\n\n\n

= 180<\/p>\n\n\n\n

6. Without adding, find the sum.<\/strong><\/p>\n\n\n\n

(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15<\/strong><\/p>\n\n\n\n

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that<\/p>\n\n\n\n

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = n2<\/sup><\/p>\n\n\n\n

Here n = 8<\/p>\n\n\n\n

So the sum = 82<\/sup> = 64<\/p>\n\n\n\n

(ii) We know that<\/p>\n\n\n\n

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = n2<\/sup><\/p>\n\n\n\n

Here n = 15<\/p>\n\n\n\n

So the sum = 152<\/sup> = 225<\/p>\n\n\n\n

7. (i) Express 64 as the sum of 8 odd numbers.<\/strong><\/p>\n\n\n\n

(ii) 121 as the sum of 11 odd numbers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that<\/p>\n\n\n\n

64 as the sum of 8 odd numbers = 82<\/sup> = n2<\/sup><\/p>\n\n\n\n

Here n = 8<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15<\/p>\n\n\n\n

(ii) We know that<\/p>\n\n\n\n

121 as the sum of 11 odd numbers = 112<\/sup> = n2<\/sup><\/p>\n\n\n\n

Here n = 11<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21<\/p>\n\n\n\n

8. Express the following as the sum of two consecutive integers.<\/strong><\/p>\n\n\n\n

(i) 192<\/sup><\/strong><\/p>\n\n\n\n

(ii) 332<\/sup><\/strong><\/p>\n\n\n\n

(iii) 472<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

n2<\/sup> = (n2<\/sup> \u2013 1)\/ 2 + (n2<\/sup> + 1)\/ 2 is the sum of two consecutive integers when n is odd<\/p>\n\n\n\n

(i) 192<\/sup> = (192<\/sup> \u2013 1)\/ 2 + (192<\/sup> + 1)\/ 2<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

192<\/sup> = 361<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (361 \u2013 1)\/ 2 + (361 + 1)\/ 2<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 180 + 181<\/p>\n\n\n\n

(ii) 332<\/sup> = (332<\/sup> \u2013 1)\/ 2 + (332<\/sup> + 1)\/ 2<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

332<\/sup> = 1089<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (1089 \u2013 1)\/ 2 + (1089 + 1)\/ 1<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 1088\/2 + 1090\/2<\/p>\n\n\n\n

= 544 + 545<\/p>\n\n\n\n

(iii) 472<\/sup> = (472<\/sup> \u2013 1)\/ 2 + (472<\/sup> + 1)\/ 2<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

472<\/sup> = 2209<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (2209 \u2013 1)\/ 2 + (2209 + 1)\/ 1<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

= 2208\/2 + 2210\/2<\/p>\n\n\n\n

= 1104 + 1105<\/p>\n\n\n\n

9. Find the squares of the following numbers without actual multiplication:<\/strong><\/p>\n\n\n\n

(i) 31<\/strong><\/p>\n\n\n\n

(ii) 42<\/strong><\/p>\n\n\n\n

(iii) 86<\/strong><\/p>\n\n\n\n

(iv) 94<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

(a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup><\/p>\n\n\n\n

(i) 312<\/sup> = (30 + 1)2<\/sup><\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 302<\/sup> + 2 \u00d7 30 \u00d7 1 + 12<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 900 + 60 + 1<\/p>\n\n\n\n

= 961<\/p>\n\n\n\n

(ii) 422<\/sup> = (40 + 2)2<\/sup><\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 402<\/sup> + 2 \u00d7 40 \u00d7 2 + 22<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 1600 + 160 + 4<\/p>\n\n\n\n

= 1764<\/p>\n\n\n\n

(iii) 862<\/sup> = (80 + 6)2<\/sup><\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 802<\/sup> + 2 \u00d7 80 \u00d7 6 + 62<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 6400 + 960 + 36<\/p>\n\n\n\n

= 7396<\/p>\n\n\n\n

(iv) 942<\/sup> = (90 + 4)2<\/sup><\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

= 902<\/sup> + 2 \u00d7 90 \u00d7 4 + 42<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 8100 + 720 + 16<\/p>\n\n\n\n

= 8836<\/p>\n\n\n\n

10. Find the squares of the following numbers containing 5 in unit\u2019s place:<\/strong><\/p>\n\n\n\n

(i) 45<\/strong><\/p>\n\n\n\n

(ii) 305<\/strong><\/p>\n\n\n\n

(iii) 525<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 452<\/sup> = n52<\/sup><\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

= n (n + 1) hundred + 52<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 4 \u00d7 5 hundred + 25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2000 + 25<\/p>\n\n\n\n

= 2025<\/p>\n\n\n\n

(ii) 3052<\/sup> = (30 \u00d7 31) hundred + 25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 93000 + 25<\/p>\n\n\n\n

= 93025<\/p>\n\n\n\n

(iii) 5252<\/sup> = (52 \u00d7 53) hundred + 25<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 275600 + 25<\/p>\n\n\n\n

= 275625<\/p>\n\n\n\n

11. Write a Pythagorean triplet whose one number is<\/strong><\/p>\n\n\n\n

(i) 8<\/strong><\/p>\n\n\n\n

(ii) 15<\/strong><\/p>\n\n\n\n

(iii) 63<\/strong><\/p>\n\n\n\n

(iv) 80<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 8<\/p>\n\n\n\n

Take n = 8<\/p>\n\n\n\n

So the triplet will be 2n, n2<\/sup> \u2013 1, n2<\/sup> + 1<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

If 2n = 8, then n = 8\/2 = 4<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

n2<\/sup> \u2013 1 = 42<\/sup> \u2013 1 = 16 \u2013 1 = 15<\/p>\n\n\n\n

n2<\/sup> + 1 = 42<\/sup> + 1 = 16 + 1 = 17<\/p>\n\n\n\n

Therefore, the triplets are 8, 15 and 17.<\/p>\n\n\n\n

(ii) 15<\/p>\n\n\n\n

Take 2n = 15<\/p>\n\n\n\n

So n = n\/2 is not possible<\/p>\n\n\n\n

n2<\/sup> \u2013 1 = 15<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

n2<\/sup> = 15 + 1 = 16 = 42<\/sup><\/p>\n\n\n\n

n = 4<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

2n = 2 \u00d7 4 = 8<\/p>\n\n\n\n

n2<\/sup> \u2013 1 = 15<\/p>\n\n\n\n

n2<\/sup> + 1 = 42<\/sup> + 1 = 16 + 1 = 17<\/p>\n\n\n\n

Therefore, the triplets are 8, 15 and 17.<\/p>\n\n\n\n

(iii) 63<\/p>\n\n\n\n

Take n2<\/sup> \u2013 1 = 63<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

n2<\/sup> = 63 + 1 = 64 = 82<\/sup><\/p>\n\n\n\n

So n = 8<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

2n = 2 \u00d7 8 = 16<\/p>\n\n\n\n

n2<\/sup> \u2013 1 = 63<\/p>\n\n\n\n

n2<\/sup> + 1 = 82<\/sup> + 1 = 64 + 1 = 65<\/p>\n\n\n\n

Therefore, the triplets are 16, 63 and 65.<\/p>\n\n\n\n

(iv) 80<\/p>\n\n\n\n

Take 2n = 80<\/p>\n\n\n\n

n = 80\/2 = 40<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

n2<\/sup> \u2013 1 = 402<\/sup> \u2013 1 = 1600 \u2013 1 = 1599<\/p>\n\n\n\n

n2<\/sup> + 1 = 402<\/sup> + 1 = 1600 + 1 = 1601<\/p>\n\n\n\n

Therefore, the triplets are 80, 1599 and 1601.<\/p>\n\n\n\n

12. Observe the following pattern and find the missing digits:<\/strong><\/p>\n\n\n\n

212<\/sup> = 441<\/strong><\/p>\n\n\n\n

2012<\/sup> = 40401<\/strong><\/p>\n\n\n\n

20012<\/sup> = 4004001<\/strong><\/p>\n\n\n\n

200012<\/sup> = 4\u20144\u20141<\/strong><\/p>\n\n\n\n

2000012<\/sup> = \u2014\u2014<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

212<\/sup> = 441<\/p>\n\n\n\n

2012<\/sup> = 40401<\/p>\n\n\n\n

20012<\/sup> = 4004001<\/p>\n\n\n\n

200012<\/sup> = 400040001<\/p>\n\n\n\n

2000012<\/sup> = 40000400001<\/p>\n\n\n\n

13. Observe the following pattern and find the missing digits:<\/strong><\/p>\n\n\n\n

92<\/sup> = 81<\/strong><\/p>\n\n\n\n

992<\/sup> = 9801<\/strong><\/p>\n\n\n\n

9992<\/sup> = 998001<\/strong><\/p>\n\n\n\n

99992<\/sup> = 99980001<\/strong><\/p>\n\n\n\n

999992<\/sup> = 9\u20148\u201401<\/strong><\/p>\n\n\n\n

9999992<\/sup> = 9\u20140\u20141<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

92<\/sup> = 81<\/p>\n\n\n\n

992<\/sup> = 9801<\/p>\n\n\n\n

9992<\/sup> = 998001<\/p>\n\n\n\n

99992<\/sup> = 99980001<\/p>\n\n\n\n

999992<\/sup> = 9999800001<\/p>\n\n\n\n

9999992<\/sup> = 9999998000001<\/p>\n\n\n\n

14. Observe the following pattern and find the missing digits:<\/strong><\/p>\n\n\n\n

72<\/sup> = 49<\/strong><\/p>\n\n\n\n

672<\/sup> = 4489<\/strong><\/p>\n\n\n\n

6672<\/sup> = 444889<\/strong><\/p>\n\n\n\n

66672<\/sup> = 44448889<\/strong><\/p>\n\n\n\n

666672<\/sup> = 4\u2014-8\u2014\u20139<\/strong><\/p>\n\n\n\n

6666672<\/sup> = 4\u2014-8\u2014-8-<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

72<\/sup> = 49<\/p>\n\n\n\n

672<\/sup> = 4489<\/p>\n\n\n\n

6672<\/sup> = 444889<\/p>\n\n\n\n

66672<\/sup> = 44448889<\/p>\n\n\n\n

666672<\/sup> = 4444488889<\/p>\n\n\n\n

6666672<\/sup> = 4444448888889<\/p>\n\n\n\n

Exercise 3.3<\/h4>\n\n\n\n

1. By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root:<\/strong><\/p>\n\n\n\n

(i) 121<\/strong><\/p>\n\n\n\n

(ii) 55<\/strong><\/p>\n\n\n\n

(iii) 36<\/strong><\/p>\n\n\n\n

(iv) 90<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that<\/p>\n\n\n\n

Square root of 121<\/p>\n\n\n\n

121 \u2013 1 = 120<\/p>\n\n\n\n

120 \u2013 3 = 117<\/p>\n\n\n\n

117 \u2013 5 = 112<\/p>\n\n\n\n

112 \u2013 7 = 105<\/p>\n\n\n\n

105 \u2013 9 = 96<\/p>\n\n\n\n

96 \u2013 11 = 85<\/p>\n\n\n\n

85 \u2013 13 = 72<\/p>\n\n\n\n

72 \u2013 15 = 57<\/p>\n\n\n\n

57 \u2013 17 = 40<\/p>\n\n\n\n

40 \u2013 19 = 21<\/p>\n\n\n\n

21 \u2013 21 = 0<\/p>\n\n\n\n

So the square root of 121 is 11<\/p>\n\n\n\n

Hence, 121 is a perfect square.<\/p>\n\n\n\n

(ii) We know that<\/p>\n\n\n\n

Square root of 55<\/p>\n\n\n\n

55 \u2013 1 = 54<\/p>\n\n\n\n

54 \u2013 3 = 51<\/p>\n\n\n\n

51 \u2013 5 = 46<\/p>\n\n\n\n

46 \u2013 7 = 39<\/p>\n\n\n\n

39 \u2013 9 = 30<\/p>\n\n\n\n

30 \u2013 11 = 19<\/p>\n\n\n\n

19 \u2013 13 = 6<\/p>\n\n\n\n

6 \u2013 15 = \u2013 9 is not possible<\/p>\n\n\n\n

Hence, 55 is not a perfect square.<\/p>\n\n\n\n

(iii) We know that<\/p>\n\n\n\n

Square root of 36<\/p>\n\n\n\n

36 \u2013 1 = 35<\/p>\n\n\n\n

35 \u2013 3 = 32<\/p>\n\n\n\n

32 \u2013 5 = 27<\/p>\n\n\n\n

27 \u2013 7 = 20<\/p>\n\n\n\n

20 \u2013 9 = 11<\/p>\n\n\n\n

11 \u2013 11 = 0<\/p>\n\n\n\n

Hence, 36 is a perfect square and its square root is 6.<\/p>\n\n\n\n

(iv) We know that<\/p>\n\n\n\n

Square root of 90<\/p>\n\n\n\n

90 \u2013 1 = 89<\/p>\n\n\n\n

89 \u2013 3 = 86<\/p>\n\n\n\n

86 \u2013 5 = 81<\/p>\n\n\n\n

81 \u2013 7 = 74<\/p>\n\n\n\n

74 \u2013 9 = 65<\/p>\n\n\n\n

65 \u2013 11 = 54<\/p>\n\n\n\n

54 \u2013 13 = 41<\/p>\n\n\n\n

41 \u2013 15 = 26<\/p>\n\n\n\n

26 \u2013 17 = 9<\/p>\n\n\n\n

9 \u2013 19 = \u2013 10 which is not possible<\/p>\n\n\n\n

Hence, 90 is not a perfect square.<\/p>\n\n\n\n

2. Find the square roots of the following numbers by prime factorization method:<\/strong><\/p>\n\n\n\n

(i) 784<\/strong><\/p>\n\n\n\n

(ii) 441<\/strong><\/p>\n\n\n\n

(iii) 1849<\/strong><\/p>\n\n\n\n

(iv) 4356<\/strong><\/p>\n\n\n\n

(v) 6241<\/strong><\/p>\n\n\n\n

(vi) 8836<\/strong><\/p>\n\n\n\n

(vii) 8281<\/strong><\/p>\n\n\n\n

(viii) 9025<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that<\/p>\n\n\n\n

Square root of 784<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 2 \u00d7 2 \u00d7 7<\/p>\n\n\n\n

= 28<\/p>\n\n\n\n

(ii) We know that<\/p>\n\n\n\n

Square root of 441<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 3 \u00d7 7<\/p>\n\n\n\n

= 21<\/p>\n\n\n\n

(iii) We know that<\/p>\n\n\n\n

Square root of 1849<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML= 43<\/p>\n\n\n\n

(iv) We know that<\/p>\n\n\n\n

Square root of 4356<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 2 \u00d7 3 \u00d7 11<\/p>\n\n\n\n

= 66<\/p>\n\n\n\n

(v) We know that<\/p>\n\n\n\n

Square root of 6241<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML= 79<\/p>\n\n\n\n

(vi) We know that<\/p>\n\n\n\n

Square root of 8836<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 2 \u00d7 47<\/p>\n\n\n\n

= 94<\/p>\n\n\n\n

(vii) We know that<\/p>\n\n\n\n

Square root of 8281<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 7 \u00d7 13<\/p>\n\n\n\n

= 91<\/p>\n\n\n\n

(viii) We know that<\/p>\n\n\n\n

Square root of 9025<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 5 \u00d7 19<\/p>\n\n\n\n

= 95<\/p>\n\n\n\n

3. Find the square roots of the following numbers by prime factorization method:<\/strong><\/p>\n\n\n\n

(i) 9 67\/121<\/strong><\/p>\n\n\n\n

(ii) 17 13\/36<\/strong><\/p>\n\n\n\n

(iii) 1.96<\/strong><\/p>\n\n\n\n

(iv) 0.0064<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 9 67\/121 = (9 \u00d7 121 + 67)\/ 121<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (1089 + 67)\/ 121<\/p>\n\n\n\n

= 1156\/121<\/p>\n\n\n\n

By squaring we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= (2 \u00d7 17)\/ 11<\/p>\n\n\n\n

= 34\/11<\/p>\n\n\n\n

= 3 1\/11<\/p>\n\n\n\n

(ii) 17 13\/36 = (17 \u00d7 36 + 13)\/ 36<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= (612 + 13)\/ 36<\/p>\n\n\n\n

= 625\/36<\/p>\n\n\n\n

By squaring we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= (5 \u00d7 5)\/ (2 \u00d7 3)<\/p>\n\n\n\n

= 25\/6<\/p>\n\n\n\n

= 4 1\/6<\/p>\n\n\n\n

(iii) 1.96 = 196\/100<\/p>\n\n\n\n

By squaring we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= (2 \u00d7 7)\/ (2 \u00d7 5)<\/p>\n\n\n\n

= 14\/10<\/p>\n\n\n\n

= 1.4<\/p>\n\n\n\n

(iv) 0.0064 = 64\/10000<\/p>\n\n\n\n

By squaring we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= (2 \u00d7 2 \u00d7 2)\/ (2 \u00d7 2 \u00d7 5 \u00d7 5)<\/p>\n\n\n\n

= 8\/100<\/p>\n\n\n\n

= 0.08<\/p>\n\n\n\n

4. For each of the following numbers, find the smallest natural number by which it should be multiplied so as to get a perfect square. Also, find the square root of the square number so obtained:<\/strong><\/p>\n\n\n\n

(i) 588<\/strong><\/p>\n\n\n\n

(ii) 720<\/strong><\/p>\n\n\n\n

(iii) 2178<\/strong><\/p>\n\n\n\n

(iv) 3042<\/strong><\/p>\n\n\n\n

(v) 6300<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 588 = 2 \u00d7 2 \u00d7 3 \u00d7 7 \u00d7 7<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 3 is left unpaired.<\/p>\n\n\n\n

So to make it a pair we must multiply it by 3<\/p>\n\n\n\n

Required least number = 3<\/p>\n\n\n\n

Square root of 588 \u00d7 3 = 1764<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

2 \u00d7 3 \u00d7 7 = 42<\/p>\n\n\n\n

(ii) 720 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 5 is left unpaired.<\/p>\n\n\n\n

So to make it a pair we must multiply it by 5<\/p>\n\n\n\n

Required least number = 5<\/p>\n\n\n\n

Square root of 720 \u00d7 5 = 3600<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

2 \u00d7 2 \u00d7 3 \u00d7 5 = 60<\/p>\n\n\n\n

(iii) 2178 = 2 \u00d7 3 \u00d7 3 \u00d7 11 \u00d7 11<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 2 is left unpaired.<\/p>\n\n\n\n

So to make it a pair we must multiply it by 2<\/p>\n\n\n\n

Required least number = 2<\/p>\n\n\n\n

Square root of 2178 \u00d7 2 = 4356<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

2 \u00d7 3 \u00d7 11 = 66<\/p>\n\n\n\n

(iv) 3042 = 2 \u00d7 3 \u00d7 3 \u00d7 13 \u00d7 13<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 2 is left unpaired<\/p>\n\n\n\n

So to make it a pair we must multiply it by 2<\/p>\n\n\n\n

Required least number = 2<\/p>\n\n\n\n

Square root of 3042 \u00d7 2 = 6084<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

2 \u00d7 3 \u00d7 13 = 78<\/p>\n\n\n\n

(v) 6300 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 7<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 7 is left unpaired<\/p>\n\n\n\n

So to make it a pair we must multiply it by 7<\/p>\n\n\n\n

Required least number = 7<\/p>\n\n\n\n

Square root of 6300 \u00d7 7 = 44100<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

2 \u00d7 3 \u00d7 5 \u00d7 7 = 210<\/p>\n\n\n\n

5. For each of the following numbers, find the smallest natural number by which it should be divided so that this quotient is a perfect square. Also, find the square root of the square number so obtained:<\/strong><\/p>\n\n\n\n

(i) 1872<\/strong><\/p>\n\n\n\n

(ii) 2592<\/strong><\/p>\n\n\n\n

(iii) 3380<\/strong><\/p>\n\n\n\n

(iv) 16224<\/strong><\/p>\n\n\n\n

(v) 61347<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 1872 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 13<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 13 is left unpaired<\/p>\n\n\n\n

Required least number = 13<\/p>\n\n\n\n

The number 1872 should be divided by 13 so that the resultant number will be a perfect square<\/p>\n\n\n\n

Resultant number = 1872 \u00f7 13 = 144<\/p>\n\n\n\n

Square root = 2 \u00d7 2 \u00d7 3 = 12<\/p>\n\n\n\n

(ii) 2592 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 2 is left unpaired<\/p>\n\n\n\n

Required least number = 2<\/p>\n\n\n\n

The number 2592 should be divided by 2 so that the resultant number will be a perfect square<\/p>\n\n\n\n

Resultant number = 2592 \u00f7 2 = 1296<\/p>\n\n\n\n

Square root = 2 \u00d7 2 \u00d7 3 \u00d7 3 = 36<\/p>\n\n\n\n

(iii) 3380 = 2 \u00d7 2 \u00d7 5 \u00d7 13 \u00d7 13<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 5 is left unpaired<\/p>\n\n\n\n

Required least number = 5<\/p>\n\n\n\n

The number 3380 should be divided by 5 so that the resultant number will be a perfect square<\/p>\n\n\n\n

Resultant number = 3380 \u00f7 5 = 676<\/p>\n\n\n\n

Square root = 2 \u00d7 13 = 26<\/p>\n\n\n\n

(iv ) 16224 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 13 \u00d7 13<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, two factors 2 and 3 is left unpaired<\/p>\n\n\n\n

Required least number = 2 \u00d7 3 = 6<\/p>\n\n\n\n

The number 16224 should be divided by 6 so that the resultant number will be a perfect square<\/p>\n\n\n\n

Resultant number = 16224 \u00f7 6 = 2704<\/p>\n\n\n\n

Square root = 2 \u00d7 2 \u00d7 13 = 52<\/p>\n\n\n\n

(v) 61347 = 3 \u00d7 11 \u00d7 11 \u00d7 13 \u00d7 13<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By pairing the same kind of factors, one factor 3 is left unpaired<\/p>\n\n\n\n

Required least number = 3<\/p>\n\n\n\n

The number 61347 should be divided by 3 so that the resultant number will be a perfect square<\/p>\n\n\n\n

Resultant number = 61347 \u00f7 3 = 20449<\/p>\n\n\n\n

Square root = 11 \u00d7 13 = 143<\/p>\n\n\n\n

6. Find the smallest square number that is divisible by each of the following numbers:<\/strong><\/p>\n\n\n\n

(i) 3, 6, 10, 15<\/strong><\/p>\n\n\n\n

(ii) 6, 9, 27, 36<\/strong><\/p>\n\n\n\n

(iii) 4, 7, 8, 16<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 3, 6, 10, 15<\/p>\n\n\n\n

Number which is divisible by<\/p>\n\n\n\n

3, 6, 10, 15 = LCM of 3, 6, 10, 15<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 2 \u00d7 3 \u00d7 5<\/p>\n\n\n\n

= 30<\/p>\n\n\n\n

(ii) 6, 9, 27, 36<\/p>\n\n\n\n

Number which is divisible by<\/p>\n\n\n\n

6, 9, 27, 36 = LCM of 6, 9, 27, 36<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 3 \u00d7 3 \u00d7 2 \u00d7 2 \u00d7 3<\/p>\n\n\n\n

= 108<\/p>\n\n\n\n

Here the smallest square<\/p>\n\n\n\n

= 108 \u00d7 3<\/p>\n\n\n\n

= 324<\/p>\n\n\n\n

(iii) 4, 7, 8, 16<\/p>\n\n\n\n

Number which is divisible by<\/p>\n\n\n\n

4, 7, 8, 16 = LCM of 4, 7, 8, 16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 7<\/p>\n\n\n\n

= 112<\/p>\n\n\n\n

Here the smallest square<\/p>\n\n\n\n

= 112 \u00d7 7<\/p>\n\n\n\n

= 784<\/p>\n\n\n\n

7. 4225 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Total number of plants = 4225<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

Number of rows = Number of plant in each row<\/p>\n\n\n\n

So the number of rows = square root of 4225<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

= 5 \u00d7 13<\/p>\n\n\n\n

= 65<\/p>\n\n\n\n

Hence, the number of rows is 65 and the number of plants in each row is 65.<\/p>\n\n\n\n

8. The area of rectangle is 1936 sq. m. If the length of the rectangle is 4 times its breadth, find the dimensions of the rectangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Area of rectangle = 1936 sq. m<\/p>\n\n\n\n

Take breadth = x m<\/p>\n\n\n\n

Length = 4x m<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

4x2<\/sup> = 1936<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x2<\/sup> = 1936\/4 = 484<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2 \u00d7 11<\/p>\n\n\n\n

= 22<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

Length = 4x = 4 \u00d7 22 = 88 m<\/p>\n\n\n\n

Breadth = x = 22 m<\/p>\n\n\n\n

9. In a school a P.T. teacher wants to arrange 2000 students in the form of rows and columns for P.T. display. If the number of rows is equal to number of columns and 64 students could not be accommodated in this arrangement. Find the number of rows.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Total number of students in a school = 2000<\/p>\n\n\n\n

The P.T. teacher arranges in such a way that<\/p>\n\n\n\n

No. of rows = no. of students in each row<\/p>\n\n\n\n

So 64 students are left<\/p>\n\n\n\n

Required number of students = 2000 \u2013 64 = 1936<\/p>\n\n\n\n

No. of rows = \u221a1936<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2 \u00d7 2 \u00d7 11<\/p>\n\n\n\n

= 44<\/p>\n\n\n\n

10. In a school, the students of class VIII collected \u20b92304 for a picnic. Each student contributed as many rupees as the number of students in the class. Find the number of students in the class.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Amount collected for picnic = \u20b92304<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

No. of students = no. of rupees contributed by each student = \u221a2304<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 34= 48<\/p>\n\n\n\n

Therefore, the number of students in class VIII is 4811.<\/p>\n\n\n\n

11. The product of two numbers is 7260. If one number is 15 times the other number, find the numbers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Product of two numbers = 7260<\/p>\n\n\n\n

Consider one number = x<\/p>\n\n\n\n

Second number = 15x<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

15x \u00d7 x = 7260<\/p>\n\n\n\n

15x2<\/sup> = 7260<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x2<\/sup> = 7260\/15 = 484<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

= 2 \u00d7 11<\/p>\n\n\n\n

= 22<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

One number = 22<\/p>\n\n\n\n

Second number = 22 \u00d7 15 = 330<\/p>\n\n\n\n

12. Find three positive numbers in the ratio 2: 3: 5, the sum of whose squares is 950.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Ratio of three positive numbers = 2: 3: 5<\/p>\n\n\n\n

Sum of their squares = 950<\/p>\n\n\n\n

Consider<\/p>\n\n\n\n

First number = 2x<\/p>\n\n\n\n

Second number = 3x<\/p>\n\n\n\n

Third number = 5x<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

(2x)2<\/sup>+ (3x)2<\/sup> + (5x)2<\/sup> = 950<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

4x2<\/sup> + 9x2<\/sup> + 25x2<\/sup> = 950<\/p>\n\n\n\n

38x2<\/sup> = 950<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

X2<\/sup> = 950\/38 = 25<\/p>\n\n\n\n

x = \u221a25 = 5<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

First number = 2 \u00d7 5 = 10<\/p>\n\n\n\n

Second number = 3 \u00d7 5 = 15<\/p>\n\n\n\n

Third number = 5 \u00d7 5 = 25<\/p>\n\n\n\n

13. The perimeter of two squares is 60 metres and 144 metres respectively. Find the perimeter of another square equal in area to the sum of the first two squares.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Perimeter of first square = 60 m<\/p>\n\n\n\n

Side = 60\/4 = 15 m<\/p>\n\n\n\n

Perimeter of second square = 144 m<\/p>\n\n\n\n

Side = 144\/4 = 36 m<\/p>\n\n\n\n

So the sum of perimeters of two squares = 60 + 144 = 204 m<\/p>\n\n\n\n

Sum of areas of these two squares = 152<\/sup> + 362<\/sup><\/p>\n\n\n\n

= 225 + 1296<\/p>\n\n\n\n

= 1521 m2<\/sup><\/p>\n\n\n\n

Here<\/p>\n\n\n\n

Area of third square = 1521 m2<\/sup><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

So we get<\/p>\n\n\n\n

Side = \u221aArea = \u221a1521<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

= 3 \u00d7 13<\/p>\n\n\n\n

= 39 m<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

Perimeter = 4 \u00d7 side<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 4 \u00d7 39<\/p>\n\n\n\n

= 156 m<\/p>\n\n\n\n

Exercise 3.4<\/h4>\n\n\n\n

1. Find the square root of each of the following by division method:<\/strong><\/p>\n\n\n\n

(i) 2401<\/strong><\/p>\n\n\n\n

(ii) 4489<\/strong><\/p>\n\n\n\n

(iii) 106929<\/strong><\/p>\n\n\n\n

(iv) 167281<\/strong><\/p>\n\n\n\n

(v) 53824<\/strong><\/p>\n\n\n\n

(vi) 213444<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) \u221a2401 = 49<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(ii) \u221a4489 = 67<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iii) \u221a106929 = 327<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iv) \u221a167281 = 409<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(v) \u221a53824 = 232<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(vi) \u221a213444 = 462<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

2. Find the number of digits in the square root of each of the following (without any calculation):<\/strong><\/p>\n\n\n\n

(i) 81<\/strong><\/p>\n\n\n\n

(ii) 169<\/strong><\/p>\n\n\n\n

(iii) 4761<\/strong><\/p>\n\n\n\n

(iv) 27889<\/strong><\/p>\n\n\n\n

(v) 525625<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 81<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

In 81, a group of two\u2019s is 1<\/p>\n\n\n\n

Therefore, its square root has one digit.<\/p>\n\n\n\n

(ii) 169<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

In 169, group of two\u2019s are 2<\/p>\n\n\n\n

Therefore, its square root has two digits.<\/p>\n\n\n\n

(iii) 4761<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

In 4761, group of two\u2019s are 2<\/p>\n\n\n\n

Therefore, its square root has two digits.<\/p>\n\n\n\n

(iv) 27889<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

In 27889, groups of two\u2019s are 3<\/p>\n\n\n\n

Therefore, its square root has three digits.<\/p>\n\n\n\n

(v) 525625<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

In 525625, groups of two\u2019s are 3<\/p>\n\n\n\n

Therefore, its square root has three digits.<\/p>\n\n\n\n

3. Find the square root of the following decimal numbers by division method:<\/strong><\/p>\n\n\n\n

(i) 51.84<\/strong><\/p>\n\n\n\n

(ii) 42.25<\/strong><\/p>\n\n\n\n

(iii) 18.4041<\/strong><\/p>\n\n\n\n

(iv) 5.774409<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) \u221a51.84 = 7.2<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(ii) \u221a42.25 = 6.5<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iii) \u221a18.4041 = 4.29<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iv) \u221a5.774409 = 2.403<\/p>\n\n\n\n

By division method<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

4. Find the square root of the following numbers correct to two decimal places:<\/strong><\/p>\n\n\n\n

(i) 645.8<\/strong><\/p>\n\n\n\n

(ii) 107.45<\/strong><\/p>\n\n\n\n

(iii) 5.462<\/strong><\/p>\n\n\n\n

(iv) 2<\/strong><\/p>\n\n\n\n

(v) 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) \u221a645.8 = 25.41<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(ii) \u221a107.45 = 10.36<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iii) \u221a5.462 = 2.337 = 2.34<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iv) \u221a2 = 1.41<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(v) \u221a3 = 1.73<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

5. Find the square root of the following fractions by division method:<\/strong><\/p>\n\n\n\n

(i) 841\/1521<\/strong><\/p>\n\n\n\n

(ii) 8 257\/529<\/strong><\/p>\n\n\n\n

(iii) 16 169\/441<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 841\/1521<\/p>\n\n\n\n

By squaring<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(ii) 8 257\/529<\/p>\n\n\n\n

By squaring<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iii) 16 169\/441<\/p>\n\n\n\n

By squaring<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

It can be written as<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

6. Find the least number which must be subtracted from each of the following numbers to make them a perfect square. Also find the square root of the perfect square number so obtained:<\/strong><\/p>\n\n\n\n

(i) 2000<\/strong><\/p>\n\n\n\n

(ii) 984<\/strong><\/p>\n\n\n\n

(iii) 8934<\/strong><\/p>\n\n\n\n

(iv) 11021<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 2000<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root, 64 is left as remainder<\/p>\n\n\n\n

Subtracting 64 from 2000<\/p>\n\n\n\n

We get 1936 which is a perfect square and its square root is 44.<\/p>\n\n\n\n

(ii) 984<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root, 23 is left as remainder<\/p>\n\n\n\n

Subtracting 23 from 984<\/p>\n\n\n\n

We get 961 which is a perfect square and its square root is 31.<\/p>\n\n\n\n

(iii) 8934<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root, 98 is left as remainder<\/p>\n\n\n\n

Subtracting 98 from 894<\/p>\n\n\n\n

We get 8934 \u2013 98 = 8836 which is a perfect square and its square root is 94.<\/p>\n\n\n\n

(iv) 11021<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root, 205 is left as remainder<\/p>\n\n\n\n

Subtracting 205 from 11021<\/p>\n\n\n\n

We get 11021 \u2013 205 = 10816 which is a perfect square and its square root is 104.<\/p>\n\n\n\n

7. Find the least number which must be added to each of the following numbers to make them a perfect square. Also, find the square root of the perfect square number so obtained:<\/strong><\/p>\n\n\n\n

(i) 1750<\/strong><\/p>\n\n\n\n

(ii) 6412<\/strong><\/p>\n\n\n\n

(iii) 6598<\/strong><\/p>\n\n\n\n

(iv) 8000<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 1750<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root<\/p>\n\n\n\n

412<\/sup> is less than 1750<\/p>\n\n\n\n

So by taking 422<\/sup><\/p>\n\n\n\n

164 \u2013 150 = 14 less<\/p>\n\n\n\n

Adding 14 we get a square of 42 which is 1764.<\/p>\n\n\n\n

(ii) 6412<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root<\/p>\n\n\n\n

802<\/sup> is less than 6412<\/p>\n\n\n\n

So by taking 812<\/sup><\/p>\n\n\n\n

161 \u2013 12 = 14 less<\/p>\n\n\n\n

Adding 149 we get a square of 81 which is 6561.<\/p>\n\n\n\n

(iii) 6598<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root<\/p>\n\n\n\n

812<\/sup> is less than 6598<\/p>\n\n\n\n

So by taking 822<\/sup><\/p>\n\n\n\n

324 \u2013 198 = 126 less<\/p>\n\n\n\n

Adding 126 we get a square of 82 which is 6724.<\/p>\n\n\n\n

(iv) 8000<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root<\/p>\n\n\n\n

892<\/sup> is less than 8000<\/p>\n\n\n\n

So by taking 902<\/sup><\/p>\n\n\n\n

8100 \u2013 8000 = 100 less<\/p>\n\n\n\n

Adding 100 we get a square of 90 which is 8100.<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

8. Find the smallest four-digit number which is a perfect square.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Smallest four \u2013 digit number = 1000<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root, we find that 39 is left.<\/p>\n\n\n\n

If we subtract any number from 1000 we get 3 digit number<\/p>\n\n\n\n

Take 322<\/sup> = 1024<\/p>\n\n\n\n

Here 1024 \u2013 1000 = 24 is to be added to get a perfect square of least 4 digit number<\/p>\n\n\n\n

Therefore, the required 4 digit smallest number is 1024.<\/p>\n\n\n\n

9. Find the greatest number of six digits which is a perfect square.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Greatest six digit number = 999999<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

By taking square root, we find that 1998 is left<\/p>\n\n\n\n

If we subtract 1998 from 999999 we get 998001 which is a perfect square.<\/p>\n\n\n\n

Therefore, required six digit greatest number is 998001.<\/p>\n\n\n\n

10. In a right triangle ABC, \u2220B = 900<\/sup>.<\/strong><\/p>\n\n\n\n

(i) If AB = 14 cm, BC = 48 cm, find AC.<\/strong><\/p>\n\n\n\n

(ii) If AC = 37 cm, BC = 35 cm, find AB.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) In a right angled triangle ABC<\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

AB = 14 cm and BC = 48 cm<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Using Pythagoras theorem<\/p>\n\n\n\n

AC2<\/sup> = AB2<\/sup> + BC2<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 142<\/sup> + 482<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 196 + 2304<\/p>\n\n\n\n

= 2500<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

AC = \u221a2500 = 50 cm<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(ii) In right triangle ABC<\/p>\n\n\n\n

B = 900<\/sup>, AC = 37 cm, BC = 35 cm<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Using Pythagoras Theorem<\/p>\n\n\n\n

AC2<\/sup> = AB2<\/sup> + BC2<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

372<\/sup> = AB2<\/sup> + 352<\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

1369 = AB2<\/sup> + 1225<\/p>\n\n\n\n

AB2<\/sup> = 1369 \u2013 1225 = 144<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

AB = \u221a144 = 12 cm<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

11. A gardener has 1400 plants. He wants to plant these in such a way that the number of rows and number of columns remains same. Find the minimum number of plants he needs more for this.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Total number of plants = 1400<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Here<\/p>\n\n\n\n

Number of columns = Number of rows<\/p>\n\n\n\n

By taking the square root of 1400<\/p>\n\n\n\n

372<\/sup> < 1400<\/p>\n\n\n\n

So take 382<\/sup> = 1444<\/p>\n\n\n\n

We need 1444 \u2013 1400 = 44 plants more<\/p>\n\n\n\n

Therefore, the minimum number of plants he needs more for this is 44.<\/p>\n\n\n\n

12. There are 1000 children in a school. For a P.T. drill they have to stand in such a way that the number of rows is equal to number of columns. How many children would be left out in this arrangement?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

No. of total children in a school = 1000<\/p>\n\n\n\n

For a P.T. drill, children have to stand in such a way that<\/p>\n\n\n\n

No. of rows = No. of columns<\/p>\n\n\n\n

Take the square root of 1000<\/p>\n\n\n\n

39 is left as remainder<\/p>\n\n\n\n

Left out children = 39<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Hence, 39 children would be left out in this arrangement.<\/p>\n\n\n\n

13. Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend\u2019s house. While returning, he walks diagonally from his friend\u2019s house to reach back to his house. What distance did he walk while returning?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Amit walks 16 m south from his house and turns east to walk 63 m to reach his friend\u2019s house<\/p>\n\n\n\n

Consider O as the house and A and B as the places<\/p>\n\n\n\n

OA = 16 m, AO = 63 m<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Using Pythagoras theorem<\/p>\n\n\n\n

OB2<\/sup> = OA2<\/sup> + AB2<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 162<\/sup> + 632<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

= 256 + 3969<\/p>\n\n\n\n

= 4225<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

OB = \u221a4225 = 65<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Therefore, Amit has to walk 65 m to reach his house.<\/p>\n\n\n\n

14. A ladder 6 m long leaned against a wall. The ladder reaches the wall to a height of 4.8 m. Find the distance between the wall and the foot of the ladder.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Length of ladder = 6 m<\/p>\n\n\n\n

Ladder reaches the wall to a height of 4.8 m<\/p>\n\n\n\n

Consider AB as the ladder and AC as the height of the wall<\/p>\n\n\n\n

AB = 6 m and AC = 4.8 m<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Distance between the foot of ladder and wall is BC<\/p>\n\n\n\n

Using Pythagoras theorem,<\/p>\n\n\n\n

AB2<\/sup> = AC2<\/sup> + BC2<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

62<\/sup> = 4.82<\/sup> + BC2<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

BC2<\/sup> = 62<\/sup> \u2013 4.82<\/sup><\/p>\n\n\n\n

BC2<\/sup> = 36 \u2013 23.04 = 12.96<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

BC = \u221a12.96 = 3.6 m<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

Hence, the distance between the wall and the foot of the ladder is 3.6 m.<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 3- Squares and Square Roots<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 8 Maths Chapter 3- Squares and Square Roots PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n