{"id":3130,"date":"2022-12-06T07:36:43","date_gmt":"2022-12-06T07:36:43","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=3130"},"modified":"2023-01-31T06:53:46","modified_gmt":"2023-01-31T06:53:46","slug":"ncert-solutions-class-8-maths-chapter-6-squares-and-square-roots-ex-6-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-6-squares-and-square-roots-ex-6-3\/","title":{"rendered":"NCERT Solutions Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 8 Mathematics\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 6 (Squares and Square)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 8 Mathematics <strong>Chapter &#8211; 6 Squares and Square <\/strong>Exercise 6.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<h4><span style=\"color: #000000;\"><strong>Chapter 6: Squares and Square Roots<\/strong><\/span><\/h4>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-6-squares-and-square-roots-ex-6-1\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 8 Maths Ex &#8211; 6.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-6-squares-and-square-roots-ex-6-2\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 8 Maths Ex &#8211; 6.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-6-squares-and-square-roots-ex-6-4\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 8 Maths Ex &#8211; 6.4<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 6.3\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. What could be the possible \u2018one\u2019s\u2019 digits of the square root of each of the following numbers?<br \/>\n<\/strong><strong>(i)<\/strong>\u00a09801<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii)<\/strong>\u00a099856<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iii)<\/strong>\u00a0998001<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iv)<\/strong> 657666025<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>(i) <\/strong>9801<\/span><br \/>\n<span style=\"color: #000000;\">1 x 1 = 1 and 9 x 9 = 81<\/span><br \/>\n<span style=\"color: #000000;\">\u2235 The possible one\u2019s digit of the square root of the number 9801 could be 1 or 9.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii) <\/strong>99856<\/span><br \/>\n<span style=\"color: #000000;\">4 x 4 = 16 and 6 x 6 = 36<\/span><br \/>\n<span style=\"color: #000000;\">\u2235 The possible one\u2019s digit of the square root of the number 99856 could be 4 or 6.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iii) <\/strong>998001<\/span><br \/>\n<span style=\"color: #000000;\">1 \u00d7 1 = 1 and 9 x 9 = 81<\/span><br \/>\n<span style=\"color: #000000;\">\u2235 The possible one\u2019s digit of the square root of the number 998001 could be 1 or 9.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iv) <\/strong>657666025<\/span><br \/>\n<span style=\"color: #000000;\">5 x 5 = 25<\/span><br \/>\n<span style=\"color: #000000;\">\u2235 The possible one\u2019s digit of the square root of the number 657666025 could be 5.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Without doing any calculation, find the numbers which are surely not perfect squares.<br \/>\n<\/strong><strong>(i)<\/strong>\u00a0153<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii)<\/strong>\u00a0257<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iii)<\/strong>\u00a0408<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iv)<\/strong> 441<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>We know that natural numbers ending with the digits 0, 2, 3, 7 and 8 are not perfect square.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) <\/strong>153<\/span><br \/>\n<span style=\"color: #000000;\">The number 153 is surely not a perfect square because it ends in 3 whereas the square numbers end with 0, 1, 4, 5, 6 or 9.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii) <\/strong>257<\/span><br \/>\n<span style=\"color: #000000;\">The number 257 is surely not a perfect square because it ends in 7 whereas the square numbers end with 0, 1, 4, 5, 6 or 9.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iii) <\/strong>408<\/span><br \/>\n<span style=\"color: #000000;\">The number 408 is surely not a perfect square because it ends in 8 whereas the square numbers end with 0, 1, 4, 5, 6 or 9.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iv) <\/strong>441<\/span><br \/>\n<span style=\"color: #000000;\">The number may be a perfect square as the square numbers end wTith 0, 1, 4, 5, 6 or 9.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Find the square roots of 100 and 169 by the method of repeated subtraction<\/strong>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n100<br \/>\n<\/strong>100 \u2013 1 = 99<br \/>\n99 \u2013 3 = 96<br \/>\n96 \u2013 5 = 91<br \/>\n91 \u2013 7 = 84<br \/>\n84 \u2013 9 = 75<br \/>\n75 \u2013 11 = 64<br \/>\n64 \u2013 13 = 51<br \/>\n51 \u2013 15 = 36<br \/>\n36 \u2013 17 = 19<br \/>\n19 \u2013 19 = 0<br \/>\nHere, we have performed subtraction ten times.<br \/>\n\u2234 \u221a100 = 10<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>169<br \/>\n<\/strong>169 \u2013 1 = 168<br \/>\n168 \u2013 3 = 165<br \/>\n165 \u2013 5 = 160<br \/>\n160 \u2013 7 = 153<br \/>\n153 \u2013 9 = 144<br \/>\n144 \u2013 11 = 133<br \/>\n133 \u2013 13 = 120<br \/>\n120 \u2013 15 = 105<br \/>\n105 \u2013 17 = 88<br \/>\n88 \u2013 19 = 69<br \/>\n69 \u2013 21 = 48<br \/>\n48 \u2013 23 = 25<br \/>\n25 \u2013 25 = 0<br \/>\nHere, we have performed subtraction thirteen times.<br \/>\n\u2234 \u221a169 = 13<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Find the square roots of the following numbers by the Prime Factorisation Method.<br \/>\n<\/strong><strong>(i)<\/strong> 729\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<strong>(ii)<\/strong> 400\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<strong>(iii)<\/strong>\u00a01764<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iv)<\/strong> 4095\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <strong>(v)<\/strong> 7744\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <strong>(vi)<\/strong>\u00a09604<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(vii)\u00a0<\/strong>5929\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<strong>(viii)<\/strong> 9216\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<strong>(ix)<\/strong>\u00a0529<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(x)<\/strong>\u00a08100<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> 729<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3139\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4i.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"65\" height=\"230\" \/><br \/>\n729 = 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3<br \/>\n\u21d2 729 = (3\u00d73)\u00d7(3\u00d73)\u00d7(3\u00d73)<br \/>\n\u21d2 729 = (3)<sup>3<br \/>\n<\/sup>\u21d2 \u221a729 = 3\u00d73\u00d73 = 27<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> 400\u00a0 \u00a0 \u00a0 \u00a0<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3140\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4ii.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"77\" height=\"226\" \/><\/span><br \/>\n<span style=\"color: #000000;\">400 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 5<br \/>\n\u21d2 400 = (2\u00d72)\u00d7(2\u00d72)\u00d7(5\u00d75)<br \/>\n\u21d2 400 = (2\u00d72\u00d75)<sup>2<\/sup> =20<sup>2<\/sup><br \/>\n\u21d2 \u221a400 = 20<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii)<\/strong> 1764<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3141\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4iii.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"67\" height=\"238\" \/><br \/>\n1764 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 7 \u00d7 7<br \/>\n\u21d2 1764 = (2\u00d72)\u00d7(3\u00d73)\u00d7(7\u00d77)<br \/>\n\u21d2 1764 = (2\u00d73\u00d77)<sup>2<\/sup> = 42<sup>2<br \/>\n<\/sup>\u21d2 \u221a1764 = 42<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv)<\/strong> 4096<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3142\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4iv.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"68\" height=\"437\" \/><br \/>\n4096 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2<br \/>\n\u21d2 4096 = (2\u00d72)\u00d7(2\u00d72)\u00d7(2\u00d72)\u00d7(2\u00d72)\u00d7(2\u00d72)\u00d7(2\u00d72)<br \/>\n\u21d2 4096 = (2\u00d72\u00d72\u00d72\u00d72\u00d72)<sup>2<\/sup> = 64<sup>2<br \/>\n<\/sup>\u21d2 \u221a4096 = 64<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v)<\/strong> 7744<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3144\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4v.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"73\" height=\"302\" \/><br \/>\n7744 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 11 \u00d7 11<br \/>\n\u21d2 7744 = (2\u00d72)\u00d7(2\u00d72)\u00d7(2\u00d72)\u00d7(11\u00d711)<br \/>\n\u21d2 7744 = (2\u00d72\u00d72\u00d711)<sup>2<\/sup> = 88<sup>2<br \/>\n<\/sup>\u21d2 \u221a7744 = 88<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) <\/strong>9604<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3145\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4vi.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"57\" height=\"234\" \/><br \/>\n9604 = 2 \u00d7 2 \u00d7 7 \u00d7 7 \u00d7 7 \u00d7 7<br \/>\n\u21d2 9604 = ( 2 \u00d7 2 ) \u00d7 ( 7 \u00d7 7 ) \u00d7 ( 7 \u00d7 7 )<br \/>\n\u21d2 9604 = ( 2\u00d77\u00d77 )<sup>2<\/sup> = 98<sup>2<br \/>\n<\/sup>\u21d2 \u221a9604 = 98<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vii)\u00a0<\/strong>5929<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3146\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4vii.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"62\" height=\"163\" \/><br \/>\n5929 = 7 \u00d7 7 \u00d7 11 \u00d7 11<br \/>\n\u21d2 5929 = (7\u00d77)\u00d7(11\u00d711)<br \/>\n\u21d2 5929 = (7\u00d711)<sup>2\u00a0<\/sup>= 77<sup>2<br \/>\n<\/sup>\u21d2 \u221a5929 = 77<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(viii)\u00a0<\/strong>9216<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3147\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4viii.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"85\" height=\"444\" \/><br \/>\n9216 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3<br \/>\n\u21d2 9216 = (2\u00d72) \u00d7 (2\u00d72) \u00d7 ( 2 \u00d7 2 ) \u00d7 ( 2 \u00d7 2 ) \u00d7 ( 2 \u00d7 2 ) \u00d7 ( 3 \u00d7 3 )<br \/>\n\u21d2 9216 = 96 \u00d7 96<br \/>\n\u21d2 9216 = (96)<sup>2<br \/>\n<\/sup>\u21d2 \u221a9216 = 96<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ix)\u00a0<\/strong>529<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3143\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4ix.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"67\" height=\"100\" \/><br \/>\n529 = 23 \u00d7 23<br \/>\n529 = (23)<sup>2<br \/>\n<\/sup>\u221a529 = 23<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(x)\u00a0<\/strong>8100<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3148\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-4x.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"72\" height=\"294\" \/><br \/>\n8100 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5<br \/>\n\u21d2 8100 = (2\u00d72) \u00d7(3\u00d73)\u00d7(3\u00d73)\u00d7(5\u00d75)<br \/>\n\u21d2 8100 = (2\u00d73\u00d73\u00d75)\u00d7(2\u00d73\u00d73\u00d75)<br \/>\n\u21d2 8100 = 90\u00d790<br \/>\n\u21d2 8100 = (90)<sup>2<br \/>\n<\/sup>\u21d2 \u221a8100 = 90<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.<br \/>\n<\/strong><strong>(i)<\/strong>\u00a0252<br \/>\n<strong>(ii)<\/strong>\u00a0180<br \/>\n<strong>(iii)<\/strong>\u00a01008<br \/>\n<strong>(iv)<\/strong>\u00a02028<br \/>\n<strong>(v)<\/strong>\u00a01458<br \/>\n<strong>(vi)<\/strong>\u00a0768<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong>\u00a0252<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3149\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-5i.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"67\" height=\"191\" \/><br \/>\n252 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 7<br \/>\n= (2\u00d72) \u00d7 (3\u00d73) \u00d7 7<br \/>\nHere, 7 cannot be paired.<br \/>\n\u2234 We will multiply 252 by 7 to get perfect square.<br \/>\nNew number = 252 \u00d7 7 = 1764<br \/>\nNow each prime factor has a pair. Therefore, 252 \u00d7 7 = 1764 is a perfect square.<br \/>\nThus the required smallest number is 7.<br \/>\n\u21d2 \u221a1764 = 2 \u00d7 3 \u00d7 7 = 42<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> 180<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3150\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-5ii.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"59\" height=\"198\" \/><br \/>\n180 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5<br \/>\n= (2\u00d72) \u00d7 (3\u00d73) \u00d7 5<br \/>\nHere, 5 cannot be paired.<br \/>\n\u2234 We will multiply 180 by 5 to get perfect square.<br \/>\nNew number = 180 \u00d7 5 = 900<br \/>\nNow each prime factor has a pair. Therefore, 180 \u00d7 5 = 900 is a perfect square.<br \/>\nThus the required smallest number is 5.<br \/>\n\u21d2 \u221a900 = 2\u00d73\u00d75 = 30<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii)<\/strong>\u00a01008<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3151\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-5iii.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"97\" height=\"266\" \/><br \/>\n1008 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 7<br \/>\n= (2\u00d72) \u00d7 (2\u00d72) \u00d7 (3\u00d73) \u00d7 7<br \/>\nHere, 7 cannot be paired.<br \/>\n\u2234 We will multiply 1008 by 7 to get perfect square.<br \/>\nNew number = 1008 \u00d7 7 = 7056<br \/>\nNow each prime factor has a pair. Therefore, 1008 \u00d7 7 = 7056 is a perfect square.<br \/>\nThus the required smallest number is 7.<br \/>\n\u21d2 \u221a7056 = 2\u00d72\u00d73\u00d77 = 84<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv)<\/strong> 2028<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3152\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-5iv.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"94\" height=\"194\" \/><br \/>\n2028 = 2 \u00d7 2 \u00d7 3 \u00d7 13 \u00d7 13<br \/>\n= (2\u00d72) \u00d7 (13\u00d713) \u00d7 3<br \/>\nHere, 3 cannot be paired.<br \/>\n\u2234 We will multiply 2028 by 3 to get perfect square.<br \/>\nNew number = 2028 \u00d7 3 = 6084<br \/>\nNow each prime factor has a pair. Therefore, 2028 \u00d7 3 = 6084 is a perfect square.<br \/>\nThus the required smallest number is 3.<br \/>\n\u21d2 \u221a6084 = 2 \u00d7 3 \u00d7 13 = 78<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v)<\/strong>\u00a01458<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3153\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-5v.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"97\" height=\"262\" \/><br \/>\n1458 = 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3<br \/>\n= (3\u00d73) \u00d7 (3\u00d73) \u00d7 (3\u00d73) \u00d7 2<br \/>\nHere, 2 cannot be paired.<br \/>\n\u2234 We will multiply 1458 by 2 to get perfect square.<br \/>\nNew number = 1458 \u00d7 2 = 2916<br \/>\nNow each prime factor has a pair. Therefore, 1458 \u00d7 2 = 2916 is a perfect square.<br \/>\nThus the required smallest number is 2.<br \/>\n\u21d2 \u221a2916 = 3\u00d73\u00d73\u00d72 = 54<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi)<\/strong>\u00a0768<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3154\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-5vi.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"97\" height=\"334\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-5vi.png 97w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-5vi-87x300.png 87w\" sizes=\"auto, (max-width: 97px) 100vw, 97px\" \/><br \/>\n768 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3<br \/>\n= (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 (2\u00d72) \u00d7 3<br \/>\nHere, 3 cannot be paired.<br \/>\n\u2234 We will multiply 768 by 3 to get perfect square.<br \/>\nNew number = 768 \u00d7 3 = 2304<br \/>\nNow each prime factor has a pair. Therefore, 768 \u00d7 3 = 2304 is a perfect square.<br \/>\nThus the required smallest number is 3.<br \/>\n\u21d2 \u221a2304 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 = 48<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.<br \/>\n<\/strong><strong>(i)<\/strong>\u00a0252<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii)<\/strong>\u00a02925<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iii)<\/strong>\u00a0396<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iv)<\/strong>\u00a02645<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(v)<\/strong>\u00a02800<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(vi)<\/strong>\u00a01620<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong>\u00a0252<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3155\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-6i.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"67\" height=\"199\" \/><br \/>\n252 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 7<br \/>\n= (2\u00d72) \u00d7 (3\u00d73) \u00d7 7<br \/>\nHere, 7 cannot be paired.<br \/>\n\u2234 We will divide 252 by 7 to get perfect square. New number = 252 \u00f7 7 = 36<br \/>\n36 = 2 \u00d7 2 \u00d7 3 \u00d7 3<br \/>\n\u21d2 36 = (2\u00d72) \u00d7 (3\u00d73)<br \/>\n\u21d2 36 = 2<sup>2<\/sup>\u00d73<sup>2<br \/>\n<\/sup>\u21d2 36 = (2\u00d73)<sup>2<br \/>\n<\/sup>\u21d2 \u221a36 = 2\u00d73 = 6<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong>\u00a02925<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3156\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-6ii.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"76\" height=\"203\" \/><br \/>\n2925 = 3 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 13<br \/>\n= (3\u00d73) \u00d7 (5\u00d75) \u00d7 13<br \/>\nHere, 13 cannot be paired.<br \/>\n\u2234 We will divide 2925 by 13 to get perfect square. New number = 2925 \u00f7 13 = 225<br \/>\n225 = 3 \u00d7 3 \u00d7 5 \u00d7 5<br \/>\n\u21d2 225 = (3\u00d73) \u00d7 (5\u00d75)<br \/>\n\u21d2 225 = 3<sup>2 <\/sup>\u00d7 5<sup>2<br \/>\n<\/sup>\u21d2 225 = (3 \u00d7 5)<sup>2<br \/>\n<\/sup>\u21d2 \u221a36 = 3 \u00d7 5 = 15<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii)<\/strong> 396<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3157\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-6iii.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"71\" height=\"197\" \/><br \/>\n396 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 11<br \/>\n= (2\u00d72) \u00d7 (3\u00d73) \u00d7 11<br \/>\nHere, 11 cannot be paired.<br \/>\n\u2234 We will divide 396 by 11 to get perfect square. New number = 396 \u00f7 11 = 36<br \/>\n36 = 2\u00d72\u00d73\u00d73<br \/>\n\u21d2 36 = (2\u00d72)\u00d7(3\u00d73)<br \/>\n\u21d2 36 = 2<sup>2<\/sup>\u00d73<sup>2<br \/>\n<\/sup>\u21d2 36 = (2\u00d73)<sup>2<br \/>\n<\/sup>\u21d2 \u221a36 = 2\u00d73 = 6<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv)<\/strong>\u00a02645<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3158\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-6iv.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"69\" height=\"134\" \/><br \/>\n2645 = 5 \u00d7 23 \u00d7 23<br \/>\n\u21d2 2645 = (23\u00d723) \u00d7 5<br \/>\nHere, 5 cannot be paired.<br \/>\n\u2234 We will divide 2645 by 5 to get perfect square.<br \/>\nNew number = 2645 \u00f7 5 = 529<br \/>\n529 = 23 \u00d7 23<br \/>\n\u21d2 529 = (23)<sup>2<br \/>\n<\/sup>\u21d2 \u221a529 = 23<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v)<\/strong>\u00a02800<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3159\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-6v.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"70\" height=\"268\" \/><\/span><br \/>\n<span style=\"color: #000000;\">2800 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 5 \u00d7 7<br \/>\n= (2\u00d72) \u00d7 (2\u00d72) \u00d7 (5\u00d75) \u00d7 7<br \/>\nHere, 7 cannot be paired.<br \/>\n\u2234 We will divide 2800 by 7 to get perfect square. New number = 2800 \u00f7 7 = 400<br \/>\n400 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 5<br \/>\n\u21d2 400 = (2\u00d72)\u00d7(2\u00d72)\u00d7(5\u00d75)<br \/>\n\u21d2 400 = (2\u00d72\u00d75)<sup>2<br \/>\n<\/sup>\u21d2 \u221a400 = 20<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi)<\/strong>\u00a01620<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3160\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-6vi.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"75\" height=\"263\" \/><br \/>\n1620 = 2\u00d72\u00d73\u00d73\u00d73\u00d73\u00d75<br \/>\n= (2\u00d72) \u00d7 (3\u00d73) \u00d7 (3\u00d73) \u00d7 5<br \/>\nHere, 5 cannot be paired.<br \/>\n\u2234 We will divide 1620 by 5 to get perfect square. New number = 1620 \u00f7 5 = 324<br \/>\n324 = 2\u00d72\u00d73\u00d73\u00d73\u00d73<br \/>\n\u21d2 324 = (2\u00d72) \u00d7 (3\u00d73) \u00d7 (3\u00d73)<br \/>\n\u21d2 324 = (2 \u00d7 3 \u00d7 3)<sup>2<br \/>\n<\/sup>\u21d2 \u221a324 = 18<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister\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><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let the number of students in the school be, x.<br \/>\nRupees donated by student = x<br \/>\nTotal many contributed by all the students= x \u00d7 x = x<sup>2<\/sup><br \/>\nGiven, x<sup>2<\/sup>\u00a0= Rs.2401<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3161\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-7.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"84\" height=\"203\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x<sup>2<\/sup>\u00a0= 7\u00d77\u00d77\u00d77<br \/>\n\u21d2 x<sup>2<\/sup>\u00a0= (7\u00d77)\u00d7(7\u00d77)<br \/>\n\u21d2 x<sup>2\u00a0<\/sup>= 49\u00d749<br \/>\n\u21d2 x = \u221a(49\u00d749)<br \/>\n\u21d2 x = 49<br \/>\n\u2234 The number of students = 49<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let the number of rows be, x.<br \/>\nnumber of plants in each row = x.<br \/>\nTotal many contributed by all the students = x \u00d7 x =x<sup>2<br \/>\n<\/sup>Given, x<sup>2<\/sup> = Rs. 2025<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3162\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-8.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"103\" height=\"218\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x<sup>2<\/sup>\u00a0= 3\u00d73\u00d73\u00d73\u00d75\u00d75<br \/>\n\u21d2 x<sup>2<\/sup>\u00a0= (3\u00d73)\u00d7(3\u00d73)\u00d7(5\u00d75)<br \/>\n\u21d2 x<sup>2<\/sup> = (3\u00d73\u00d75)<sup>2<\/sup><br \/>\n\u21d2 x = \u221a45 \u00d7 45<br \/>\n\u21d2 x = 45<br \/>\n\u2234 The number of rows = 45 and the number of plants in each rows = 45.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3163\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-9.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"92\" height=\"70\" \/><br \/>\n<\/strong>L.C.M of 4, 9 and 10 is (2 \u00d7 2 \u00d7 9 \u00d7 5) = 180.<br \/>\n180 = 2\u00d72\u00d79\u00d75<br \/>\n= (2\u00d72)\u00d7(3\u00d73)\u00d75<br \/>\nHere, 5 cannot be paired.<br \/>\n\u2234 we will multiply 180 by 5 to get perfect square.<br \/>\nHence, the smallest square number divisible by 4, 9 and 10 = 180\u00d75 = 900<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3164\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-6.3-Q-10.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"98\" height=\"134\" \/><br \/>\n<\/strong>L.C.M of 8, 15 and 20 is (2\u00d72\u00d75\u00d72\u00d73) = 120.<br \/>\n120 = 2\u00d72\u00d73\u00d75\u00d72<br \/>\n= (2\u00d72)\u00d73\u00d75\u00d72<br \/>\nHere, 3, 5 and 2 cannot be paired.<br \/>\n\u2234 We will multiply 120 by (3\u00d75\u00d72) 30 to get perfect square.<br \/>\nHence, the smallest square number divisible by 8, 15 and 20 =120\u00d730 = 3600<\/span><\/p>\n<p>&nbsp;<\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr style=\"height: 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palatino, serif; font-size: 12pt;\">NCERT Solutions Class 8 Sanskrit<\/span><\/strong><\/span><\/a><\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 100%; text-align: left; height: 28px;\"><span style=\"font-size: 14pt;\"><strong><span style=\"font-family: georgia, palatino, serif; font-size: 12pt;\"><a class=\"row-title\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-science\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Social Science\u201d (Edit)\">NCERT Solutions Class 8 Science<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 100%; text-align: left; height: 28px;\"><span style=\"font-size: 14pt;\"><strong><span style=\"font-family: georgia, palatino, serif; font-size: 12pt;\"><a class=\"row-title\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-social-science\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Social Science\u201d (Edit)\">NCERT Solutions Class 8 Social Science<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 8 Mathematics\u00a0 Chapter &#8211; 6 (Squares and Square)\u00a0 The NCERT Solutions in English Language for Class 8 Mathematics Chapter &#8211; 6 Squares and Square Exercise 6.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 6: Squares and Square Roots NCERT Solution Class 8 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[442],"tags":[454,455,445,5],"class_list":["post-3130","post","type-post","status-publish","format-standard","hentry","category-class-8-maths","tag-ncert-class-8-maths-chapter-6-squares-and-square-roots-in-english","tag-ncert-solutions-class-8-maths-chapter-6-in-english","tag-ncert-solutions-class-8-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 8 Mathematics\u00a0Chapter - 6 (Squares and Square)\u00a0The NCERT Solutions in English Language for Class 8 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