{"id":546014,"date":"2021-10-06T04:28:54","date_gmt":"2021-10-06T04:28:54","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=546014"},"modified":"2021-10-07T04:56:39","modified_gmt":"2021-10-07T04:56:39","slug":"rd-sharma-solutions-for-class-8-maths-chapter-3-squares-and-square-roots","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-8-maths-chapter-3-squares-and-square-roots\/","title":{"rendered":"RD Sharma Solutions for Class 8 Maths Chapter 3\u2013Squares and Square Roots"},"content":{"rendered":"\n

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

RD Sharma Solutions for Class 8 Maths Chapter 3\u2013Squares and Square Roots<\/h2>\n\n\n\n

RD Sharma 8th Maths Chapter 3, Class 8 Maths Chapter 3 solutions<\/p>\n\n\n\n

EXERCISE 3.1 PAGE NO: 3.4<\/h4>\n\n\n\n

1. Which of the following numbers are perfect squares?<\/strong><\/p>\n\n\n\n

(i) 484<\/strong><\/p>\n\n\n\n

(ii) 625<\/strong><\/p>\n\n\n\n

(iii) 576<\/strong><\/p>\n\n\n\n

(iv) 941<\/strong><\/p>\n\n\n\n

(v) 961<\/strong><\/p>\n\n\n\n

(vi) 2500<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>484<\/p>\n\n\n\n

First find the prime factors for 484<\/p>\n\n\n\n

484 = 2\u00d72\u00d711\u00d711<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (11\u00d711)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 484 is a perfect square.<\/p>\n\n\n\n

(ii)<\/strong> 625<\/p>\n\n\n\n

First find the prime factors for 625<\/p>\n\n\n\n

625 = 5\u00d75\u00d75\u00d75<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (5\u00d75) \u00d7 (5\u00d75)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 625 is a perfect square.<\/p>\n\n\n\n

(iii)<\/strong> 576<\/p>\n\n\n\n

First find the prime factors for 576<\/p>\n\n\n\n

576 = 2\u00d72\u00d72\u00d72\u00d72\u00d72\u00d73\u00d73<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (3\u00d73)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 576 is a perfect square.<\/p>\n\n\n\n

(iv)<\/strong> 941<\/p>\n\n\n\n

First find the prime factors for 941<\/p>\n\n\n\n

941 = 941 \u00d7 1<\/p>\n\n\n\n

We know that 941 itself is a prime factor.<\/p>\n\n\n\n

\u2234 941 is not a perfect square.<\/p>\n\n\n\n

(v)<\/strong> 961<\/p>\n\n\n\n

First find the prime factors for 961<\/p>\n\n\n\n

961 = 31\u00d731<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (31\u00d731)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 961 is a perfect square.<\/p>\n\n\n\n

(vi)<\/strong> 2500<\/p>\n\n\n\n

First find the prime factors for 2500<\/p>\n\n\n\n

2500 = 2\u00d72\u00d75\u00d75\u00d75\u00d75<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (5\u00d75) \u00d7 (5\u00d75)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 2500 is 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 in each case:
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 1156<\/p>\n\n\n\n

First find the prime factors for 1156<\/p>\n\n\n\n

1156 = 2\u00d72\u00d717\u00d717<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (17\u00d717)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 1156 is a perfect square.<\/p>\n\n\n\n

To find the square of the given number<\/p>\n\n\n\n

1156 = (2\u00d717) \u00d7 (2\u00d717)<\/p>\n\n\n\n

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

= (34)2<\/sup><\/p>\n\n\n\n

\u2234 1156 is a square of 34.<\/p>\n\n\n\n

(ii)<\/strong> 2025<\/p>\n\n\n\n

First find the prime factors for 2025<\/p>\n\n\n\n

2025 = 3\u00d73\u00d73\u00d73\u00d75\u00d75<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (3\u00d73) \u00d7 (3\u00d73) \u00d7 (5\u00d75)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 2025 is a perfect square.<\/p>\n\n\n\n

To find the square of the given number<\/p>\n\n\n\n

2025 = (3\u00d73\u00d75) \u00d7 (3\u00d73\u00d75)<\/p>\n\n\n\n

= 45 \u00d7 45<\/p>\n\n\n\n

= (45)2<\/sup><\/p>\n\n\n\n

\u2234 2025 is a square of 45.<\/p>\n\n\n\n

(iii)<\/strong> 14641<\/p>\n\n\n\n

First find the prime factors for 14641<\/p>\n\n\n\n

14641 = 11\u00d711\u00d711\u00d711<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (11\u00d711) \u00d7 (11\u00d711)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 14641 is a perfect square.<\/p>\n\n\n\n

To find the square of the given number<\/p>\n\n\n\n

14641 = (11\u00d711) \u00d7 (11\u00d711)<\/p>\n\n\n\n

= 121 \u00d7 121<\/p>\n\n\n\n

= (121)2<\/sup><\/p>\n\n\n\n

\u2234 14641 is a square of 121.<\/p>\n\n\n\n

(iv)<\/strong> 4761<\/p>\n\n\n\n

First find the prime factors for 4761<\/p>\n\n\n\n

4761 = 3\u00d73\u00d723\u00d723<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (3\u00d73) \u00d7 (23\u00d723)<\/p>\n\n\n\n

By observation, none of the prime factors are left out.<\/p>\n\n\n\n

\u2234 4761 is a perfect square.<\/p>\n\n\n\n

To find the square of the given number<\/p>\n\n\n\n

4761 = (3\u00d723) \u00d7 (3\u00d723)<\/p>\n\n\n\n

= 69 \u00d7 69<\/p>\n\n\n\n

= (69)2<\/sup><\/p>\n\n\n\n

\u2234 4761 is a square of 69.<\/p>\n\n\n\n

3. Find the smallest number by which the given number must be multiplied so that the product is a perfect square:
(i) 23805
(ii) 12150
(iii) 7688<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 23805<\/p>\n\n\n\n

First find the prime factors for 23805<\/p>\n\n\n\n

23805 = 3\u00d73\u00d723\u00d723\u00d75<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (3\u00d73) \u00d7 (23\u00d723) \u00d7 5<\/p>\n\n\n\n

By observation, prime factor 5 is left out.<\/p>\n\n\n\n

So, multiply by 5 we get,<\/p>\n\n\n\n

23805 \u00d7 5 = (3\u00d73) \u00d7 (23\u00d723) \u00d7 (5\u00d75)<\/p>\n\n\n\n

= (3\u00d75\u00d723) \u00d7 (3\u00d75\u00d723)<\/p>\n\n\n\n

= 345 \u00d7 345<\/p>\n\n\n\n

= (345)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 345.<\/p>\n\n\n\n

(ii)<\/strong> 12150<\/p>\n\n\n\n

First find the prime factors for 12150<\/p>\n\n\n\n

12150 = 2\u00d73\u00d73\u00d73\u00d73\u00d73\u00d75\u00d75<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= 2\u00d73 \u00d7 (3\u00d73) \u00d7 (3\u00d73) \u00d7 (5\u00d75)<\/p>\n\n\n\n

By observation, prime factor 2 and 3 are left out.<\/p>\n\n\n\n

So, multiply by 2\u00d73 = 6 we get,<\/p>\n\n\n\n

12150 \u00d7 6 = 2\u00d73 \u00d7 (3\u00d73) \u00d7 (3\u00d73) \u00d7 (5\u00d75) \u00d7 2 \u00d7 3<\/p>\n\n\n\n

= (2\u00d73\u00d73\u00d73\u00d75) \u00d7 (2\u00d73\u00d73\u00d73\u00d75)<\/p>\n\n\n\n

= 270 \u00d7 270<\/p>\n\n\n\n

= (270)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 270.<\/p>\n\n\n\n

(iii)<\/strong> 7688<\/p>\n\n\n\n

First find the prime factors for 7688<\/p>\n\n\n\n

7688 = 2\u00d72\u00d731\u00d731\u00d72<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (31\u00d731) \u00d7 2<\/p>\n\n\n\n

By observation, prime factor 2 is left out.<\/p>\n\n\n\n

So, multiply by 2 we get,<\/p>\n\n\n\n

7688 \u00d7 2 = (2\u00d72) \u00d7 (31\u00d731)\u00d7 (2\u00d72)<\/p>\n\n\n\n

= (2\u00d731\u00d72) \u00d7 (2\u00d731\u00d72)<\/p>\n\n\n\n

= 124 \u00d7 124<\/p>\n\n\n\n

= (124)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 124.<\/p>\n\n\n\n

4. Find the smallest number by which the given number must be divided so that the resulting number is a perfect square:
(i) 14283
(ii) 1800
(iii) 2904<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 14283<\/p>\n\n\n\n

First find the prime factors for 14283<\/p>\n\n\n\n

14283 = 3\u00d73\u00d73\u00d723\u00d723<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (3\u00d73) \u00d7 (23\u00d723) \u00d7 3<\/p>\n\n\n\n

By observation, prime factor 3 is left out.<\/p>\n\n\n\n

So, divide by 3 to eliminate 3 we get,<\/p>\n\n\n\n

14283\/3 = (3\u00d73) \u00d7 (23\u00d723)<\/p>\n\n\n\n

= (3\u00d723) \u00d7 (3\u00d723)<\/p>\n\n\n\n

= 69 \u00d7 69<\/p>\n\n\n\n

= (69)2<\/sup><\/p>\n\n\n\n

\u2234 Resultant is the square of 69.<\/p>\n\n\n\n

(ii)<\/strong> 1800<\/p>\n\n\n\n

First find the prime factors for 1800<\/p>\n\n\n\n

1800 = 2\u00d72\u00d75\u00d75\u00d73\u00d73\u00d72<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (5\u00d75) \u00d7 (3\u00d73) \u00d7 2<\/p>\n\n\n\n

By observation, prime factor 2 is left out.<\/p>\n\n\n\n

So, divide by 2 to eliminate 2 we get,<\/p>\n\n\n\n

1800\/2 = (2\u00d72) \u00d7 (5\u00d75) \u00d7 (3\u00d73)<\/p>\n\n\n\n

= (2\u00d75\u00d73) \u00d7 (2\u00d75\u00d73)<\/p>\n\n\n\n

= 30 \u00d7 30<\/p>\n\n\n\n

= (30)2<\/sup><\/p>\n\n\n\n

\u2234 Resultant is the square of 30.<\/p>\n\n\n\n

(iii)<\/strong> 2904<\/p>\n\n\n\n

First find the prime factors for 2904<\/p>\n\n\n\n

2904 = 2\u00d72\u00d711\u00d711\u00d72\u00d73<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (11\u00d711) \u00d7 2 \u00d7 3<\/p>\n\n\n\n

By observation, prime factor 2 and 3 are left out.<\/p>\n\n\n\n

So, divide by 6 to eliminate 2 and 3 we get,<\/p>\n\n\n\n

2904\/6 = (2\u00d72) \u00d7 (11\u00d711)<\/p>\n\n\n\n

= (2\u00d711) \u00d7 (2\u00d711)<\/p>\n\n\n\n

= 22 \u00d7 22<\/p>\n\n\n\n

= (22)2<\/sup><\/p>\n\n\n\n

\u2234 Resultant is the square of 22.<\/p>\n\n\n\n

5. Which of the following numbers are perfect squares?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121<\/strong><\/p>\n\n\n\n

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

11 it is a prime number by itself.<\/p>\n\n\n\n

So it is not a perfect square.<\/p>\n\n\n\n

12 is not a perfect square.<\/p>\n\n\n\n

16= (4)2<\/sup><\/p>\n\n\n\n

16 is a perfect square.<\/p>\n\n\n\n

32 is not a perfect square.<\/p>\n\n\n\n

36= (6)2<\/sup><\/p>\n\n\n\n

36 is a perfect square.<\/p>\n\n\n\n

50 is not a perfect square.<\/p>\n\n\n\n

64= (8)2<\/sup><\/p>\n\n\n\n

64 is a perfect square.<\/p>\n\n\n\n

79 it is a prime number.<\/p>\n\n\n\n

So it is not a perfect square.<\/p>\n\n\n\n

81= (9)2<\/sup><\/p>\n\n\n\n

81 is a perfect square.<\/p>\n\n\n\n

111 it is a prime number.<\/p>\n\n\n\n

So it is not a perfect square.<\/p>\n\n\n\n

121= (11)2<\/sup><\/p>\n\n\n\n

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

6. Using prime factorization method, find which of the following numbers are perfect squares?
189, 225, 2048, 343, 441, 2961, 11025, 3549<\/strong><\/p>\n\n\n\n

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

189 prime factors are<\/p>\n\n\n\n

189 = 32<\/sup>\u00d73\u00d77<\/p>\n\n\n\n

Since it does not have equal pair of factors 189 is not a perfect square.<\/p>\n\n\n\n

225 prime factors are<\/p>\n\n\n\n

225 = (5\u00d75) \u00d7 (3\u00d73)<\/p>\n\n\n\n

Since 225 has equal pair of factors. \u2234 It is a perfect square.<\/p>\n\n\n\n

2048 prime factors are<\/p>\n\n\n\n

2048 = (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 2<\/p>\n\n\n\n

Since it does not have equal pair of factors 2048 is not a perfect square.<\/p>\n\n\n\n

343 prime factors are<\/p>\n\n\n\n

343 = (7\u00d77) \u00d7 7<\/p>\n\n\n\n

Since it does not have equal pair of factors 2048 is not a perfect square.<\/p>\n\n\n\n

441 prime factors are<\/p>\n\n\n\n

441 = (7\u00d77) \u00d7 (3\u00d73)<\/p>\n\n\n\n

Since 441 has equal pair of factors. \u2234 It is a perfect square.<\/p>\n\n\n\n

2961 prime factors are<\/p>\n\n\n\n

2961 = (3\u00d73) \u00d7 (3\u00d73) \u00d7 (3\u00d73) \u00d7 (2\u00d72)<\/p>\n\n\n\n

Since 2961 has equal pair of factors. \u2234 It is a perfect square.<\/p>\n\n\n\n

11025 prime factors are<\/p>\n\n\n\n

11025 = (3\u00d73) \u00d7 (5\u00d75) \u00d7 (7\u00d77)<\/p>\n\n\n\n

Since 11025 has equal pair of factors. \u2234 It is a perfect square.<\/p>\n\n\n\n

3549 prime factors are<\/p>\n\n\n\n

3549 = (13\u00d713) \u00d7 3 \u00d7 7<\/p>\n\n\n\n

Since it does not have equal pair of factors 3549 is not a perfect square.<\/p>\n\n\n\n

7. By what number should each of the following numbers by multiplied to get a perfect square in each case? Also find the number whose square is the new number.
<\/strong>
<\/a>(i) 8820<\/strong><\/p>\n\n\n\n

(ii) 3675
(iii) 605<\/strong><\/p>\n\n\n\n

(iv) 2880
(v) 4056<\/strong><\/p>\n\n\n\n

(vi) 3468
(vii) 7776<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 8820<\/p>\n\n\n\n

First find the prime factors for 8820<\/p>\n\n\n\n

8820 = 2\u00d72\u00d73\u00d73\u00d77\u00d77\u00d75<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (3\u00d73) \u00d7 (7\u00d77) \u00d7 5<\/p>\n\n\n\n

By observation, prime factor 5 is left out.<\/p>\n\n\n\n

So, multiply by 5 we get,<\/p>\n\n\n\n

8820 \u00d7 5 = (2\u00d72) \u00d7 (3\u00d73) \u00d7 (7\u00d77) \u00d7 (5\u00d75)<\/p>\n\n\n\n

= (2\u00d73\u00d77\u00d75) \u00d7 (2\u00d73\u00d77\u00d75)<\/p>\n\n\n\n

= 210 \u00d7 210<\/p>\n\n\n\n

= (210)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 210.<\/p>\n\n\n\n

(ii)<\/strong> 3675<\/p>\n\n\n\n

First find the prime factors for 3675<\/p>\n\n\n\n

3675 = 5\u00d75\u00d77\u00d77\u00d73<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (5\u00d75) \u00d7 (7\u00d77) \u00d7 3<\/p>\n\n\n\n

By observation, prime factor 3 is left out.<\/p>\n\n\n\n

So, multiply by 3 we get,<\/p>\n\n\n\n

3675 \u00d7 3 = (5\u00d75) \u00d7 (7\u00d77) \u00d7 (3\u00d73)<\/p>\n\n\n\n

= (5\u00d77\u00d73) \u00d7 (5\u00d77\u00d73)<\/p>\n\n\n\n

= 105 \u00d7 105<\/p>\n\n\n\n

= (105)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 105.<\/p>\n\n\n\n

(iii)<\/strong> 605<\/p>\n\n\n\n

First find the prime factors for 605<\/p>\n\n\n\n

605 = 5\u00d711\u00d711<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (11\u00d711) \u00d7 5<\/p>\n\n\n\n

By observation, prime factor 5 is left out.<\/p>\n\n\n\n

So, multiply by 5 we get,<\/p>\n\n\n\n

605 \u00d7 5 = (11\u00d711) \u00d7 (5\u00d75)<\/p>\n\n\n\n

= (11\u00d75) \u00d7 (11\u00d75)<\/p>\n\n\n\n

= 55 \u00d7 55<\/p>\n\n\n\n

= (55)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 55.<\/p>\n\n\n\n

(iv)<\/strong> 2880<\/p>\n\n\n\n

First find the prime factors for 2880<\/p>\n\n\n\n

2880 = 5\u00d73\u00d73\u00d72\u00d72\u00d72\u00d72\u00d72\u00d72<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (3\u00d73) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 5<\/p>\n\n\n\n

By observation, prime factor 5 is left out.<\/p>\n\n\n\n

So, multiply by 5 we get,<\/p>\n\n\n\n

2880 \u00d7 5 = (3\u00d73) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (5\u00d75)<\/p>\n\n\n\n

= (3\u00d72\u00d72\u00d72\u00d75) \u00d7 (3\u00d72\u00d72\u00d72\u00d75)<\/p>\n\n\n\n

= 120 \u00d7 120<\/p>\n\n\n\n

= (120)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 120.<\/p>\n\n\n\n

(v)<\/strong> 4056<\/p>\n\n\n\n

First find the prime factors for 4056<\/p>\n\n\n\n

4056 = 2\u00d72\u00d713\u00d713\u00d72\u00d73<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (13\u00d713) \u00d7 2 \u00d7 3<\/p>\n\n\n\n

By observation, prime factors 2 and 3 are left out.<\/p>\n\n\n\n

So, multiply by 6 we get,<\/p>\n\n\n\n

4056 \u00d7 6 = (2\u00d72) \u00d7 (13\u00d713) \u00d7 (2\u00d72) \u00d7 (3\u00d73)<\/p>\n\n\n\n

= (2\u00d713\u00d72\u00d73) \u00d7 (2\u00d713\u00d72\u00d73)<\/p>\n\n\n\n

= 156 \u00d7 156<\/p>\n\n\n\n

= (156)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 156.<\/p>\n\n\n\n

(vi)<\/strong> 3468<\/p>\n\n\n\n

First find the prime factors for 3468<\/p>\n\n\n\n

3468 = 2\u00d72\u00d717\u00d717\u00d73<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (17\u00d717) \u00d7 3<\/p>\n\n\n\n

By observation, prime factor 3 is left out.<\/p>\n\n\n\n

So, multiply by 3 we get,<\/p>\n\n\n\n

3468 \u00d7 3 = (2\u00d72) \u00d7 (17\u00d717) \u00d7 (3\u00d73)<\/p>\n\n\n\n

= (2\u00d717\u00d73) \u00d7 (2\u00d717\u00d73)<\/p>\n\n\n\n

= 102 \u00d7 102<\/p>\n\n\n\n

= (102)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 102.<\/p>\n\n\n\n

(vii)<\/strong> 7776<\/p>\n\n\n\n

First find the prime factors for 7776<\/p>\n\n\n\n

7776 = 2\u00d72\u00d72\u00d72\u00d73\u00d73\u00d73\u00d73\u00d72\u00d73<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (2\u00d72) \u00d7 (3\u00d73) \u00d7 (3\u00d73) \u00d7 2 \u00d7 3<\/p>\n\n\n\n

By observation, prime factors 2 and 3 are left out.<\/p>\n\n\n\n

So, multiply by 6 we get,<\/p>\n\n\n\n

7776 \u00d7 6 = (2\u00d72) \u00d7 (2\u00d72) \u00d7 (3\u00d73) \u00d7 (3\u00d73) \u00d7 (2\u00d72) \u00d7 (3\u00d73)<\/p>\n\n\n\n

= (2\u00d72\u00d73\u00d73\u00d72\u00d73) \u00d7 (2\u00d72\u00d73\u00d73\u00d72\u00d73)<\/p>\n\n\n\n

= 216 \u00d7 216<\/p>\n\n\n\n

= (216)2<\/sup><\/p>\n\n\n\n

\u2234 Product is the square of 216.<\/p>\n\n\n\n

8. By What numbers should each of the following be divided to get a perfect square in each case? Also, find the number whose square is the new number.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 16562<\/p>\n\n\n\n

First find the prime factors for 16562<\/p>\n\n\n\n

16562 = 7\u00d77\u00d713\u00d713\u00d72<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (7\u00d77) \u00d7 (13\u00d713) \u00d7 2<\/p>\n\n\n\n

By observation, prime factor 2 is left out.<\/p>\n\n\n\n

So, divide by 2 to eliminate 2 we get,<\/p>\n\n\n\n

16562\/2 = (7\u00d77) \u00d7 (13\u00d713)<\/p>\n\n\n\n

= (7\u00d713) \u00d7 (7\u00d713)<\/p>\n\n\n\n

= 91 \u00d7 91<\/p>\n\n\n\n

= (91)2<\/sup><\/p>\n\n\n\n

\u2234 Resultant is the square of 91.<\/p>\n\n\n\n

(ii)<\/strong> 3698<\/p>\n\n\n\n

First find the prime factors for 3698<\/p>\n\n\n\n

3698 = 2\u00d743\u00d743<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (43\u00d743) \u00d7 2<\/p>\n\n\n\n

By observation, prime factor 2 is left out.<\/p>\n\n\n\n

So, divide by 2 to eliminate 2 we get,<\/p>\n\n\n\n

3698\/2 = (43\u00d743)<\/p>\n\n\n\n

= (43)2<\/sup><\/p>\n\n\n\n

\u2234 Resultant is the square of 43.<\/p>\n\n\n\n

(iii)<\/strong> 5103<\/p>\n\n\n\n

First find the prime factors for 5103<\/p>\n\n\n\n

5103 = 3\u00d73\u00d73\u00d73\u00d73\u00d73\u00d77<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (3\u00d73) \u00d7 (3\u00d73) \u00d7 (3\u00d73) \u00d7 7<\/p>\n\n\n\n

By observation, prime factor 7 is left out.<\/p>\n\n\n\n

So, divide by 7 to eliminate 7 we get,<\/p>\n\n\n\n

5103\/7 = (3\u00d73) \u00d7 (3\u00d73) \u00d7 (3\u00d73)<\/p>\n\n\n\n

= (3\u00d73\u00d73) \u00d7 (3\u00d73\u00d73)<\/p>\n\n\n\n

= 27 \u00d7 27<\/p>\n\n\n\n

= (27)2<\/sup><\/p>\n\n\n\n

\u2234 Resultant is the square of 27.<\/p>\n\n\n\n

(iv)<\/strong> 3174<\/p>\n\n\n\n

First find the prime factors for 3174<\/p>\n\n\n\n

3174 = 2\u00d73\u00d723\u00d723<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (23\u00d723) \u00d7 2 \u00d7 3<\/p>\n\n\n\n

By observation, prime factor 2 and 3 are left out.<\/p>\n\n\n\n

So, divide by 6 to eliminate 2 and 3 we get,<\/p>\n\n\n\n

3174\/6 = (23\u00d723)<\/p>\n\n\n\n

= (23)2<\/sup><\/p>\n\n\n\n

\u2234 Resultant is the square of 23.<\/p>\n\n\n\n

(v)<\/strong> 1575<\/p>\n\n\n\n

First find the prime factors for 1575<\/p>\n\n\n\n

1575 = 3\u00d73\u00d75\u00d75\u00d77<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (3\u00d73) \u00d7 (5\u00d75) \u00d7 7<\/p>\n\n\n\n

By observation, prime factor 7 is left out.<\/p>\n\n\n\n

So, divide by 7 to eliminate 7 we get,<\/p>\n\n\n\n

1575\/7 = (3\u00d73) \u00d7 (5\u00d75)<\/p>\n\n\n\n

= (3\u00d75) \u00d7 (3\u00d75)<\/p>\n\n\n\n

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

= (15)2<\/sup><\/p>\n\n\n\n

\u2234 Resultant is the square of 15.<\/p>\n\n\n\n

9. Find the greatest number of two digits which is a perfect square.<\/strong><\/p>\n\n\n\n

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

We know that the two digit greatest number is 99<\/p>\n\n\n\n

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

\u2234 Greatest two digit perfect square number is 99-18 = 81<\/p>\n\n\n\n

10. Find the least number of three digits which is perfect square.<\/strong><\/p>\n\n\n\n

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

We know that the three digit greatest number is 100<\/p>\n\n\n\n

To find the square root of 100<\/p>\n\n\n\n

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

\u2234 the least number of three digits which is a perfect square is 100 itself.<\/p>\n\n\n\n

11. Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.<\/strong><\/p>\n\n\n\n

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

First find the prime factors for 4851<\/p>\n\n\n\n

4851 = 3\u00d73\u00d77\u00d77\u00d711<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (3\u00d73) \u00d7 (7\u00d77) \u00d7 11<\/p>\n\n\n\n

\u2234 The smallest number by which 4851 must be multiplied so that the product becomes a perfect square is 11.<\/p>\n\n\n\n

12. Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.<\/strong><\/p>\n\n\n\n

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

First find the prime factors for 28812<\/p>\n\n\n\n

28812 = 2\u00d72\u00d73\u00d77\u00d77\u00d77\u00d77<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 3 \u00d7 (7\u00d77) \u00d7 (7\u00d77)<\/p>\n\n\n\n

\u2234 The smallest number by which 28812 must be divided so that the quotient becomes a perfect square is 3.<\/p>\n\n\n\n

13. Find the smallest number by which 1152 must be divided so that it becomes 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

First find the prime factors for 1152<\/p>\n\n\n\n

1152 = 2\u00d72\u00d72\u00d72\u00d72\u00d72\u00d72\u00d73\u00d73<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (3\u00d73) \u00d7 2<\/p>\n\n\n\n

\u2234 The smallest number by which 1152 must be divided so that the quotient becomes a perfect square is 2.<\/p>\n\n\n\n

The number after division, 1152\/2 = 576<\/p>\n\n\n\n

prime factors for 576 = 2\u00d72\u00d72\u00d72\u00d72\u00d72\u00d73\u00d73<\/p>\n\n\n\n

By grouping the prime factors in equal pairs we get,<\/p>\n\n\n\n

= (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (3\u00d73)<\/p>\n\n\n\n

= 26<\/sup> \u00d7 32<\/sup><\/p>\n\n\n\n

= 242<\/sup><\/p>\n\n\n\n

\u2234 The resulting number is the square of 24.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.2 PAGE NO: 3.18<\/h4>\n\n\n\n

1. The following numbers are not perfect squares. Give reason.
(i) 1547<\/strong><\/p>\n\n\n\n

(ii) 45743
(iii)8948<\/strong><\/p>\n\n\n\n

(iv) 333333<\/strong><\/p>\n\n\n\n

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

The numbers ending with 2, 3, 7 or 8 is not a perfect square.<\/p>\n\n\n\n

So, (i) <\/strong>1547<\/p>\n\n\n\n

(ii) <\/strong>45743
(iii) <\/strong>8948<\/p>\n\n\n\n

(iv) <\/strong>333333<\/p>\n\n\n\n

Are not perfect squares.<\/p>\n\n\n\n

2. Show that the following numbers are not, perfect squares:
(i) 9327<\/strong><\/p>\n\n\n\n

(ii) 4058
(iii)22453<\/strong><\/p>\n\n\n\n

(iv) 743522<\/strong><\/p>\n\n\n\n

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

The numbers ending with 2, 3, 7 or 8 is not a perfect square.<\/p>\n\n\n\n

So, (i)<\/strong> 9327<\/p>\n\n\n\n

(ii)<\/strong> 4058<\/p>\n\n\n\n

(iii)<\/strong> 22453<\/p>\n\n\n\n

(iv)<\/strong> 743522<\/p>\n\n\n\n

Are not perfect squares.<\/p>\n\n\n\n

3. The square of which of the following numbers would be an old number?
(i) 731<\/strong><\/p>\n\n\n\n

(ii) 3456
(iii)5559<\/strong><\/p>\n\n\n\n

(iv) 42008<\/strong><\/p>\n\n\n\n

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

We know that square of an even number is even number.<\/p>\n\n\n\n

Square of an odd number is odd number.<\/p>\n\n\n\n

(i)<\/strong> 731<\/p>\n\n\n\n

Since 731 is an odd number, the square of the given number is also odd.<\/p>\n\n\n\n

(ii)<\/strong> 3456<\/p>\n\n\n\n

Since 3456 is an even number, the square of the given number is also even.<\/p>\n\n\n\n

(iii)<\/strong> 5559<\/p>\n\n\n\n

Since 5559 is an odd number, the square of the given number is also odd.<\/p>\n\n\n\n

(iv)<\/strong> 42008<\/p>\n\n\n\n

Since 42008 is an even number, the square of the given number is also even.<\/p>\n\n\n\n

4. What will be the unit\u2019s digit of the squares of the following numbers?
(i) 52<\/strong><\/p>\n\n\n\n

(ii) 977
(iii) 4583<\/strong><\/p>\n\n\n\n

(iv) 78367
(v) 52698<\/strong><\/p>\n\n\n\n

(vi) 99880
(vii) 12796<\/strong><\/p>\n\n\n\n

(viii) 55555
(ix) 53924<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 52<\/p>\n\n\n\n

Unit digit of (52)2<\/sup> = (22<\/sup>) = 4<\/p>\n\n\n\n

(ii)<\/strong> 977<\/p>\n\n\n\n

Unit digit of (977)2<\/sup> = (72<\/sup>) = 49 = 9<\/p>\n\n\n\n

(iii)<\/strong> 4583<\/p>\n\n\n\n

Unit digit of (4583)2<\/sup> = (32<\/sup>) = 9<\/p>\n\n\n\n

(iv)<\/strong> 78367<\/p>\n\n\n\n

Unit digit of (78367)2<\/sup> = (72<\/sup>) = 49 = 9<\/p>\n\n\n\n

(v)<\/strong> 52698<\/p>\n\n\n\n

Unit digit of (52698)2<\/sup> = (82<\/sup>) = 64 = 4<\/p>\n\n\n\n

(vi)<\/strong> 99880<\/p>\n\n\n\n

Unit digit of (99880)2<\/sup> = (02<\/sup>) = 0<\/p>\n\n\n\n

(vii)<\/strong> 12796<\/p>\n\n\n\n

Unit digit of (12796)2<\/sup> = (62<\/sup>) = 36 = 6<\/p>\n\n\n\n

(viii)<\/strong> 55555<\/p>\n\n\n\n

Unit digit of (55555)2<\/sup> = (52<\/sup>) = 25 = 5<\/p>\n\n\n\n

(ix)<\/strong> 53924<\/p>\n\n\n\n

Unit digit of (53924)2<\/sup> = (42<\/sup>) = 16 = 6<\/p>\n\n\n\n

5. Observe the following pattern
1+3 = 22<\/sup><\/strong><\/p>\n\n\n\n

1+3+5 = 32<\/sup><\/strong><\/p>\n\n\n\n

1+3+5+7 = 42<\/sup>
And write the value of 1+3+5+7+9+\u2026\u2026\u2026 up to n terms.<\/strong><\/p>\n\n\n\n

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

We know that the pattern given is the square of the given number on the right hand side is equal to the sum of the given numbers on the left hand side.<\/p>\n\n\n\n

\u2234 The value of 1+3+5+7+9+\u2026\u2026\u2026 up to n terms = n2<\/sup> (as there are only n terms).<\/p>\n\n\n\n

6. Observe the following pattern<\/strong><\/p>\n\n\n\n

22<\/sup> -12<\/sup> = 2 + 1<\/strong><\/p>\n\n\n\n

32<\/sup> \u2013 22<\/sup> = 3 + 2<\/strong><\/p>\n\n\n\n

42<\/sup> \u2013 32<\/sup> = 4 + 3<\/strong><\/p>\n\n\n\n

52<\/sup> \u2013 42<\/sup> = 5 + 4<\/strong><\/p>\n\n\n\n

And find the value of<\/strong><\/p>\n\n\n\n

(i) 1002<\/sup> -992<\/sup><\/strong><\/p>\n\n\n\n

(ii)1112<\/sup> \u2013 1092<\/sup><\/strong><\/p>\n\n\n\n

(iii) 992<\/sup> \u2013 962<\/sup><\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 1002<\/sup> -992<\/sup><\/p>\n\n\n\n

100 + 99 = 199<\/p>\n\n\n\n

(ii)<\/strong> 1112<\/sup> \u2013 1092<\/sup><\/p>\n\n\n\n

(1112<\/sup> \u2013 1102<\/sup>) + (1102<\/sup> \u2013 1092<\/sup>)<\/p>\n\n\n\n

(111 + 110) + (100 + 109)<\/p>\n\n\n\n

440<\/p>\n\n\n\n

(iii)<\/strong> 992<\/sup> \u2013 962<\/sup><\/p>\n\n\n\n

(992<\/sup> \u2013 982<\/sup>) + (982<\/sup> \u2013 972<\/sup>) + (972<\/sup> \u2013 962<\/sup>)<\/p>\n\n\n\n

(99 + 98) + (98 + 97) + (97 + 96)<\/p>\n\n\n\n

585<\/p>\n\n\n\n

7. Which of the following triplets are Pythagorean?
(i) (8, 15, 17)
(ii) (18, 80, 82)
(iii) (14, 48, 51)
(iv) (10, 24, 26)
(v) (16, 63, 65)
(vi) (12, 35, 38)<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> (8, 15, 17)<\/p>\n\n\n\n

LHS = 82<\/sup> + 152<\/sup><\/p>\n\n\n\n

= 289<\/p>\n\n\n\n

RHS = 172<\/sup><\/p>\n\n\n\n

= 289<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

\u2234 The given triplet is a Pythagorean.<\/p>\n\n\n\n

(ii)<\/strong> (18, 80, 82)<\/p>\n\n\n\n

LHS = 182<\/sup> + 802<\/sup><\/p>\n\n\n\n

= 6724<\/p>\n\n\n\n

RHS = 822<\/sup><\/p>\n\n\n\n

= 6724<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

\u2234 The given triplet is a Pythagorean.<\/p>\n\n\n\n

(iii)<\/strong> (14, 48, 51)<\/p>\n\n\n\n

LHS = 142<\/sup> + 482<\/sup><\/p>\n\n\n\n

= 2500<\/p>\n\n\n\n

RHS = 512<\/sup><\/p>\n\n\n\n

= 2601<\/p>\n\n\n\n

LHS \u2260 RHS<\/p>\n\n\n\n

\u2234 The given triplet is not a Pythagorean.<\/p>\n\n\n\n

(iv)<\/strong> (10, 24, 26)<\/p>\n\n\n\n

LHS = 102<\/sup> + 242<\/sup><\/p>\n\n\n\n

= 676<\/p>\n\n\n\n

RHS = 262<\/sup><\/p>\n\n\n\n

= 676<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

\u2234 The given triplet is a Pythagorean.<\/p>\n\n\n\n

(v)<\/strong> (16, 63, 65)<\/p>\n\n\n\n

LHS = 162<\/sup> + 632<\/sup><\/p>\n\n\n\n

= 4225<\/p>\n\n\n\n

RHS = 652<\/sup><\/p>\n\n\n\n

= 4225<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

\u2234 The given triplet is a Pythagorean.<\/p>\n\n\n\n

(vi)<\/strong> (12, 35, 38)<\/p>\n\n\n\n

LHS = 122<\/sup> + 352<\/sup><\/p>\n\n\n\n

= 1369<\/p>\n\n\n\n

RHS = 382<\/sup><\/p>\n\n\n\n

= 1444<\/p>\n\n\n\n

LHS \u2260 RHS<\/p>\n\n\n\n

\u2234 The given triplet is not a Pythagorean.<\/p>\n\n\n\n

8. Observe the following pattern<\/strong><\/p>\n\n\n\n

(1\u00d72) + (2\u00d73) = (2\u00d73\u00d74)\/3<\/strong><\/p>\n\n\n\n

(1\u00d72) + (2\u00d73) + (3\u00d74) = (3\u00d74\u00d75)\/3<\/strong><\/p>\n\n\n\n

(1\u00d72) + (2\u00d73) + (3\u00d74) + (4\u00d75) = (4\u00d75\u00d76)\/3<\/strong><\/p>\n\n\n\n

And find the value of<\/strong><\/p>\n\n\n\n

(1\u00d72) + (2\u00d73) + (3\u00d74) + (4\u00d75) + (5\u00d76)<\/strong><\/p>\n\n\n\n

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

(1\u00d72) + (2\u00d73) + (3\u00d74) + (4\u00d75) + (5\u00d76) = (5\u00d76\u00d77)\/3 = 70<\/p>\n\n\n\n

9. Observe the following pattern<\/strong><\/p>\n\n\n\n

1 = 1\/2 (1\u00d7(1+1))<\/strong><\/p>\n\n\n\n

1+2 = 1\/2 (2\u00d7(2+1))<\/strong><\/p>\n\n\n\n

1+2+3 = 1\/2 (3\u00d7(3+1))<\/strong><\/p>\n\n\n\n

1+2+3+4 = 1\/2 (4\u00d7(4+1))<\/strong><\/p>\n\n\n\n

And find the values of each of the following:<\/strong><\/p>\n\n\n\n

(i) 1+2+3+4+5+\u2026+50<\/strong><\/p>\n\n\n\n

(ii) 31+32+\u2026.+50<\/strong><\/p>\n\n\n\n

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

We know that R.H.S = 1\/2 [No. of terms in L.H.S \u00d7 (No. of terms + 1)] (if only when L.H.S starts with 1)<\/p>\n\n\n\n

(i)<\/strong> 1+2+3+4+5+\u2026+50 = 1\/2 (5\u00d7(5+1))<\/p>\n\n\n\n

25 \u00d7 51 = 1275<\/p>\n\n\n\n

(ii)<\/strong> 31+32+\u2026.+50 = (1+2+3+4+5+\u2026+50) \u2013 (1+2+3+\u2026+30)<\/p>\n\n\n\n

1275 \u2013 1\/2 (30\u00d7(30+1))<\/p>\n\n\n\n

1275 \u2013 465<\/p>\n\n\n\n

810<\/p>\n\n\n\n

10. Observe the following pattern<\/strong><\/p>\n\n\n\n

12<\/sup> = 1\/6 (1\u00d7(1+1)\u00d7(2\u00d71+1))<\/strong><\/p>\n\n\n\n

12<\/sup>+22<\/sup> = 1\/6 (2\u00d7(2+1)\u00d7(2\u00d72+1)))<\/strong><\/p>\n\n\n\n

12<\/sup>+22<\/sup>+32<\/sup> = 1\/6 (3\u00d7(3+1)\u00d7(2\u00d73+1)))<\/strong><\/p>\n\n\n\n

12<\/sup>+22<\/sup>+32<\/sup>+42<\/sup> = 1\/6 (4\u00d7(4+1)\u00d7(2\u00d74+1)))<\/strong><\/p>\n\n\n\n

And find the values of each of the following:<\/strong><\/p>\n\n\n\n

(i) 12<\/sup>+22<\/sup>+32<\/sup>+42<\/sup>+\u2026+102<\/sup><\/strong><\/p>\n\n\n\n

(ii) 52<\/sup>+62<\/sup>+72<\/sup>+82<\/sup>+92<\/sup>+102<\/sup>+112<\/sup>+122<\/sup><\/strong><\/p>\n\n\n\n

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

RHS = 1\/6 [(No. of terms in L.H.S) \u00d7 (No. of terms + 1) \u00d7 (2 \u00d7 No. of terms + 1)]<\/p>\n\n\n\n

(i)<\/strong> 12<\/sup>+22<\/sup>+32<\/sup>+42<\/sup>+\u2026+102<\/sup> = 1\/6 (10\u00d7(10+1)\u00d7(2\u00d710+1))<\/p>\n\n\n\n

= 1\/6 (2310)<\/p>\n\n\n\n

= 385<\/p>\n\n\n\n

(ii)<\/strong> 52<\/sup>+62<\/sup>+72<\/sup>+82<\/sup>+92<\/sup>+102<\/sup>+112<\/sup>+122<\/sup> = 12<\/sup>+22<\/sup>+32<\/sup>+\u2026+122<\/sup> \u2013 (12<\/sup>+22<\/sup>+32<\/sup>+42<\/sup>)<\/p>\n\n\n\n

1\/6 (12\u00d7(12+1)\u00d7(2\u00d712+1)) \u2013 1\/6 (4\u00d7(4+1)\u00d7(2\u00d74+1))<\/p>\n\n\n\n

650-30<\/p>\n\n\n\n

620<\/p>\n\n\n\n

11. Which of the following numbers are squares of even numbers?
121, 225, 256, 324, 1296, 6561, 5476, 4489, 373758<\/strong><\/p>\n\n\n\n

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

We know that only even numbers be the squares of even numbers.<\/p>\n\n\n\n

So, 256, 324, 1296, 5476, 373758 are even numbers, since 373758 is not a perfect square<\/p>\n\n\n\n

\u2234 256, 324, 1296, 5476 are squares of even numbers.<\/p>\n\n\n\n

12. By just examining the units digits, can you tell which of the following cannot be whole squares?<\/strong><\/p>\n\n\n\n

(i) 1026<\/p>\n\n\n\n

(ii) 1028
(iii)1024<\/strong><\/p>\n\n\n\n

(iv) 1022
(v) 1023<\/strong><\/p>\n\n\n\n

(vi) 1027<\/strong><\/p>\n\n\n\n

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

We know that numbers ending with 2, 3, 7, 8 cannot be a perfect square.<\/p>\n\n\n\n

\u2234 1028, 1022, 1023, and 1027 cannot be whole squares.<\/p>\n\n\n\n

13. Which of the numbers for which you cannot decide whether they are squares.<\/strong><\/p>\n\n\n\n

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

We know that the natural numbers such as 0, 1, 4, 5, 6 or 9 cannot be decided surely whether they are squares or not.<\/p>\n\n\n\n

14. Write five numbers which you cannot decide whether they are square just by looking at the unit\u2019s digit.<\/strong><\/p>\n\n\n\n

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

We know that any natural number ending with 0, 1, 4, 5, 6 or 9 can be or cannot be a square number.<\/p>\n\n\n\n

Here are the five examples which you cannot decide whether they are square or not just by looking at the units place:<\/p>\n\n\n\n

(i)<\/strong> 2061<\/p>\n\n\n\n

The unit digit is 1. So, it may or may not be a square number<\/p>\n\n\n\n

(ii)<\/strong> 1069<\/p>\n\n\n\n

The unit digit is 9. So, it may or may not be a square number<\/p>\n\n\n\n

(iii)<\/strong> 1234<\/p>\n\n\n\n

The unit digit is 4. So, it may or may not be a square number<\/p>\n\n\n\n

(iv)<\/strong> 56790<\/p>\n\n\n\n

The unit digit is 0. So, it may or may not be a square number<\/p>\n\n\n\n

(v)<\/strong> 76555<\/p>\n\n\n\n

The unit digit is 5. So, it may or may not be a square number<\/p>\n\n\n\n

15.<\/strong> Write true (T) or false (F) for the following statements.
(i) The number of digits in a square number is even.
(ii) The square of a prime number is prime.
(iii) The sum of two square numbers is a square number.
(iv) The difference of two square numbers is a square number.
(v) The product of two square numbers is a square number.
(vi) No square number is negative.
(vii) There is no square number between 50 and 60.
(viii) There are fourteen square number up to 200.<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> False, because 169 is a square number with odd digit.<\/p>\n\n\n\n

(ii)<\/strong> False, because square of 3(which is prime) is 9(which is not prime).<\/p>\n\n\n\n

(iii)<\/strong> False, because sum of 22<\/sup> and 32<\/sup> is 13 which is not square number.<\/p>\n\n\n\n

(iv)<\/strong> False, because difference of 32<\/sup> and 22<\/sup> is 5, which is not square number.<\/p>\n\n\n\n

(v)<\/strong> True, because the square of 22<\/sup> and 32<\/sup> is 36 which is square of 6<\/p>\n\n\n\n

(vi)<\/strong> True, because (-2)2<\/sup> is 4, which is not negative.<\/p>\n\n\n\n

(vii)<\/strong> True, because as there is no square number between them.<\/p>\n\n\n\n

(viii)<\/strong> True, because the fourteen numbers up to 200 are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.3 PAGE NO: 3.32<\/h4>\n\n\n\n

1. Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication:
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>25<\/p>\n\n\n\n

So here, a = 2 and b = 5<\/p>\n\n\n\n

Column I<\/td>Column II<\/td>Column III<\/td><\/tr>
a2<\/sup>4+26<\/td>2ab20+222<\/td>b2<\/sup>25<\/td><\/tr>
6<\/td>2<\/td>5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

\u2234 252<\/sup> = 625<\/p>\n\n\n\n

Where, it can be expressed as<\/p>\n\n\n\n

252<\/sup> = 25\u00d7 25 = 625<\/p>\n\n\n\n

(ii)<\/strong> 37<\/p>\n\n\n\n

So here, a = 3 and b = 7<\/p>\n\n\n\n

Column I<\/td>Column II<\/td>Column III<\/td><\/tr>
a2<\/sup>9+413<\/td>2ab42+446<\/td>b2<\/sup>49<\/td><\/tr>
13<\/td>6<\/td>9<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

\u2234 372<\/sup> = 1369<\/p>\n\n\n\n

Where, it can be expressed as<\/p>\n\n\n\n

252<\/sup> = 37\u00d7 37 = 1369<\/p>\n\n\n\n

(iii) 54<\/strong><\/p>\n\n\n\n

So here, a = 5 and b = 4<\/p>\n\n\n\n

Column I<\/td>Column II<\/td>Column III<\/td><\/tr>
a2<\/sup>25+429<\/td>2ab40+141<\/td>b2<\/sup>16<\/td><\/tr>
29<\/td>1<\/td>6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

\u2234 542<\/sup> = 2916<\/p>\n\n\n\n

Where, it can be expressed as<\/p>\n\n\n\n

542<\/sup> = 54 \u00d7 54 = 2916<\/p>\n\n\n\n

(iv) <\/strong>71<\/p>\n\n\n\n

So here, a = 7 and b = 1<\/p>\n\n\n\n

Column I<\/td>Column II<\/td>Column III<\/td><\/tr>
a2<\/sup>49+150<\/td>2ab14+014<\/td>b2<\/sup>01<\/td><\/tr>
50<\/td>4<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

\u2234 712<\/sup> = 5041<\/p>\n\n\n\n

Where, it can be expressed as<\/p>\n\n\n\n

712<\/sup> = 71 \u00d7 71 = 5041<\/p>\n\n\n\n

(v)<\/strong> 96<\/p>\n\n\n\n

So here, a = 9 and b = 6<\/p>\n\n\n\n

Column I<\/td>Column II<\/td>Column III<\/td><\/tr>
a2<\/sup>81+1192<\/td>2ab108+3111<\/td>b2<\/sup>36<\/td><\/tr>
92<\/td>1<\/td>6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

\u2234 962<\/sup> = 9216<\/p>\n\n\n\n

Where, it can be expressed as<\/p>\n\n\n\n

962<\/sup> = 96 \u00d7 96 = 9216<\/p>\n\n\n\n

2. Find the squares of the following numbers using diagonal method:
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 98<\/p>\n\n\n\n

Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.<\/p>\n\n\n\n

Step 2: Draw square and divide it into n2<\/sup> sub-squares of the same size by drawing (n \u2013 1) horizontal and (n \u2013 1) vertical lines.<\/p>\n\n\n\n

Step 3: Draw the diagonals of each sub-square.<\/p>\n\n\n\n

Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.<\/p>\n\n\n\n

Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.<\/p>\n\n\n\n

Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.<\/p>\n\n\n\n

Step 7: Obtain the required square by writing the digits from the left-most side.<\/p>\n\n\n\n

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

\u2234 982<\/sup> = 9604<\/p>\n\n\n\n

(ii)<\/strong> 273<\/p>\n\n\n\n

Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.<\/p>\n\n\n\n

Step 2: Draw square and divide it into n2<\/sup> sub-squares of the same size by drawing (n \u2013 1) horizontal and (n \u2013 1) vertical lines.<\/p>\n\n\n\n

Step 3: Draw the diagonals of each sub-square.<\/p>\n\n\n\n

Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.<\/p>\n\n\n\n

Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.<\/p>\n\n\n\n

Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.<\/p>\n\n\n\n

Step 7: Obtain the required square by writing the digits from the left-most side.<\/p>\n\n\n\n

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

\u2234 2732<\/sup> = 74529<\/p>\n\n\n\n

(iii)<\/strong> 348<\/p>\n\n\n\n

Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.<\/p>\n\n\n\n

Step 2: Draw square and divide it into n2<\/sup> sub-squares of the same size by drawing (n \u2013 1) horizontal and (n \u2013 1) vertical lines.<\/p>\n\n\n\n

Step 3: Draw the diagonals of each sub-square.<\/p>\n\n\n\n

Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.<\/p>\n\n\n\n

Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.<\/p>\n\n\n\n

Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.<\/p>\n\n\n\n

Step 7: Obtain the required square by writing the digits from the left-most side.<\/p>\n\n\n\n

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

\u2234 3482<\/sup> = 121104<\/p>\n\n\n\n

(iv)<\/strong> 295<\/p>\n\n\n\n

Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.<\/p>\n\n\n\n

Step 2: Draw square and divide it into n2<\/sup> sub-squares of the same size by drawing (n \u2013 1) horizontal and (n \u2013 1) vertical lines.<\/p>\n\n\n\n

Step 3: Draw the diagonals of each sub-square.<\/p>\n\n\n\n

Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.<\/p>\n\n\n\n

Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.<\/p>\n\n\n\n

Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.<\/p>\n\n\n\n

Step 7: Obtain the required square by writing the digits from the left-most side.<\/p>\n\n\n\n

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

\u2234 2952<\/sup> = 87025<\/p>\n\n\n\n

(v)<\/strong> 171<\/p>\n\n\n\n

Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.<\/p>\n\n\n\n

Step 2: Draw square and divide it into n2<\/sup> sub-squares of the same size by drawing (n \u2013 1) horizontal and (n \u2013 1) vertical lines.<\/p>\n\n\n\n

Step 3: Draw the diagonals of each sub-square.<\/p>\n\n\n\n

Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.<\/p>\n\n\n\n

Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.<\/p>\n\n\n\n

Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.<\/p>\n\n\n\n

Step 7: Obtain the required square by writing the digits from the left-most side.<\/p>\n\n\n\n

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

\u2234 1712<\/sup> = 29241<\/p>\n\n\n\n

3. Find the squares of the following numbers:
(i) 127<\/strong><\/p>\n\n\n\n

(ii) 503
(iii) 450<\/strong><\/p>\n\n\n\n

(iv) 862
(v) 265<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>127<\/p>\n\n\n\n

1272<\/sup> = 127 \u00d7 127 = 16129<\/p>\n\n\n\n

(ii) <\/strong>503<\/p>\n\n\n\n

5032<\/sup> = 503 \u00d7 503 = 253009<\/p>\n\n\n\n


(iii) <\/strong>450<\/p>\n\n\n\n

4502<\/sup> = 450 \u00d7 450 = 203401<\/p>\n\n\n\n

(iv) <\/strong>862<\/p>\n\n\n\n

8622<\/sup> = 862 \u00d7 862 = 743044<\/p>\n\n\n\n


(v) <\/strong>265<\/p>\n\n\n\n

2652<\/sup> = 265 \u00d7 265 = 70225<\/p>\n\n\n\n

4. Find the squares of the following numbers:
(i) 425<\/strong><\/p>\n\n\n\n

(ii) 575
(iii)405<\/strong><\/p>\n\n\n\n

(iv) 205
(v) 95<\/strong><\/p>\n\n\n\n

(vi) 745
(vii) 512<\/strong><\/p>\n\n\n\n

(viii) 995<\/strong><\/p>\n\n\n\n

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

(i)<\/strong>425<\/p>\n\n\n\n

4252<\/sup> = 425 \u00d7 425 = 180625<\/p>\n\n\n\n

(ii) 575<\/strong><\/p>\n\n\n\n

5752<\/sup> = 575 \u00d7 575 = 330625<\/p>\n\n\n\n


(iii)405<\/strong><\/p>\n\n\n\n

4052<\/sup> = 405 \u00d7 405 = 164025<\/p>\n\n\n\n

(iv) 205<\/strong><\/p>\n\n\n\n

2052<\/sup> = 205 \u00d7 205 = 42025<\/p>\n\n\n\n


(v) 95<\/strong><\/p>\n\n\n\n

952<\/sup> = 95 \u00d7 95 = 9025<\/p>\n\n\n\n

(vi) 745<\/strong><\/p>\n\n\n\n

7452<\/sup> = 745 \u00d7 745 = 555025<\/p>\n\n\n\n


(vii) 512<\/strong><\/p>\n\n\n\n

5122<\/sup> = 512 \u00d7 512 = 262144<\/p>\n\n\n\n

(viii) 995<\/strong><\/p>\n\n\n\n

9952<\/sup> = 995 \u00d7 995 = 990025<\/p>\n\n\n\n

5. Find the squares of the following numbers using the identity (a+b) 2<\/sup>= a2<\/sup>+2ab+b2<\/sup>:
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 405<\/p>\n\n\n\n

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

405 = (400+5)2<\/sup><\/p>\n\n\n\n

= (400)2<\/sup> + 52<\/sup> + 2 (400) (5)<\/p>\n\n\n\n

= 160000 + 25 + 4000<\/p>\n\n\n\n

= 164025<\/p>\n\n\n\n

(ii) <\/strong>510<\/p>\n\n\n\n

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

510 = (500+10)2<\/sup><\/p>\n\n\n\n

= (500)2<\/sup> + 102<\/sup> + 2 (500) (10)<\/p>\n\n\n\n

= 250000 + 100 + 10000<\/p>\n\n\n\n

= 260100<\/p>\n\n\n\n


(iii) <\/strong>1001<\/p>\n\n\n\n

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

1001 = (1000+1)2<\/sup><\/p>\n\n\n\n

= (1000)2<\/sup> + 12<\/sup> + 2 (1000) (1)<\/p>\n\n\n\n

= 1000000 + 1 + 2000<\/p>\n\n\n\n

= 1002001<\/p>\n\n\n\n


(iv) <\/strong>209<\/p>\n\n\n\n

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

209 = (200+9)2<\/sup><\/p>\n\n\n\n

= (200)2<\/sup> + 92<\/sup> + 2 (200) (9)<\/p>\n\n\n\n

= 40000 + 81 + 3600<\/p>\n\n\n\n

= 43681<\/p>\n\n\n\n


(v)<\/strong> 605<\/p>\n\n\n\n

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

605 = (600+5)2<\/sup><\/p>\n\n\n\n

= (600)2<\/sup> + 52<\/sup> + 2 (600) (5)<\/p>\n\n\n\n

= 360000 + 25 + 6000<\/p>\n\n\n\n

= 366025<\/p>\n\n\n\n

6. Find the squares of the following numbers using the identity (a-b) 2<\/sup>= a2<\/sup>-2ab+b2<\/sup>
(i) 395<\/strong><\/p>\n\n\n\n

(ii) 995
(iii)495<\/strong><\/p>\n\n\n\n

(iv) 498
(v) 99<\/strong><\/p>\n\n\n\n

(vi) 999
(vii)599<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>395<\/p>\n\n\n\n

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

395 = (400-5)2<\/sup><\/p>\n\n\n\n

= (400)2<\/sup> + 52<\/sup> \u2013 2 (400) (5)<\/p>\n\n\n\n

= 160000 + 25 \u2013 4000<\/p>\n\n\n\n

= 156025<\/p>\n\n\n\n

(ii) <\/strong>995<\/p>\n\n\n\n

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

995 = (1000-5)2<\/sup><\/p>\n\n\n\n

= (1000)2<\/sup> + 52<\/sup> \u2013 2 (1000) (5)<\/p>\n\n\n\n

= 1000000 + 25 \u2013 10000<\/p>\n\n\n\n

= 990025<\/p>\n\n\n\n


(iii) <\/strong>495<\/p>\n\n\n\n

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

495 = (500-5)2<\/sup><\/p>\n\n\n\n

= (500)2<\/sup> + 52<\/sup> \u2013 2 (500) (5)<\/p>\n\n\n\n

= 250000 + 25 \u2013 5000<\/p>\n\n\n\n

= 245025<\/p>\n\n\n\n

(iv) <\/strong>498<\/p>\n\n\n\n

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

498 = (500-2)2<\/sup><\/p>\n\n\n\n

= (500)2<\/sup> + 22<\/sup> \u2013 2 (500) (2)<\/p>\n\n\n\n

= 250000 + 4 \u2013 2000<\/p>\n\n\n\n

= 248004<\/p>\n\n\n\n


(v) <\/strong>99<\/p>\n\n\n\n

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

99 = (100-1)2<\/sup><\/p>\n\n\n\n

= (100)2<\/sup> + 12<\/sup> \u2013 2 (100) (1)<\/p>\n\n\n\n

= 10000 + 1 \u2013 200<\/p>\n\n\n\n

= 9801<\/p>\n\n\n\n

(vi) <\/strong>999<\/p>\n\n\n\n

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

999 = (1000-1)2<\/sup><\/p>\n\n\n\n

= (1000)2<\/sup> + 12<\/sup> \u2013 2 (1000) (1)<\/p>\n\n\n\n

= 1000000 + 1 \u2013 2000<\/p>\n\n\n\n

= 998001<\/p>\n\n\n\n


(vii) <\/strong>599<\/p>\n\n\n\n

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

599 = (600-1)2<\/sup><\/p>\n\n\n\n

= (600)2<\/sup> + 12<\/sup> \u2013 2 (600) (1)<\/p>\n\n\n\n

= 360000 + 1 \u2013 1200<\/p>\n\n\n\n

= 358801<\/p>\n\n\n\n

7. Find the squares of the following numbers by visual method:
(i) 52<\/strong><\/p>\n\n\n\n

(ii) 95
(iii) 505<\/strong><\/p>\n\n\n\n

(iv) 702
(v) 99<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>52<\/p>\n\n\n\n

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

52 = (50+2)2<\/sup><\/p>\n\n\n\n

= (50)2<\/sup> + 22<\/sup> + 2 (50) (2)<\/p>\n\n\n\n

= 2500 + 4 + 200<\/p>\n\n\n\n

= 2704<\/p>\n\n\n\n

(ii) <\/strong>95<\/p>\n\n\n\n

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

95 = (100-5)2<\/sup><\/p>\n\n\n\n

= (100)2<\/sup> + 52<\/sup> \u2013 2 (100) (5)<\/p>\n\n\n\n

= 10000 + 25 \u2013 1000<\/p>\n\n\n\n

= 9025<\/p>\n\n\n\n


(iii) <\/strong>505<\/p>\n\n\n\n

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

505 = (500+5)2<\/sup><\/p>\n\n\n\n

= (500)2<\/sup> + 52<\/sup> + 2 (500) (5)<\/p>\n\n\n\n

= 250000 + 25 + 5000<\/p>\n\n\n\n

= 255025<\/p>\n\n\n\n

(iv) <\/strong>702<\/p>\n\n\n\n

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

702 = (700+2)2<\/sup><\/p>\n\n\n\n

= (700)2<\/sup> + 22<\/sup> + 2 (700) (2)<\/p>\n\n\n\n

= 490000 + 4 + 2800<\/p>\n\n\n\n

= 492804<\/p>\n\n\n\n


(v) <\/strong>99<\/p>\n\n\n\n

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

99 = (100-1)2<\/sup><\/p>\n\n\n\n

= (100)2<\/sup> + 12<\/sup> \u2013 2 (100) (1)<\/p>\n\n\n\n

= 10000 + 1 \u2013 200<\/p>\n\n\n\n

= 9801<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.4 PAGE NO: 3.38<\/h4>\n\n\n\n

1.Write the possible unit\u2019s digits of the square root of the following numbers. Which of these numbers are odd square roots?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 9801<\/p>\n\n\n\n

We know that unit digit of 9801 is 1<\/p>\n\n\n\n

Unit digit of square root = 1 or 9<\/p>\n\n\n\n

Since the number is odd, square root is also odd<\/p>\n\n\n\n

(ii)<\/strong> 99856<\/p>\n\n\n\n

We know that unit digit of 99856 = 6<\/p>\n\n\n\n

Unit digit of square root = 4 or 6<\/p>\n\n\n\n

Since the number is even, square root is also even<\/p>\n\n\n\n

(iii)<\/strong> 998001<\/p>\n\n\n\n

We know that unit digit of 998001 = 1<\/p>\n\n\n\n

Unit digit of square root = 1 or 9<\/p>\n\n\n\n

Since the number is odd, square root is also odd<\/p>\n\n\n\n

(iv)<\/strong> 657666025<\/p>\n\n\n\n

We know that unit digit of 657666025 = 5<\/p>\n\n\n\n

Unit digit of square root = 5<\/p>\n\n\n\n

Since the number is odd, square root is also odd<\/p>\n\n\n\n

2. Find the square root of each of the following by prime factorization.
(i) 441 (ii) 196
(iii) 529 (iv) 1764
(v) 1156 (vi) 4096
(vii) 7056 (viii) 8281
(ix) 11664 (x) 47089
(xi) 24336 (xii) 190969
(xiii) 586756 (xiv) 27225
(xv) 3013696<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>441<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

441 = 3\u00d73\u00d77\u00d77<\/p>\n\n\n\n

= 32<\/sup>\u00d772<\/sup><\/p>\n\n\n\n

\u221a441 = 3\u00d77<\/p>\n\n\n\n

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

(ii)<\/strong> 196<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

196 = 2\u00d72\u00d77\u00d77<\/p>\n\n\n\n

= 22<\/sup>\u00d772<\/sup><\/p>\n\n\n\n

\u221a196 = 2\u00d77<\/p>\n\n\n\n

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


(iii) <\/strong>529<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

529 = 23\u00d723<\/p>\n\n\n\n

= 232<\/sup><\/p>\n\n\n\n

\u221a529 = 23<\/p>\n\n\n\n

(iv) <\/strong>1764<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

1764 = 2\u00d72\u00d73\u00d73\u00d77\u00d77<\/p>\n\n\n\n

= 22<\/sup>\u00d732<\/sup>\u00d772<\/sup><\/p>\n\n\n\n

\u221a1764 = 2\u00d73\u00d77<\/p>\n\n\n\n

= 42<\/p>\n\n\n\n


(v) <\/strong>1156<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

1156 = 2\u00d72\u00d717\u00d717<\/p>\n\n\n\n

= 22<\/sup>\u00d7172<\/sup><\/p>\n\n\n\n

\u221a1156 = 2\u00d717<\/p>\n\n\n\n

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

(vi) <\/strong>4096<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

4096 = 2\u00d72\u00d72\u00d72\u00d72\u00d72\u00d72\u00d72\u00d72\u00d72\u00d72\u00d72<\/p>\n\n\n\n

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

\u221a4096 = 26<\/sup><\/p>\n\n\n\n

= 64<\/p>\n\n\n\n

(vii) <\/strong>7056<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

7056 = 2\u00d72\u00d72\u00d72\u00d721\u00d721<\/p>\n\n\n\n

= 22<\/sup>\u00d722<\/sup>\u00d7212<\/sup><\/p>\n\n\n\n

\u221a7056 = 2\u00d72\u00d721<\/p>\n\n\n\n

= 84<\/p>\n\n\n\n

(viii) <\/strong>8281<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

8281 = 91\u00d791<\/p>\n\n\n\n

= 912<\/sup><\/p>\n\n\n\n

\u221a8281 = 91<\/p>\n\n\n\n


(ix) <\/strong>11664<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

11664 = 2\u00d72\u00d72\u00d72\u00d73\u00d73\u00d73\u00d73\u00d73\u00d73<\/p>\n\n\n\n

= 22<\/sup>\u00d722<\/sup>\u00d732<\/sup>\u00d732<\/sup>\u00d732<\/sup><\/p>\n\n\n\n

\u221a11664 = 2\u00d72\u00d73\u00d73\u00d73<\/p>\n\n\n\n

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

(x) <\/strong>47089<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

47089 = 217\u00d7217<\/p>\n\n\n\n

= 2172<\/sup><\/p>\n\n\n\n

\u221a47089 = 217<\/p>\n\n\n\n


(xi)<\/strong> 24336<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

24336 = 2\u00d72\u00d72\u00d72\u00d73\u00d73\u00d713\u00d713<\/p>\n\n\n\n

= 22<\/sup>\u00d722<\/sup>\u00d732<\/sup>\u00d7132<\/sup><\/p>\n\n\n\n

\u221a24336 = 2\u00d72\u00d73\u00d713<\/p>\n\n\n\n

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

(xii) <\/strong>190969<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

190969 = 23\u00d723\u00d719\u00d719<\/p>\n\n\n\n

= 232<\/sup>\u00d7192<\/sup><\/p>\n\n\n\n

\u221a190969 = 23\u00d719<\/p>\n\n\n\n

= 437<\/p>\n\n\n\n


(xiii) <\/strong>586756<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

586756 = 2\u00d72\u00d7383\u00d7383<\/p>\n\n\n\n

= 22<\/sup>\u00d73832<\/sup><\/p>\n\n\n\n

\u221a586756 = 2\u00d7383<\/p>\n\n\n\n

= 766<\/p>\n\n\n\n

(xiv) <\/strong>27225<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

27225 = 5\u00d75\u00d73\u00d73\u00d711\u00d711<\/p>\n\n\n\n

= 52<\/sup>\u00d732<\/sup>\u00d7112<\/sup><\/p>\n\n\n\n

\u221a27225 = 5\u00d73\u00d711<\/p>\n\n\n\n

= 165<\/p>\n\n\n\n


(xv) <\/strong>3013696<\/p>\n\n\n\n

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

3013696 = 2\u00d72\u00d72\u00d72\u00d72\u00d72\u00d7217\u00d7217<\/p>\n\n\n\n

= 26<\/sup>\u00d72172<\/sup><\/p>\n\n\n\n

\u221a3013696 = 23<\/sup>\u00d7217<\/p>\n\n\n\n

= 1736<\/p>\n\n\n\n

3.Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.<\/strong><\/p>\n\n\n\n

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

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

180 = (2 \u00d7 2) \u00d7 (3 \u00d7 3) \u00d7 5<\/p>\n\n\n\n

=22<\/sup> \u00d7 32<\/sup> \u00d7 5<\/p>\n\n\n\n

To make the unpaired 5 into paired, multiply the number with 5<\/p>\n\n\n\n

180 \u00d7 5 = 22<\/sup> \u00d7 32<\/sup> \u00d7 52<\/sup><\/p>\n\n\n\n

\u2234 Square root of \u221a (180 \u00d7 5) = 2 \u00d7 3 \u00d7 5<\/p>\n\n\n\n

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

4. Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.<\/strong><\/p>\n\n\n\n

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

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

147 = (7 \u00d7 7) \u00d7 3<\/p>\n\n\n\n

=72<\/sup> \u00d7 3<\/p>\n\n\n\n

To make the unpaired 3 into paired, multiply the number with 3<\/p>\n\n\n\n

147 \u00d7 3 = 72<\/sup> \u00d7 32<\/sup><\/p>\n\n\n\n

\u2234 Square root of \u221a (147 \u00d7 3) = 7 \u00d7 3<\/p>\n\n\n\n

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

5. Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.<\/strong><\/p>\n\n\n\n

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

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

3645 = (3 \u00d7 3) \u00d7 (3 \u00d7 3) \u00d7 (3 \u00d7 3) \u00d7 5<\/p>\n\n\n\n

=32<\/sup> \u00d7 32<\/sup> \u00d7 32<\/sup> \u00d7 5<\/p>\n\n\n\n

To make the unpaired 5 into paired, the number 3645 has to be divided by 5<\/p>\n\n\n\n

3645 \u00f7 5 = 32<\/sup> \u00d7 32<\/sup> \u00d7 32<\/sup><\/p>\n\n\n\n

\u2234 Square root of \u221a (3645 \u00f7 5) = 3 \u00d7 3 \u00d7 3<\/p>\n\n\n\n

= 27<\/p>\n\n\n\n

6. Find the smallest number by which 1152 must be divided so that it becomes a square. Also, find the square root of the number so obtained.<\/strong><\/p>\n\n\n\n

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

Firstly let\u2019s find the prime factors for<\/p>\n\n\n\n

1152 = (2 \u00d7 2) \u00d7 (2 \u00d7 2) \u00d7 (2 \u00d7 2) \u00d7 2 \u00d7 (3 \u00d7 3)<\/p>\n\n\n\n

=22<\/sup> \u00d7 22<\/sup> \u00d7 22<\/sup> \u00d7 32<\/sup> \u00d7 2<\/p>\n\n\n\n

To make the unpaired 2 into paired, the number 1152 has to be divided by 2<\/p>\n\n\n\n

1152 \u00f7 2 = 22<\/sup> \u00d7 22 <\/sup>\u00d7 22<\/sup> \u00d7 32<\/sup><\/p>\n\n\n\n

\u2234 Square root of \u221a (1152 \u00f7 2) = 2 \u00d7 2 \u00d7 2 \u00d7 3<\/p>\n\n\n\n

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

7. The product of two numbers is 1296. If one number is 16 times the other, find the numbers.<\/strong><\/p>\n\n\n\n

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

Let us consider two numbers a and b<\/p>\n\n\n\n

So we know that one of the number, a =16b<\/p>\n\n\n\n

a \u00d7 b = 1296<\/p>\n\n\n\n

16b \u00d7 b = 1296<\/p>\n\n\n\n

16b2<\/sup> = 1296<\/p>\n\n\n\n

b2<\/sup> = 1296\/16 = 81<\/p>\n\n\n\n

b = 9<\/p>\n\n\n\n

a = 16b<\/p>\n\n\n\n

= 16(9)<\/p>\n\n\n\n

= 144<\/p>\n\n\n\n

\u2234 a =144 and b =9<\/p>\n\n\n\n

8. A welfare association collected Rs 202500 as donation from the residents. If each paid as many rupees as there were residents, find the number of residents.<\/strong><\/p>\n\n\n\n

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

Let us consider total residents as a<\/p>\n\n\n\n

So, each paid Rs. a<\/p>\n\n\n\n

Total collection = a (a) = a2<\/sup><\/p>\n\n\n\n

We know that the total Collection = 202500<\/p>\n\n\n\n

a = \u221a 202500<\/p>\n\n\n\n

a = \u221aa(2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 5 \u00d7 5)<\/p>\n\n\n\n

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

= 450<\/p>\n\n\n\n

\u2234 Total residents = 450<\/p>\n\n\n\n

9. A society collected Rs 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?<\/strong><\/p>\n\n\n\n

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

Let us consider there were few members, each attributed a paise<\/p>\n\n\n\n

a (a), i.e. total cost collected = 9216 paise<\/p>\n\n\n\n

a2<\/sup> = 9216<\/p>\n\n\n\n

a = \u221a9216<\/p>\n\n\n\n

= 2 \u00d7 2 \u00d7 2 \u00d7 12<\/p>\n\n\n\n

= 96<\/p>\n\n\n\n

\u2234 There were 96 members in the society and each contributed 96 paise<\/p>\n\n\n\n

10. A society collected Rs 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school?<\/strong><\/p>\n\n\n\n

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

Let us consider number of school students as a<\/p>\n\n\n\n

each student contributed a paise<\/p>\n\n\n\n

Total money obtained = a2<\/sup>paise<\/p>\n\n\n\n

= 2304 paise<\/p>\n\n\n\n

a = \u221a2304<\/p>\n\n\n\n

a = \u221a2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3<\/p>\n\n\n\n

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

a = 48<\/p>\n\n\n\n

\u2234 There were 48 students in the school<\/p>\n\n\n\n

11. The area of a square field is 5184 m2<\/sup>. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.<\/strong><\/p>\n\n\n\n

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

Let us consider the side of square field as a<\/p>\n\n\n\n

a2<\/sup> = 5184 m2<\/sup><\/p>\n\n\n\n

a = \u221a5184m<\/p>\n\n\n\n

a = 2 \u00d7 2 \u00d72 \u00d7 9<\/p>\n\n\n\n

= 72 m<\/p>\n\n\n\n

Perimeter of square = 4a<\/p>\n\n\n\n

= 4(72)<\/p>\n\n\n\n

= 288 m<\/p>\n\n\n\n

Perimeter of rectangle = 2 (l + b) = perimeter of the square field<\/p>\n\n\n\n

= 288 m<\/p>\n\n\n\n

2 (2b + b) = 288<\/p>\n\n\n\n

b = 48 and l = 96<\/p>\n\n\n\n

Area of rectangle = 96 \u00d7 48 m2<\/sup><\/p>\n\n\n\n

= 4608 m2<\/sup><\/p>\n\n\n\n

12. Find the least square number, exactly divisible by each one of the numbers: (i) 6, 9, 15 and 20 (ii) 8, 12, 15 and 20<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 6, 9, 15 and 20<\/p>\n\n\n\n

Firstly take L.C.M for 6, 9, 15, 20 which is 180<\/p>\n\n\n\n

So the prime factors of 180 = 22<\/sup> \u00d7 32<\/sup> \u00d7 5<\/p>\n\n\n\n

To make it a perfect square, we have to multiply the number with 5<\/p>\n\n\n\n

180 \u00d7 5 = 22<\/sup> \u00d7 32<\/sup> \u00d7 52<\/sup><\/p>\n\n\n\n

\u2234 900 is the least square number divisible by 6, 9, 15 and 20<\/p>\n\n\n\n

(ii)<\/strong> 8, 2, 15 and 20<\/p>\n\n\n\n

Firstly take L.C.M for 8, 2, 15, 20 which is 360<\/p>\n\n\n\n

So the prime factors of 360 = 22<\/sup> \u00d7 32<\/sup> \u00d72 \u00d7 5<\/p>\n\n\n\n

To make it a perfect square, we have to multiply the number with 2 \u00d7 5 = 10<\/p>\n\n\n\n

360 \u00d7 10 = 22<\/sup> \u00d7 32<\/sup> \u00d7 52<\/sup> \u00d7 22<\/sup><\/p>\n\n\n\n

\u2234 3600 is the least square number divisible by 8, 12, 15 and 20<\/p>\n\n\n\n

13. Find the square roots of 121 and 169 by the method of repeated subtraction.<\/strong><\/p>\n\n\n\n

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

Let us find the square roots of 121 and 169 by the method of repeated subtraction<\/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

Clearly, we have performed operation 11 times<\/p>\n\n\n\n

\u2234 \u221a121 = 11<\/p>\n\n\n\n

169 \u2013 1 = 168<\/p>\n\n\n\n

168 \u2013 3 = 165<\/p>\n\n\n\n

165 \u2013 5 = 160<\/p>\n\n\n\n

160 \u2013 7 = 153<\/p>\n\n\n\n

153 \u2013 9 = 144<\/p>\n\n\n\n

144 \u2013 11 = 133<\/p>\n\n\n\n

133 \u2013 13 = 120<\/p>\n\n\n\n

120 \u2013 15 = 105<\/p>\n\n\n\n

105 \u2013 17 = 88<\/p>\n\n\n\n

88 \u2013 19 = 69<\/p>\n\n\n\n

69 \u2013 21 = 48<\/p>\n\n\n\n

48 \u2013 23 = 25<\/p>\n\n\n\n

25 \u2013 25 = 0<\/p>\n\n\n\n

Clearly, we have performed subtraction 13 times<\/p>\n\n\n\n

\u2234 \u221a169 = 13<\/p>\n\n\n\n

14. Write the prime factorization of the following numbers and hence find their square roots.
(i) 7744<\/strong><\/p>\n\n\n\n

(ii) 9604
(iii) 5929
(iv) 7056<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> 7744<\/p>\n\n\n\n

Prime factors of 7744 is<\/p>\n\n\n\n

7744 = 22<\/sup> \u00d7 22<\/sup> \u00d7 22<\/sup> \u00d7 112<\/sup><\/p>\n\n\n\n

\u2234 The square root of 7744 is<\/p>\n\n\n\n

\u221a7744 = 2 \u00d7 2 \u00d7 2 \u00d7 11<\/p>\n\n\n\n

= 88<\/p>\n\n\n\n

(ii)<\/strong> 9604<\/p>\n\n\n\n

Prime factors of 9604 is<\/p>\n\n\n\n

9604 = 22<\/sup> \u00d7 72<\/sup> \u00d7 72<\/sup><\/p>\n\n\n\n

\u2234 The square root of 9604 is<\/p>\n\n\n\n

\u221a9604 = 2 \u00d7 7 \u00d7 7<\/p>\n\n\n\n

= 98<\/p>\n\n\n\n

(iii)<\/strong> 5929<\/p>\n\n\n\n

Prime factors of 5929 is<\/p>\n\n\n\n

5929 = 112<\/sup> \u00d7 72<\/sup><\/p>\n\n\n\n

\u2234 The square root of 5929 is<\/p>\n\n\n\n

\u221a5929 = 11 \u00d7 7<\/p>\n\n\n\n

= 77<\/p>\n\n\n\n

(iv)<\/strong> 7056<\/p>\n\n\n\n

Prime factors of 7056 is<\/p>\n\n\n\n

7056 = 22<\/sup> \u00d7 22<\/sup> \u00d7 72<\/sup> \u00d7 32<\/sup><\/p>\n\n\n\n

\u2234 The square root of 7056 is<\/p>\n\n\n\n

\u221a7056 = 2 \u00d7 2 \u00d7 7 \u00d7 3<\/p>\n\n\n\n

= 84<\/p>\n\n\n\n

15. The students of class VIII of a school donated Rs 2401 for PM\u2019s National Relief Fund. Each student donated 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

Let us consider number of students as a<\/p>\n\n\n\n

Each student denoted a rupee<\/p>\n\n\n\n

So, total amount collected is a \u00d7 a rupees = 2401<\/p>\n\n\n\n

a2<\/sup> = 2401<\/p>\n\n\n\n

a = \u221a2401<\/p>\n\n\n\n

a = 49<\/p>\n\n\n\n

\u2234 There are 49 students in the class.<\/p>\n\n\n\n

16. A PT teacher wants to arrange maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.<\/strong><\/p>\n\n\n\n

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

Let us consider number of rows as a<\/p>\n\n\n\n

No. of columns = a<\/p>\n\n\n\n

Total number of students who sat in the field = a2<\/sup><\/p>\n\n\n\n

Total students a2<\/sup> + 71 = 6000<\/p>\n\n\n\n

a2<\/sup> = 5929<\/p>\n\n\n\n

a = \u221a5929<\/p>\n\n\n\n

a = 77<\/p>\n\n\n\n

\u2234 total number of rows are 77.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.5 PAGE NO: 3.43<\/h4>\n\n\n\n

1.Find the square root of each of the following by long division method:
(i) 12544 (ii) 97344
(iii) 286225 (iv) 390625
(v) 363609 (vi) 974169
(vii) 120409 (viii) 1471369
(ix) 291600 (x) 9653449
(xi) 1745041 (xii) 4008004
(xiii) 20657025 (xiv) 152547201
(xv) 20421361 (xvi)62504836
(xvii) 82264900 (xviii) 3226694416
(xix) 6407522209 (xx) 3915380329<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>12544<\/p>\n\n\n\n

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

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

\u2234 the square root of 12544<\/p>\n\n\n\n

\u221a12544 = 112<\/p>\n\n\n\n

(ii) <\/strong>97344<\/p>\n\n\n\n

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

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

\u2234 the square root of 97344<\/p>\n\n\n\n

\u221a97344 = 312<\/p>\n\n\n\n


(iii) <\/strong>286225<\/p>\n\n\n\n

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

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

\u2234 the square root of 286225<\/p>\n\n\n\n

\u221a286225 = 535<\/p>\n\n\n\n

(iv) <\/strong>390625<\/p>\n\n\n\n

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

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

\u2234 the square root of 390625<\/p>\n\n\n\n

\u221a390625 = 625<\/p>\n\n\n\n


(v) <\/strong>363609<\/p>\n\n\n\n

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

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

\u2234 the square root of 363609<\/p>\n\n\n\n

\u221a36369 = 603<\/p>\n\n\n\n

(vi)<\/strong> 974169<\/p>\n\n\n\n

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

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

\u2234 the square root of 974169<\/p>\n\n\n\n

\u221a974169 = 987<\/p>\n\n\n\n

(vii) <\/strong>120409<\/p>\n\n\n\n

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

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

\u2234 the square root of 120409<\/p>\n\n\n\n

\u221a120409 = 347<\/p>\n\n\n\n

(viii) <\/strong>1471369<\/p>\n\n\n\n

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

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

\u2234 the square root of 1471369<\/p>\n\n\n\n

\u221a1471369 = 1213<\/p>\n\n\n\n


(ix) <\/strong>291600<\/p>\n\n\n\n

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

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

\u2234 the square root of 291600<\/p>\n\n\n\n

\u221a291600 = 540<\/p>\n\n\n\n

(x) <\/strong>9653449<\/p>\n\n\n\n

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

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

\u2234 the square root of 9653449<\/p>\n\n\n\n

\u221a9653449 = 3107<\/p>\n\n\n\n


(xi) <\/strong>1745041<\/p>\n\n\n\n

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

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

\u2234 the square root of 1745041<\/p>\n\n\n\n

\u221a1745041 = 1321<\/p>\n\n\n\n

(xii) <\/strong>4008004<\/p>\n\n\n\n

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

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

\u2234 the square root of 4008004<\/p>\n\n\n\n

\u221a4008004 = 2002<\/p>\n\n\n\n


(xiii) <\/strong>20657025<\/p>\n\n\n\n

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

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

\u2234 the square root of 20657025<\/p>\n\n\n\n

\u221a20657025 = 4545<\/p>\n\n\n\n

(xiv) <\/strong>152547201<\/p>\n\n\n\n

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

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

\u2234 the square root of 152547201<\/p>\n\n\n\n

\u221a152547201 = 12351<\/p>\n\n\n\n


(xv) <\/strong>20421361<\/p>\n\n\n\n

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

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

\u2234 the square root of 20421361<\/p>\n\n\n\n

\u221a20421361 = 4519<\/p>\n\n\n\n

(xvi) <\/strong>62504836<\/p>\n\n\n\n

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

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

\u2234 the square root of 62504836<\/p>\n\n\n\n

\u221a62504836 = 7906<\/p>\n\n\n\n


(xvii) <\/strong>82264900<\/p>\n\n\n\n

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

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

\u2234 the square root of 82264900<\/p>\n\n\n\n

\u221a82264900 = 9070<\/p>\n\n\n\n

(xviii) <\/strong>3226694416<\/p>\n\n\n\n

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

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

\u2234 the square root of 3226694416<\/p>\n\n\n\n

\u221a3226694416 = 56804<\/p>\n\n\n\n


(xix) <\/strong>6407522209<\/p>\n\n\n\n

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

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

\u2234 the square root of 6407522209<\/p>\n\n\n\n

\u221a6407522209 = 80047<\/p>\n\n\n\n

(xx) <\/strong>3915380329<\/p>\n\n\n\n

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

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

\u2234 the square root of 3915380329<\/p>\n\n\n\n

\u221a3915380329 = 62573<\/p>\n\n\n\n

2. Find the least number which must be subtracted from the following numbers to make them a perfect square:
(i) 2361
(ii) 194491
(iii) 26535
(iv) 161605
(v) 4401624<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>2361<\/p>\n\n\n\n

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

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

\u2234 57 has to be subtracted from 2361 to get a perfect square.<\/p>\n\n\n\n


(ii) <\/strong>194491<\/p>\n\n\n\n

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

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

\u2234 10 has to be subtracted from 194491 to get a perfect square.<\/p>\n\n\n\n


(iii) <\/strong>26535<\/p>\n\n\n\n

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

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

\u2234 291 has to be subtracted from 26535 to get a perfect square.<\/p>\n\n\n\n


(iv) <\/strong>161605<\/p>\n\n\n\n

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

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

\u2234 1 has to be subtracted from 161605 to get a perfect square.<\/p>\n\n\n\n


(v)<\/strong> 4401624<\/p>\n\n\n\n

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

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

\u2234 20 has to be subtracted from 4401624 to get a perfect square.<\/p>\n\n\n\n

3. Find the least number which must be added to the following numbers to make them a perfect square:
(i) 5607
(ii)4931
(iii) 4515600
(iv) 37460
(v) 506900<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>5607<\/p>\n\n\n\n

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

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

The remainder is 131<\/p>\n\n\n\n

Since, (74)2<\/sup> < 5607<\/p>\n\n\n\n

We take, the next perfect square number i.e., (75)2<\/sup><\/p>\n\n\n\n

(75)2<\/sup> = 5625 > 5607<\/p>\n\n\n\n

So, the number to be added = 5625 \u2013 5607 = 18<\/p>\n\n\n\n


(ii) <\/strong>4931<\/p>\n\n\n\n

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

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

The remainder is 31<\/p>\n\n\n\n

Since, (70)2<\/sup> < 4931<\/p>\n\n\n\n

We take, the next perfect square number i.e., (71)2<\/sup><\/p>\n\n\n\n

(71)2<\/sup> = 5041 > 4931<\/p>\n\n\n\n

So, the number to be added = 5041 \u2013 4931 = 110<\/p>\n\n\n\n


(iii) <\/strong>4515600<\/p>\n\n\n\n

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

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

The remainder is 4224<\/p>\n\n\n\n

Since, (2124)2<\/sup> < 4515600<\/p>\n\n\n\n

We take, the next perfect square number i.e., (2125)2<\/sup><\/p>\n\n\n\n

(2125)2<\/sup> = 4515625 > 4515600<\/p>\n\n\n\n

So, the number to be added = 4515625 \u2013 4515600 = 25<\/p>\n\n\n\n


(iv)<\/strong> 37460<\/p>\n\n\n\n

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

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

The remainder is 211<\/p>\n\n\n\n

Since, (193)2<\/sup> < 37460<\/p>\n\n\n\n

We take, the next perfect square number i.e., (194)2<\/sup><\/p>\n\n\n\n

(194)2<\/sup> = 37636 > 37460<\/p>\n\n\n\n

So, the number to be added = 37636 \u2013 37460 = 176<\/p>\n\n\n\n


(v)<\/strong> 506900<\/p>\n\n\n\n

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

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

The remainder is 1379<\/p>\n\n\n\n

Since, (711)2<\/sup> < 506900<\/p>\n\n\n\n

We take, the next perfect square number i.e., (712)2<\/sup><\/p>\n\n\n\n

(712)2<\/sup> = 506944 > 506900<\/p>\n\n\n\n

So, the number to be added = 506944 \u2013 506900 = 44<\/p>\n\n\n\n

4. Find the greatest number of 5 digits which is a perfect square.<\/strong><\/p>\n\n\n\n

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

We know that the greatest 5 digit number is 99999<\/p>\n\n\n\n

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

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

The remainder is 143<\/p>\n\n\n\n

So, the greatest 5 digit perfect square number is:<\/p>\n\n\n\n

99999 \u2013 143 = 99856<\/p>\n\n\n\n

\u2234 99856 is the required greatest 5 digit perfect square number.<\/p>\n\n\n\n

5. Find the least number of 4 digits which is a perfect square.<\/strong><\/p>\n\n\n\n

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

We know that the least 4 digit number is 1000<\/p>\n\n\n\n

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

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

The remainder is 39<\/p>\n\n\n\n

Since, (31)2<\/sup> < 1000<\/p>\n\n\n\n

We take, the next perfect square number i.e., (32)2<\/sup><\/p>\n\n\n\n

(32)2<\/sup> = 1024 > 1000<\/p>\n\n\n\n

\u2234 1024 is the required least number 4 digit number which is a perfect square.<\/p>\n\n\n\n

6. Find the least number of six digits which is a perfect square.<\/strong><\/p>\n\n\n\n

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

We know that the least 6 digit number is 100000<\/p>\n\n\n\n

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

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

The remainder is 144<\/p>\n\n\n\n

Since, (316)2<\/sup> < 100000<\/p>\n\n\n\n

We take, the next perfect square number i.e., (317)2<\/sup><\/p>\n\n\n\n

(317)2<\/sup> = 100489 > 100000<\/p>\n\n\n\n

\u2234 100489 is the required least number 6 digit number which is a perfect square.<\/p>\n\n\n\n

7. Find the greatest number of 4 digits which is a perfect square.<\/strong><\/p>\n\n\n\n

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

We know that the greatest 4 digit number is 9999<\/p>\n\n\n\n

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

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

The remainder is 198<\/p>\n\n\n\n

So, the greatest 4 digit perfect square number is:<\/p>\n\n\n\n

9999 \u2013 198 = 9801<\/p>\n\n\n\n

\u2234 9801 is the required greatest 4 digit perfect square number.<\/p>\n\n\n\n

8. A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row<\/strong><\/p>\n\n\n\n

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

We know that the total number of soldiers = 8160<\/p>\n\n\n\n

Number of soldiers left out = 60<\/p>\n\n\n\n

Number of soldiers arranged in rows to form a perfect square = 8160 \u2013 60 = 8100<\/p>\n\n\n\n

\u2234 number of soldiers in each row = \u221a8100<\/p>\n\n\n\n

= \u221a (9\u00d79\u00d710\u00d710)<\/p>\n\n\n\n

= 9\u00d710<\/p>\n\n\n\n

= 90<\/p>\n\n\n\n

9. The area of a square field is 60025m2<\/sup>. A man cycles along its boundary at 18 Km\/hr. In how much time will he return at the starting point?<\/strong><\/p>\n\n\n\n

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

We know that the area of square field = 60025 m2<\/sup><\/p>\n\n\n\n

Speed of cyclist = 18 km\/h<\/p>\n\n\n\n

= 18 \u00d7 (1000\/60\u00d760)<\/p>\n\n\n\n

= 5 m\/s2<\/sup><\/p>\n\n\n\n

Area = 60025 m2<\/sup><\/p>\n\n\n\n

Side2<\/sup> = 60025<\/p>\n\n\n\n

Side = \u221a60025<\/p>\n\n\n\n

= 245<\/p>\n\n\n\n

We know, Total length of boundary = 4 \u00d7 Side<\/p>\n\n\n\n

= 4 \u00d7 245<\/p>\n\n\n\n

= 980 m<\/p>\n\n\n\n

\u2234 Time taken to return to the starting point = 980\/5<\/p>\n\n\n\n

= 196 seconds<\/p>\n\n\n\n

= 3 minutes 16 seconds<\/p>\n\n\n\n

10. The cost of levelling and turning a square lawn at Rs 2.50 per m2<\/sup> is Rs13322.50 Find the cost of fencing it at Rs 5 per metre.<\/strong><\/p>\n\n\n\n

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

We know that the cost of levelling and turning a square lawn = 2.50 per m2<\/sup><\/p>\n\n\n\n

Total cost of levelling and turning = Rs. 13322.50<\/p>\n\n\n\n

Total area of square lawn = 13322.50\/2.50<\/p>\n\n\n\n

= 5329 m2<\/sup><\/p>\n\n\n\n

Side2<\/sup> = 5329<\/p>\n\n\n\n

Side of square lawn = \u221a5329<\/p>\n\n\n\n

= 73 m<\/p>\n\n\n\n

So, total length of lawn = 4 \u00d7 73<\/p>\n\n\n\n

= 292 m<\/p>\n\n\n\n

\u2234 Cost of fencing the lawn at Rs 5 per metre = 292 \u00d7 5<\/p>\n\n\n\n

= Rs. 1460<\/p>\n\n\n\n

11. Find the greatest number of three digits which is a perfect square.<\/strong><\/p>\n\n\n\n

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

We know that the greatest 3 digit number is 999<\/p>\n\n\n\n

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

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

The remainder is 38<\/p>\n\n\n\n

So, the greatest 3 digit perfect square number is:<\/p>\n\n\n\n

999 \u2013 38 = 961<\/p>\n\n\n\n

\u2234 961 is the required greatest 3 digit perfect square number.<\/p>\n\n\n\n

12. Find the smallest number which must be added to 2300 so that it becomes a perfect square.<\/strong><\/p>\n\n\n\n

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

By using long division method let\u2019s find the square root of 2300<\/p>\n\n\n\n

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

The remainder is 91<\/p>\n\n\n\n

Since, (47)2<\/sup> < 2300<\/p>\n\n\n\n

We take, the next perfect square number i.e., (48)2<\/sup><\/p>\n\n\n\n

(48)2<\/sup> = 2304 > 2300<\/p>\n\n\n\n

\u2234 The smallest number required to be added to 2300 to get a perfect square is<\/p>\n\n\n\n

2304 \u2013 2300 = 4<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.6 PAGE NO: 3.48<\/h4>\n\n\n\n

1. Find the square root of:<\/strong><\/p>\n\n\n\n

(i) 441\/961<\/strong><\/p>\n\n\n\n

(ii) 324\/841<\/strong><\/p>\n\n\n\n

(iii) 4 29\/29<\/strong><\/p>\n\n\n\n

(iv) 2 14\/25<\/strong><\/p>\n\n\n\n

(v) 2 137\/196<\/strong><\/p>\n\n\n\n

(vi) 23 26\/121<\/strong><\/p>\n\n\n\n

(vii) 25 544\/729<\/strong><\/p>\n\n\n\n

(viii) 75 46\/49<\/strong><\/p>\n\n\n\n

(ix) 3 942\/2209<\/strong><\/p>\n\n\n\n

(x) 3 334\/3025<\/strong><\/p>\n\n\n\n

(xi) 21 2797\/3364<\/strong><\/p>\n\n\n\n

(xii) 38 11\/25<\/strong><\/p>\n\n\n\n

(xiii) 23 394\/729<\/strong><\/p>\n\n\n\n

(xiv) 21 51\/169<\/strong><\/p>\n\n\n\n

(xv) 10 151\/225<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>441\/961<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a441\/961 = 21\/31<\/p>\n\n\n\n

(ii) <\/strong>324\/841<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a324\/841= 18\/29<\/p>\n\n\n\n

(iii) <\/strong>4 29\/29<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(4 29\/29) = \u221a(225\/49) = 15\/7<\/p>\n\n\n\n

(iv) <\/strong>2 14\/25<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(2 14\/25) = \u221a(64\/25) = 8\/5<\/p>\n\n\n\n

(v) <\/strong>2 137\/196<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a2 137\/196 = \u221a (529\/196) = 23\/14<\/p>\n\n\n\n

(vi) <\/strong>23 26\/121<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(23 26\/121) = \u221a(2809\/121) = 53\/11<\/p>\n\n\n\n

(vii) <\/strong>25 544\/729<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(25 544\/729) = \u221a(18769\/729) = 137\/27<\/p>\n\n\n\n

(viii) <\/strong>75 46\/49<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(75 46\/49) = \u221a(3721\/49) = 61\/7<\/p>\n\n\n\n

(ix) <\/strong>3 942\/2209<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(3 942\/2209) = \u221a(7569\/2209) = 87\/47<\/p>\n\n\n\n

(x) <\/strong>3 334\/3025<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(3 334\/3025) = \u221a(9409\/3025) = 97\/55<\/p>\n\n\n\n

(xi) <\/strong>21 2797\/3364<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(21 2797\/3364) = \u221a(73441\/3364) = 271\/58<\/p>\n\n\n\n

(xii) <\/strong>38 11\/25<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(38 11\/25) = \u221a(961\/25) = 31\/5<\/p>\n\n\n\n

(xiii) <\/strong>23 394\/729<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(23 394\/729) = \u221a(17161\/729) = 131\/27 = 4 23\/27<\/p>\n\n\n\n

(xiv) <\/strong>21 51\/169<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(21 51\/169) = \u221a(3600\/169) = 60\/13 = 4 8\/13<\/p>\n\n\n\n

(xv) <\/strong>10 151\/225<\/p>\n\n\n\n

The square root of<\/p>\n\n\n\n

\u221a(10 151\/225) = \u221a(2401\/225) = 49\/15 = 3 4\/15<\/p>\n\n\n\n

2. Find the value of:<\/strong><\/p>\n\n\n\n

(i) \u221a80\/\u221a405<\/strong><\/p>\n\n\n\n

(ii) \u221a441\/\u221a625<\/strong><\/p>\n\n\n\n

(iii) \u221a1587\/\u221a1728<\/strong><\/p>\n\n\n\n

(iv) \u221a72 \u00d7\u221a338<\/strong><\/p>\n\n\n\n

(v) \u221a45 \u00d7 \u221a20<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>\u221a80\/\u221a405<\/p>\n\n\n\n

\u221a80\/\u221a405 = \u221a16\/\u221a81 = 4\/9<\/p>\n\n\n\n

(ii) <\/strong>\u221a441\/\u221a625<\/p>\n\n\n\n

\u221a441\/\u221a625 = 21\/25<\/p>\n\n\n\n

(iii) <\/strong>\u221a1587\/\u221a1728<\/p>\n\n\n\n

\u221a1587\/\u221a1728 = \u221a529\/\u221a576 = 23\/24<\/p>\n\n\n\n

(iv) <\/strong>\u221a72 \u00d7\u221a338<\/p>\n\n\n\n

\u221a72 \u00d7\u221a338 = \u221a(2\u00d72\u00d72\u00d73\u00d73) \u00d7\u221a(2\u00d713\u00d713)<\/p>\n\n\n\n

By using the formula \u221aa \u00d7 \u221ab = \u221a(a\u00d7b)<\/p>\n\n\n\n

= \u221a(2\u00d72\u00d72\u00d73\u00d73\u00d72\u00d713\u00d713)<\/p>\n\n\n\n

= 22<\/sup> \u00d7 3 \u00d7 13<\/p>\n\n\n\n

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

(v) <\/strong>\u221a45 \u00d7 \u221a20<\/p>\n\n\n\n

\u221a45 \u00d7 \u221a20 = \u221a(5\u00d73\u00d73) \u00d7 \u221a(5\u00d72\u00d72)<\/p>\n\n\n\n

By using the formula \u221aa \u00d7 \u221ab = \u221a(a\u00d7b)<\/p>\n\n\n\n

= \u221a(5\u00d73\u00d73\u00d75\u00d72\u00d72)<\/p>\n\n\n\n

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

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

3. The area of a square field is 80 244\/729 square metres. Find the length of each side of the field.<\/strong><\/p>\n\n\n\n

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

We know that the given area = 80 244\/729 m2<\/sup><\/p>\n\n\n\n

= 58564\/729 m2<\/sup><\/p>\n\n\n\n

If L is length of each side<\/p>\n\n\n\n

L2<\/sup> = 58564\/729<\/p>\n\n\n\n

L = \u221a (58564\/729) = \u221a58564\/\u221a729<\/p>\n\n\n\n

= 242\/27<\/p>\n\n\n\n

= 8 26\/27<\/p>\n\n\n\n

\u2234 Length is 8 26\/27<\/p>\n\n\n\n

4. The area of a square field is 30 1\/4m2<\/sup>. Calculate the length of the side of the square.<\/strong><\/p>\n\n\n\n

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

We know that the given area = 30 1\/4 m2<\/sup><\/p>\n\n\n\n

= 121\/4 m2<\/sup><\/p>\n\n\n\n

If L is length of each side then,<\/p>\n\n\n\n

L2<\/sup> = 121\/4<\/p>\n\n\n\n

L = \u221a(121\/4)  = \u221a121\/\u221a4<\/p>\n\n\n\n

= 11\/2<\/p>\n\n\n\n

\u2234 Length is 11\/2<\/p>\n\n\n\n

5. Find the length of a side of a square playground whose area is equal to the area of a rectangular field of dimensions 72m and 338 m.<\/strong><\/p>\n\n\n\n

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

By using the formula<\/p>\n\n\n\n

Area of rectangular field = l \u00d7 b<\/p>\n\n\n\n

= 72 \u00d7 338 m2<\/sup><\/p>\n\n\n\n

= 24336 m2<\/sup><\/p>\n\n\n\n

Area of square, L2<\/sup> = 24336 m2<\/sup><\/p>\n\n\n\n

L = \u221a24336<\/p>\n\n\n\n

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

\u2234 Length of side of square playground is 156 m.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.7 PAGE NO: 3.52<\/h4>\n\n\n\n

Find the square root of the following numbers in decimal form:
1. 84.8241<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 84.8241<\/p>\n\n\n\n

\u221a84.8241 = 9.21<\/p>\n\n\n\n

2. 0.7225<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 0.7225<\/p>\n\n\n\n

\u221a0.7225 = 0.85<\/p>\n\n\n\n

3. 0.813604<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 0.813604<\/p>\n\n\n\n

\u221a0.813604 = 0.902<\/p>\n\n\n\n

4. 0.00002025<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 0.00002025<\/p>\n\n\n\n

\u221a0.00002025 = 0.0045<\/p>\n\n\n\n

5. 150.0625<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 150.0625<\/p>\n\n\n\n

\u221a150.0625 = 12.25<\/p>\n\n\n\n

6. 225.6004<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 225.6004<\/p>\n\n\n\n

\u221a225.6004 = 15.02<\/p>\n\n\n\n

7. 3600.720036<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 3600.720036<\/p>\n\n\n\n

\u221a3600.720036 = 60.006<\/p>\n\n\n\n

8. 236.144689<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 236.144689<\/p>\n\n\n\n

\u221a236.144689 = 15.367<\/p>\n\n\n\n

9. 0.00059049<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 0.00059049<\/p>\n\n\n\n

\u221a0.00059049 = 0.0243<\/p>\n\n\n\n

10. 176.252176<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 176.252176<\/p>\n\n\n\n

\u221a176.252176 = 13.276<\/p>\n\n\n\n

11. 9998.0001<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 9998.0001<\/p>\n\n\n\n

\u221a9998.0001 = 99.99<\/p>\n\n\n\n

12. 0.00038809<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 0.00038809<\/p>\n\n\n\n

\u221a0.00038809 = 0.0197<\/p>\n\n\n\n

13. What is that fraction which when multiplied by itself gives 227.798649?<\/strong><\/p>\n\n\n\n

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

Let us consider a number a<\/p>\n\n\n\n

Where, a = \u221a227.798649<\/p>\n\n\n\n

= 15.093<\/p>\n\n\n\n

By using long division method let us verify<\/p>\n\n\n\n

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

\u2234 15.093 is the fraction which when multiplied by itself gives 227.798649.<\/p>\n\n\n\n

14. The area of a square playground is 256.6404 square meter. Find the length of one side of the playground.<\/strong><\/p>\n\n\n\n

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

We know that the given area of a square playground = 256.6404<\/p>\n\n\n\n

i.e., L2<\/sup> = 256.6404 m2<\/sup><\/p>\n\n\n\n

L = \u221a256.6404<\/p>\n\n\n\n

= 16.02m<\/p>\n\n\n\n

By using long division method let us verify<\/p>\n\n\n\n

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

\u2234 length of one side of the playground is 16.02m.<\/p>\n\n\n\n

15. What is the fraction which when multiplied by itself gives 0.00053361?<\/strong><\/p>\n\n\n\n

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

Let us consider a number a<\/p>\n\n\n\n

Where, a = \u221a0.00053361<\/p>\n\n\n\n

= 0.0231<\/p>\n\n\n\n

By using long division method let us verify<\/p>\n\n\n\n

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

\u2234 0.0231 is the fraction which when multiplied by itself gives 0.00053361.<\/p>\n\n\n\n

16. Simplify:<\/strong><\/p>\n\n\n\n

(i) (\u221a59.29 \u2013 \u221a5.29)\/ (\u221a59.29 + \u221a5.29)<\/strong><\/p>\n\n\n\n

(ii) (\u221a0.2304 + \u221a0.1764)\/ (\u221a0.2304 \u2013 \u221a0.1764)<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>(\u221a59.29 \u2013 \u221a5.29)\/ (\u221a59.29 + \u221a5.29)<\/p>\n\n\n\n

Firstly let us find the square root \u221a59.29 and \u221a5.29<\/p>\n\n\n\n

\u221a59.29 = \u221a5929\/ \u221a100<\/p>\n\n\n\n

= 77\/10<\/p>\n\n\n\n

= 7.7<\/p>\n\n\n\n

\u221a5.29 = \u221a5.29\/ \u221a100<\/p>\n\n\n\n

= 23\/10<\/p>\n\n\n\n

= 2.3<\/p>\n\n\n\n

So, (7.7 \u2013 2.3)\/ (7.7 + 2.3)<\/p>\n\n\n\n

= 54\/10<\/p>\n\n\n\n

= 0.54<\/p>\n\n\n\n

(ii)<\/strong> (\u221a0.2304 + \u221a0.1764)\/ (\u221a0.2304 \u2013 \u221a0.1764)<\/p>\n\n\n\n

Firstly let us find the square root \u221a0.2304 and \u221a0.1764<\/p>\n\n\n\n

\u221a0.2304 = \u221a2304\/ \u221a10000<\/p>\n\n\n\n

= 48\/100<\/p>\n\n\n\n

= 0.48<\/p>\n\n\n\n

\u221a0.1764 = \u221a1764\/ \u221a10000<\/p>\n\n\n\n

= 42\/100<\/p>\n\n\n\n

= 0.42<\/p>\n\n\n\n

So, (0.48 + 0.42)\/ (0.48 \u2013 0.42)<\/p>\n\n\n\n

= 0.9\/0.06<\/p>\n\n\n\n

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

17. Evaluate \u221a50625 and hence find the value of \u221a506.25 + \u221a5.0625<\/strong><\/p>\n\n\n\n

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

By using long division method let us find the \u221a50625 <\/strong><\/p>\n\n\n\n

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

So now, \u221a<\/strong>506.25 = <\/strong>\u221a50625\/ \u221a100<\/p>\n\n\n\n

= 225\/10<\/p>\n\n\n\n

= 22.5<\/p>\n\n\n\n

\u221a5.0625 = \u221a50625\/ \u221a10000<\/p>\n\n\n\n

= 225\/100<\/p>\n\n\n\n

= 2.25<\/p>\n\n\n\n

So equating in the above equation we get,<\/p>\n\n\n\n

\u221a506.25 + \u221a5.0625 = 22.5 + 2.25<\/p>\n\n\n\n

= 24.75<\/p>\n\n\n\n

18. Find the value of \u221a103.0225 and hence find the value of
(i) \u221a10302.25
(ii) \u221a1.030225<\/strong><\/p>\n\n\n\n

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

By using long division method let us find the<\/p>\n\n\n\n

\u221a103.0225 = \u221a(1030225\/10000) = \u221a1030225\/\u221a10000<\/p>\n\n\n\n

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

So now, (i)\u221a<\/strong>10302.25 = <\/strong>\u221a(1030225\/ 100)<\/p>\n\n\n\n

= 1015\/ 10<\/p>\n\n\n\n

= 101.5<\/p>\n\n\n\n

(ii)\u221a1.030225 = \u221a1030225\/ \u221a1000000<\/p>\n\n\n\n

= 1015\/1000<\/p>\n\n\n\n

= 1.015<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.8 PAGE NO: 3.56<\/h4>\n\n\n\n

1. Find the square root of each of the following correct to three places of decimal.
(i) 5 (ii) 7
(iii) 17 (iv) 20
(v) 66 (vi) 427
(vii) 1.7 (viii) 23.1
(ix) 2.5 (x) 237.615
(xi) 15.3215 (xii) 0.9
(xiii) 0.1 (xiv) 0.016
(xv) 0.00064 (xvi) 0.019
(xvii) 7\/8 (xviii) 5\/12
(xix) 2 1\/2 (xx) 287 5\/8<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>5<\/p>\n\n\n\n

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

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

\u2234 the square root of 5 is 2.236<\/p>\n\n\n\n

(ii) <\/strong>7<\/p>\n\n\n\n

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

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

\u2234 the square root of 7 is 2.646<\/p>\n\n\n\n


(iii) <\/strong>17<\/p>\n\n\n\n

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

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

\u2234 the square root of 17 is 4.123<\/p>\n\n\n\n

(iv) <\/strong>20<\/p>\n\n\n\n

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

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

\u2234 the square root of 20 is 4.472<\/p>\n\n\n\n


(v) <\/strong>66<\/p>\n\n\n\n

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

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

\u2234 the square root of 66 is 8.124<\/p>\n\n\n\n

(vi) <\/strong>427<\/p>\n\n\n\n

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

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

\u2234 the square root of 427 is 20.664<\/p>\n\n\n\n


(vii) <\/strong>1.7<\/p>\n\n\n\n

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

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

\u2234 the square root of 1.7 is 1.304<\/p>\n\n\n\n

(viii) <\/strong>23.1<\/p>\n\n\n\n

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

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

\u2234 the square root of 23.1 is 4.806<\/p>\n\n\n\n


(ix) <\/strong>2.5<\/p>\n\n\n\n

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

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

\u2234 the square root of 2.5 is 1.581<\/p>\n\n\n\n

(x) <\/strong>237.615<\/p>\n\n\n\n

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

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

\u2234 the square root of 237.615 is 15.415<\/p>\n\n\n\n


(xi) <\/strong>15.3215<\/p>\n\n\n\n

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

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

\u2234 the square root of 15.3215 is 3.914<\/p>\n\n\n\n

(xii) <\/strong>0.9<\/p>\n\n\n\n

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

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

\u2234 the square root of 0.9 is 0.949<\/p>\n\n\n\n


(xiii) <\/strong>0.1<\/p>\n\n\n\n

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

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

\u2234 the square root of 0.1 is 0.316<\/p>\n\n\n\n

(xiv) <\/strong>0.016<\/p>\n\n\n\n

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

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

\u2234 the square root of 0.016 is 0.126<\/p>\n\n\n\n


(xv) <\/strong>0.00064<\/p>\n\n\n\n

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

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

\u2234 the square root of 0.00064 is 0.025<\/p>\n\n\n\n

(xvi) <\/strong>0.019<\/p>\n\n\n\n

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

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

\u2234 the square root of 0.019 is 0.138<\/p>\n\n\n\n


(xvii) <\/strong>7\/8<\/p>\n\n\n\n

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

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

\u2234 the square root of 7\/8 is 0.935<\/p>\n\n\n\n

(xviii) <\/strong>5\/12<\/p>\n\n\n\n

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

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

\u2234 the square root of 5\/12 is 0.645<\/p>\n\n\n\n


(xix) <\/strong>2 1\/2 <\/strong><\/p>\n\n\n\n

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

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

\u2234 the square root of 5\/2 is 1.581<\/p>\n\n\n\n

(xx) <\/strong>287 5\/8<\/p>\n\n\n\n

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

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

\u2234 the square root of 2301\/8 is 16.960<\/p>\n\n\n\n

2. Find the square root of 12.0068 correct to four decimal places.<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 12.0068 is 3.4651<\/p>\n\n\n\n

3. Find the square root of 11 correct to five decimal places.<\/strong><\/p>\n\n\n\n

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

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

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

\u2234 the square root of 11 is 3.31662<\/p>\n\n\n\n

4. Give that: \u221a2 = 1.414, \u221a3 = 1.732, \u221a5 = 2.236 and \u221a7 = 2.646, evaluate each of the following:
(i) \u221a (144\/7)
(ii) \u221a (2500\/3)<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>\u221a (144\/7)<\/p>\n\n\n\n

Now let us simplify the given equation<\/p>\n\n\n\n

\u221a (144\/7) = \u221a (12\u00d712)\/ \u221a7<\/p>\n\n\n\n

= 12\/ 2.646<\/p>\n\n\n\n

= 4.535<\/p>\n\n\n\n

(ii) <\/strong>\u221a (2500\/3)<\/p>\n\n\n\n

Now let us simplify the given equation<\/p>\n\n\n\n

\u221a (2500\/3) = \u221a (5\u00d75\u00d710\u00d710) \/\u221a3<\/p>\n\n\n\n

= 5\u00d710\/ 1.732<\/p>\n\n\n\n

= 50\/1.732<\/p>\n\n\n\n

= 28.867<\/p>\n\n\n\n

5. Given that \u221a2 = 1.414, \u221a3 = 1.732, \u221a5 = 2.236 and \u221a7 = 2.646 find the square roots of the following:
(i) 196\/75
(ii) 400\/63
(iii) 150\/7
(iv) 256\/5
(v) 27\/50<\/strong><\/p>\n\n\n\n

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

(i) <\/strong>196\/75<\/p>\n\n\n\n

Let us find the square root for196\/75<\/p>\n\n\n\n

\u221a<\/strong>(196\/75) = \u221a<\/strong>(196)\/ \u221a<\/strong> (75)<\/p>\n\n\n\n

= \u221a<\/strong>(14\u00d714)\/ \u221a<\/strong> (5\u00d75\u00d73)<\/p>\n\n\n\n

= 14\/ (5\u221a<\/strong>3)<\/p>\n\n\n\n

= 14\/ (5\u00d71.732)<\/p>\n\n\n\n

= 14\/8.66<\/p>\n\n\n\n

= 1.617<\/p>\n\n\n\n


(ii) <\/strong>400\/63<\/p>\n\n\n\n

Let us find the square root for400\/63<\/p>\n\n\n\n

\u221a<\/strong>(400\/63) = \u221a<\/strong>(400)\/ \u221a<\/strong> (63)<\/p>\n\n\n\n

= \u221a<\/strong>(20\u00d720)\/ \u221a<\/strong> (3\u00d73\u00d77)<\/p>\n\n\n\n

= 20\/ (3\u221a<\/strong>7)<\/p>\n\n\n\n

= 20\/ (3\u00d72.646)<\/p>\n\n\n\n

= 20\/7.938<\/p>\n\n\n\n

= 2.520<\/p>\n\n\n\n

(iii) <\/strong>150\/7<\/p>\n\n\n\n

Let us find the square root for150\/7<\/p>\n\n\n\n

\u221a<\/strong>(150\/7) = \u221a<\/strong>(150)\/ \u221a<\/strong> (7)<\/p>\n\n\n\n

= \u221a<\/strong>(3\u00d75\u00d75\u00d72)\/ \u221a<\/strong> (7)<\/p>\n\n\n\n

= (5\u221a<\/strong>3\u00d7\u221a<\/strong>2)\/ (\u221a<\/strong>7)<\/p>\n\n\n\n

= 5\u00d71.732\u00d71.414\/ (2.646)<\/p>\n\n\n\n

= 12.245\/2.646<\/p>\n\n\n\n

= 4.628<\/p>\n\n\n\n

(iv) <\/strong>256\/5<\/p>\n\n\n\n

Let us find the square root for256\/5<\/p>\n\n\n\n

\u221a<\/strong>(256\/5) = \u221a<\/strong>(256)\/ \u221a<\/strong> (5)<\/p>\n\n\n\n

= \u221a<\/strong>(16\u00d716)\/ \u221a<\/strong> (5)<\/p>\n\n\n\n

= 16\/ (\u221a<\/strong>5)<\/p>\n\n\n\n

= 16\/2.236<\/p>\n\n\n\n

= 7.155<\/p>\n\n\n\n


(v) <\/strong>27\/50<\/p>\n\n\n\n

Let us find the square root for27\/50<\/p>\n\n\n\n

\u221a<\/strong>(27\/50) = \u221a<\/strong>(27)\/ \u221a<\/strong> (50)<\/p>\n\n\n\n

= \u221a<\/strong>(3\u00d73\u00d73)\/ \u221a<\/strong> (5\u00d75\u00d72)<\/p>\n\n\n\n

= (3\u221a<\/strong>3)\/ (5\u221a<\/strong>2)<\/p>\n\n\n\n

= (3\u00d71.732)\/ (5\u00d71.414)<\/p>\n\n\n\n

= 5.196\/7.07<\/p>\n\n\n\n

= 0.735<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 3.9 PAGE NO: 3.61<\/h4>\n\n\n\n
Square Root Table<\/td><\/tr>
Number<\/strong><\/td>Square Root(\u221a)<\/strong><\/td>Number<\/strong><\/td>Square Root(\u221a)<\/strong><\/td>Number<\/strong><\/td>Square Root(\u221a)<\/strong><\/td>Number<\/strong><\/td>Square Root(\u221a)<\/strong><\/td>Number<\/strong><\/td>Square Root(\u221a)<\/strong><\/td><\/tr>
1<\/td>1<\/td>21<\/td>4.583<\/td>41<\/td>6.403<\/td>61<\/td>7.81<\/td>81<\/td>9<\/td><\/tr>
2<\/td>1.414<\/td>22<\/td>4.69<\/td>42<\/td>6.481<\/td>62<\/td>7.874<\/td>82<\/td>9.055<\/td><\/tr>
3<\/td>1.732<\/td>23<\/td>4.796<\/td>43<\/td>6.557<\/td>63<\/td>7.937<\/td>83<\/td>9.11<\/td><\/tr>
4<\/td>2<\/td>24<\/td>4.899<\/td>44<\/td>6.633<\/td>64<\/td>8<\/td>84<\/td>9.165<\/td><\/tr>
5<\/td>2.236<\/td>25<\/td>5<\/td>45<\/td>6.708<\/td>65<\/td>8.062<\/td>85<\/td>9.22<\/td><\/tr>
6<\/td>2.449<\/td>26<\/td>5.099<\/td>46<\/td>6.782<\/td>66<\/td>8.124<\/td>86<\/td>9.274<\/td><\/tr>
7<\/td>2.646<\/td>27<\/td>5.196<\/td>47<\/td>6.856<\/td>67<\/td>8.185<\/td>87<\/td>9.327<\/td><\/tr>
8<\/td>2.828<\/td>28<\/td>5.292<\/td>48<\/td>6.928<\/td>68<\/td>8.246<\/td>88<\/td>9.381<\/td><\/tr>
9<\/td>3<\/td>29<\/td>5.385<\/td>49<\/td>7<\/td>69<\/td>8.307<\/td>89<\/td>9.434<\/td><\/tr>
10<\/td>3.162<\/td>30<\/td>5.477<\/td>50<\/td>7.071<\/td>70<\/td>8.367<\/td>90<\/td>9.487<\/td><\/tr>
11<\/td>3.317<\/td>31<\/td>5.568<\/td>51<\/td>7.141<\/td>71<\/td>8.426<\/td>91<\/td>9.539<\/td><\/tr>
12<\/td>3.464<\/td>32<\/td>5.657<\/td>52<\/td>7.211<\/td>72<\/td>8.485<\/td>92<\/td>9.592<\/td><\/tr>
13<\/td>3.606<\/td>33<\/td>5.745<\/td>53<\/td>7.28<\/td>73<\/td>8.544<\/td>93<\/td>9.644<\/td><\/tr>
14<\/td>3.742<\/td>34<\/td>5.831<\/td>54<\/td>7.348<\/td>74<\/td>8.602<\/td>94<\/td>9.695<\/td><\/tr>
15<\/td>3.873<\/td>35<\/td>5.916<\/td>55<\/td>7.416<\/td>75<\/td>8.66<\/td>95<\/td>9.747<\/td><\/tr>
16<\/td>4<\/td>36<\/td>6<\/td>56<\/td>7.483<\/td>76<\/td>8.718<\/td>96<\/td>9.798<\/td><\/tr>
17<\/td>4.123<\/td>37<\/td>6.083<\/td>57<\/td>7.55<\/td>77<\/td>8.775<\/td>97<\/td>9.849<\/td><\/tr>
18<\/td>4.243<\/td>28<\/td>6.164<\/td>58<\/td>7.616<\/td>78<\/td>8.832<\/td>98<\/td>9.899<\/td><\/tr>
19<\/td>4.359<\/td>29<\/td>6.245<\/td>59<\/td>7.681<\/td>79<\/td>8.888<\/td>99<\/td>9.95<\/td><\/tr>
20<\/td>4.472<\/td>40<\/td>6.325<\/td>60<\/td>7.746<\/td>80<\/td>8.944<\/td>100<\/td>10<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Using square root table, find the square roots of the following:<\/strong><\/p>\n\n\n\n

1. 7<\/strong><\/p>\n\n\n\n

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

From square root table we know,<\/p>\n\n\n\n

Square root of 7 is:<\/p>\n\n\n\n

\u221a7 = 2.645<\/p>\n\n\n\n

\u2234 The square root of 7 is 2.645<\/p>\n\n\n\n

2. 15<\/strong><\/p>\n\n\n\n

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

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

15 = 3 x 5<\/p>\n\n\n\n

So, \u221a15 = \u221a3 x \u221a5<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 3 and 5 are:<\/p>\n\n\n\n

\u221a3 = 1.732 and \u221a5 = 2.236<\/p>\n\n\n\n

\u21d2 \u221a15 = 1.732 x 2.236 = 3.873<\/p>\n\n\n\n

\u2234 The square root of 15 is 3.873<\/p>\n\n\n\n

3. 74<\/strong><\/p>\n\n\n\n

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

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

74 = 2 x 37<\/p>\n\n\n\n

So, \u221a74 = \u221a2 x \u221a37<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 2 and 37 are:<\/p>\n\n\n\n

\u221a2 = 1.414 and \u221a37 = 6.083<\/p>\n\n\n\n

\u21d2 \u221a74 = 1.414 x 6.083 = 8.602<\/p>\n\n\n\n

\u2234 The square root of 74 is 8.602<\/p>\n\n\n\n

4. 82<\/strong><\/p>\n\n\n\n

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

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

82 = 2 x 41<\/p>\n\n\n\n

So, \u221a82 = \u221a2 x \u221a41<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 2 and 41 are:<\/p>\n\n\n\n

\u221a2 = 1.414 and \u221a41 = 6.403<\/p>\n\n\n\n

\u21d2 \u221a82 = 1.414 x 6.403 = 9.055<\/p>\n\n\n\n

\u2234 The square root of 82 is 9.055<\/p>\n\n\n\n

5. 198<\/strong><\/p>\n\n\n\n

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

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

198 = 2 x 9 x 11<\/p>\n\n\n\n

So, \u221a198 = \u221a2 x \u221a9 x \u221a11<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 2, 9 and 11 are:<\/p>\n\n\n\n

\u221a2 = 1.414, \u221a9 = 3 and \u221a11 = 3.317<\/p>\n\n\n\n

\u21d2 \u221a198 = 1.414 x 3 x 3.317 = 14.071<\/p>\n\n\n\n

\u2234 The square root of 198 is 14.071<\/p>\n\n\n\n

6. 540<\/strong><\/p>\n\n\n\n

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

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

540 = 6 x 9 x 10<\/p>\n\n\n\n

So, \u221a540 = \u221a6 x \u221a9 x \u221a10<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 6, 9 and 10 are:<\/p>\n\n\n\n

\u221a6 = 2.449, \u221a9 = 3 and \u221a10 = 3.162<\/p>\n\n\n\n

\u21d2 \u221a540 = 2.449 x 3 x 3.162 = 23.24<\/p>\n\n\n\n

\u2234 The square root of 540 is 23.24<\/p>\n\n\n\n

7. 8700<\/strong><\/p>\n\n\n\n

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

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

8700 = 87 x 100<\/p>\n\n\n\n

So, \u221a8700 = \u221a87 x \u221a100<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 87 and 100 are:<\/p>\n\n\n\n

\u221a8700 = 9.327 and \u221a100 = 10<\/p>\n\n\n\n

\u21d2 \u221a8700 = 9.327 x 10 = 93.27<\/p>\n\n\n\n

\u2234 The square root of 8700 is 93.27<\/p>\n\n\n\n

8. 3509<\/strong><\/p>\n\n\n\n

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

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

3509 = 121 x 29<\/p>\n\n\n\n

So, \u221a3509 = \u221a121 x \u221a29<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 121 and 29 are:<\/p>\n\n\n\n

\u221a121 = 11 and \u221a29 = 5.385<\/p>\n\n\n\n

\u21d2 \u221a3509 = 11 x 10 = 5.385<\/p>\n\n\n\n

\u2234 The square root of 3509 is 59.235<\/p>\n\n\n\n

9. 6929<\/strong><\/p>\n\n\n\n

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

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

6929 = 169 x 41<\/p>\n\n\n\n

So, \u221a6929 = \u221a169 x \u221a41<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 169 and 41 are:<\/p>\n\n\n\n

\u221a169 = 13 and \u221a41 = 6.403<\/p>\n\n\n\n

\u21d2 \u221a6929 = 13 x 6.403 = 83.239<\/p>\n\n\n\n

\u2234 The square root of 6929 is 83.239<\/p>\n\n\n\n

10. 25725<\/strong><\/p>\n\n\n\n

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

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

25725 = 3 x 7 x 25 x 49<\/p>\n\n\n\n

So, \u221a25725 = \u221a3 x \u221a7 x \u221a25 x \u221a49<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 3, 7, 25 and 49 are:<\/p>\n\n\n\n

\u221a3 = 1.732, \u221a7 = 2.646, \u221a25 = 5 and \u221a49 = 7<\/p>\n\n\n\n

\u21d2 \u221a25725 = 1.732 x 2.646 x 5 x 7 = 160.41<\/p>\n\n\n\n

\u2234 The square root of 25725 is 160.41<\/p>\n\n\n\n

11. 1312.<\/strong><\/p>\n\n\n\n

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

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

1312 = 2 x 16 x 41<\/p>\n\n\n\n

So, \u221a1312 = \u221a2 x \u221a16 x \u221a41<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 2, 16 and 41 are:<\/p>\n\n\n\n

\u221a2 = 1.414, \u221a16 = 4 and \u221a41 = 6.403<\/p>\n\n\n\n

\u21d2 \u221a1312 = 1.414 x 4 x 6.403 = 36.22<\/p>\n\n\n\n

\u2234 The square root of 1312 is 36.22<\/p>\n\n\n\n

12. 4192<\/strong><\/p>\n\n\n\n

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

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

4192 = 2 x 16 x 131<\/p>\n\n\n\n

So, \u221a4192 = \u221a2 x \u221a16 x \u221a131<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 2 and16 are:<\/p>\n\n\n\n

\u221a2 = 1.414 and \u221a16 = 4<\/p>\n\n\n\n

The square root of 131 is not listed in the table<\/p>\n\n\n\n

Thus, let\u2019s apply long division to find it<\/p>\n\n\n\n

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

So, square root of 131 is 11.445<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

\u21d2 \u221a4192 = 1.414 x 4 x 11.445 = 64.75<\/p>\n\n\n\n

\u2234 The square root of 4192 is 64.75<\/p>\n\n\n\n

13. 4955<\/strong><\/p>\n\n\n\n

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

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

4955 = 5 x 991<\/p>\n\n\n\n

So, \u221a4955 = \u221a5 x \u221a991<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 5 is:<\/p>\n\n\n\n

\u221a5 = 2.236<\/p>\n\n\n\n

The square root of 991 is not listed in the table<\/p>\n\n\n\n

Thus, let\u2019s apply long division to find it<\/p>\n\n\n\n

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

So, square root of 991 is 31.480<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

\u21d2 \u221a4955 = 2.236 x 31.480 = 70.39<\/p>\n\n\n\n

\u2234 The square root of 4955 is 70.39<\/p>\n\n\n\n

14. 99\/144<\/strong><\/p>\n\n\n\n

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

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

99\/144 = (9 x 11) \/ (12 x 12)<\/p>\n\n\n\n

So, \u221a(99\/144) = \u221a[(9 x 11) x (12 x 12)]<\/p>\n\n\n\n

= 3\/12 x \u221a11<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 11 is:<\/p>\n\n\n\n

\u221a11 = 3.317<\/p>\n\n\n\n

\u21d2 \u221a(99\/144) = 3\/12 x 3.317 = 3.317\/4 = 0.829<\/p>\n\n\n\n

\u2234 The square root of 99\/144 is 0.829<\/p>\n\n\n\n

15. 57\/169<\/strong><\/p>\n\n\n\n

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

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

57\/169 = (3 x 19) \/ (13 x 13)<\/p>\n\n\n\n

So, \u221a(57\/169) = \u221a[(3 x 19) x (13 x 13)]<\/p>\n\n\n\n

= \u221a3 x \u221a19 x 1\/13<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 3 and 19 is:<\/p>\n\n\n\n

\u221a3 = 1.732 and \u221a19 = 4.359<\/p>\n\n\n\n

\u21d2 \u221a(57\/169) = 1.732 x 4.359 x 1\/13 = 0.581<\/p>\n\n\n\n

\u2234 The square root of 57\/169 is 0.581<\/p>\n\n\n\n

16. 101\/169<\/strong><\/p>\n\n\n\n

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

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

101\/169 = 101 \/ (13 x 13)<\/p>\n\n\n\n

So, \u221a(101\/169) = \u221a[101 \/ (13 x 13)]<\/p>\n\n\n\n

= \u221a101\/13<\/p>\n\n\n\n

From square root table we don\u2019t have the square root of 101<\/p>\n\n\n\n

Thus, we have to manipulate the number such that we get the square root of a number less than 100<\/p>\n\n\n\n

\u221a101 = \u221a(1.01 x 100)<\/p>\n\n\n\n

= \u221a1.01 x 10<\/p>\n\n\n\n

Now, we have to find the square of 1.01<\/p>\n\n\n\n

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

\u221a1 = 1 and \u221a2 = 1.414 (From the square root table)<\/p>\n\n\n\n

Their difference = 1.414 \u2013 1 = 0.414<\/p>\n\n\n\n

Hence, for a difference of 1 (2 \u2013 1), the difference in the value of the square root is 0.414<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

For the difference of 0.01, the difference in the value of the square roots will be<\/p>\n\n\n\n

0.01 x 1.414 = 0.00414<\/p>\n\n\n\n

\u2234\u221a1.01 = 1 + 0.00414 = 1.00414<\/p>\n\n\n\n

Then, \u221a101 = 1.00414 x 10 = 10.0414<\/p>\n\n\n\n

\u21d2 \u221a(101\/169) = \u221a101\/13 = 10.0414\/13<\/p>\n\n\n\n

\u2234 The square root of 101\/169 is 0.773<\/p>\n\n\n\n

17. 13.21<\/strong><\/p>\n\n\n\n

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

We need to find \u221a13.21<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 13 and 14 are:<\/p>\n\n\n\n

\u221a13 = 3.606 and \u221a14 = 3.742<\/p>\n\n\n\n

Their difference = 3.742 \u2013 3.606 = 0.136<\/p>\n\n\n\n

Hence, for a difference of 1 (14 \u2013 13), the difference in the value of the square root is 0.136<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

For the difference of 0.21, the difference in the value of the square roots will be<\/p>\n\n\n\n

0.136 x 0.21 = 0.0286<\/p>\n\n\n\n

\u21d2 \u221a13.21 = 3.606 + 0.0286 = 3.635<\/p>\n\n\n\n

\u2234 The square root of 13.21 is 3.635<\/p>\n\n\n\n

18. 21.97<\/strong><\/p>\n\n\n\n

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

We need to find \u221a21.97<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 21 and 22 are:<\/p>\n\n\n\n

\u221a21 = 4.583 and \u221a22 = 4.690<\/p>\n\n\n\n

Their difference = 4.690 \u2013 4.583 = 0.107<\/p>\n\n\n\n

Hence, for a difference of 1 (23 \u2013 22), the difference in the value of the square root is 0.107<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

For the difference of 0.97, the difference in the value of the square roots will be<\/p>\n\n\n\n

0.107 x 0.97 = 0.104<\/p>\n\n\n\n

\u21d2 \u221a21.97 = 4.583 + 0.104 = 4.687<\/p>\n\n\n\n

\u2234 The square root of 21.97 is 4.687<\/p>\n\n\n\n

19. 110<\/strong><\/p>\n\n\n\n

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

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

110 = 11 x 10<\/p>\n\n\n\n

So, \u221a110 = \u221a11 x \u221a10<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 11 and 10 are:<\/p>\n\n\n\n

\u221a11 = 3.317 and \u221a10 = 3.162<\/p>\n\n\n\n

\u21d2 \u221a110 = 3.317 x 3.162 = 10.488<\/p>\n\n\n\n

\u2234 The square root of 110 is 10.488<\/p>\n\n\n\n

20. 1110<\/strong><\/p>\n\n\n\n

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

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

1110 = 37 x 30<\/p>\n\n\n\n

So, \u221a1110 = \u221a37 x \u221a30<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 37 and 30 are:<\/p>\n\n\n\n

\u221a37 = 6.083 and \u221a30 = 5.477<\/p>\n\n\n\n

\u21d2 \u221a1110 = 6.083 x 5.477 = 33.317<\/p>\n\n\n\n

\u2234 The square root of 1110 is 33.317<\/p>\n\n\n\n

21. 11.11<\/strong><\/p>\n\n\n\n

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

We need to find \u221a11.11<\/p>\n\n\n\n

From square root table we know,<\/p>\n\n\n\n

Square root of 11 and 12 are:<\/p>\n\n\n\n

\u221a11 = 3.317 and \u221a12 = 3.464<\/p>\n\n\n\n

Their difference = 3.464 \u2013 3.317 = 0.147<\/p>\n\n\n\n

Hence, for a difference of 1 (12 \u2013 11), the difference in the value of the square root is 0.147<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

For the difference of 0.11, the difference in the value of the square roots will be<\/p>\n\n\n\n

0.11 x 0.147 = 0.01617<\/p>\n\n\n\n

\u21d2 \u221a11.11 = 3.317 + 0.0162 = 3.333<\/p>\n\n\n\n

\u2234 The square root of 11.11 is 3.333<\/p>\n\n\n\n

22. The area of a square field is 325m2<\/sup>. Find the approximate length of one side of the field.<\/strong><\/p>\n\n\n\n

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

We know that the given area of the field = 325 m2<\/sup><\/p>\n\n\n\n

To find the approximate length of the side of the field we will have to calculate the square root of 325<\/p>\n\n\n\n

\u221a325 = \u221a25 x \u221a13<\/p>\n\n\n\n

From the square root table, we know<\/p>\n\n\n\n

\u221a25 = 5 and \u221a13 = 3.606

\u21d2 \u221a325 = 5 x 3.606 = 18.030<\/p>\n\n\n\n

\u2234 The approximate length of one side of the field is 18.030 m<\/p>\n\n\n\n

23. Find the length of a side of a square, whose area is equal to the area of a rectangle with sides 240 m and 70 m.<\/strong><\/p>\n\n\n\n

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

We know that from the question,<\/p>\n\n\n\n

Area of square = Area of rectangle<\/p>\n\n\n\n

Side2<\/sup> = 240 \u00d7 70<\/p>\n\n\n\n

Side = \u221a(240 \u00d7 70)<\/p>\n\n\n\n

= \u221a(10 \u00d7 10 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 7)<\/p>\n\n\n\n

= 20\u221a42<\/p>\n\n\n\n

Now, from the square root table, we know \u221a42 = 6.481<\/p>\n\n\n\n

= 20 \u00d7 6.48<\/p>\n\n\n\n

= 129.60 m<\/p>\n\n\n\n

\u2234 The length of side of the square is 129.60 m<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 3: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 3\u2013Squares and Square Roots<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 8 Maths Chapter 3\u2013Squares and Square Roots PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n