{"id":27382,"date":"2021-06-30T15:12:24","date_gmt":"2021-06-30T09:42:24","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27382"},"modified":"2021-06-30T15:16:34","modified_gmt":"2021-06-30T09:46:34","slug":"ncert-solutions-for-class-8-maths-chapter-6-ex-6-4","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-6-ex-6-4\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 6 Square and Square Roots Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.4<\/h2>\n

Question 1.
\nFind the square root of each of the following numbers by division method.
\n(i) 2304
\n(ii) 4489
\n(iii) 3481
\n(iv) 529
\n(v) 3249
\n(vi) 1369
\n(vii) 5776
\n(viii) 7921
\n(ix) 576
\n(x) 1024
\n(xi) 3136
\n(xii) 900
\nSolution:
\n(i) 2304
\n\u2234 \u221a2304 = 48
\n\"NCERT<\/p>\n

(ii) 4489
\n\u2234 \u221a4489 = 67
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(iii) 3481
\n\u2234 \u221a3481 = 59
\n\"NCERT<\/p>\n

(iv) 529
\n\u2234 \u221a529 = 23
\n\"NCERT<\/p>\n

(v) 3249
\n\u2234 \u221a3249 = 57
\n\"NCERT<\/p>\n

(vi) 1369
\n\u2234 \u221a1369 = 37
\n\"NCERT<\/p>\n

(vii) 5776
\n\u2234 \u221a5776 = 76
\n\"NCERT<\/p>\n

(viii) 7921
\n\u2234 \u221a7921 = 89
\n\"NCERT<\/p>\n

(ix) 576
\n\u2234 \u221a576 = 24
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(x) 1024
\n\u2234 \u221a1024 = 32
\n\"NCERT<\/p>\n

(xi) 3136
\n\u2234 \u221a3136 = 56
\n\"NCERT<\/p>\n

(xii) 900
\n\u2234 \u221a900 = 30
\n\"NCERT<\/p>\n

Question 2.
\nFind the number of digits in the square root of each of the following numbers. (without any calculation).
\n(i) 64
\n(ii) 144
\n(iii) 4489
\n(iv) 27225
\n(v) 390625
\nSolution:
\nIf n stands fnr nnmher of dibits in the given number then
\n(i) For 64
\nHere, n = 2 (even number)
\nNumber of digits in the square root = \\(\\frac{n}{2}=\\frac{2}{2}=1\\)<\/p>\n

(ii) For 144
\nHere, n = 3 (odd number)
\nNumber of digits in the, square root = \\(\\frac{n+1}{2}=\\frac{3+4}{2}=\\frac{4}{2}=2\\)<\/p>\n

(iii) For 4489
\nHere, n = 4 (even number)
\nNumber of digits in the square root = \\(\\frac{\\mathrm{n}}{2}=\\frac{4}{2}=2\\)<\/p>\n

\"NCERT<\/p>\n

(iv) For 27225
\nHere, n = 5 (odd number)
\nNumber of digits in the square root = \\(\\frac{\\mathrm{n}+1}{2}=\\frac{5+1}{2}=\\frac{6}{2}=3\\)<\/p>\n

(v) For 390625
\nHere, n = 6 (even number)
\nNumber of digits in the square root = \\(\\frac{n}{2}=\\frac{6}{2}=3\\)<\/p>\n

Question 3.
\nFind the square root of the following decimal numbers.
\n(i) 2.56
\n(ii) 7.29
\n(iii) 51.84
\n(iv) 42.25
\n(v) 31.36
\nSolution:
\n(i) 2.56
\n\u2234 \u221a2.56 = 1.6
\n\"NCERT<\/p>\n

(ii) 7.29
\n\u2234 \u221a7.29 = 2.7
\n\"NCERT<\/p>\n

(iii) 51.84
\n\u2234 \u221a51.84 = 7.2
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(iv) 42.25
\n\u2234 \u221a42.25 = 6.5
\n\"NCERT<\/p>\n

(v) 31.36
\n\u2234 \u221a31.36 = 5.6
\n\"NCERT<\/p>\n

Question 4.
\nFind the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.
\n(i) 402
\n(ii) 1989
\n(iii) 3250
\n(iv) 825
\n(v) 4000
\nSolution:
\n(i) 402
\nThe required least number to be subtracted from 402 is 2.
\n\u2234 402 – 2 = 400
\n\u2234 \u221a400 = 20
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(ii) 1989
\nThe remainder is 53
\n\u2234 The required least numbers to be subtracted from the given number is 53.
\n1989 – 53 = 1936
\n\u2234 \u221a1936 = 44
\n\"NCERT<\/p>\n

(iii) 3250
\nThe remainder is 1
\n\u2234 The required least number to be subtracted from the given number is 1.
\n3250 – 1 = 3249
\n\u2234 \u221a3249 = 57
\n\"NCERT<\/p>\n

(iv) 825
\nThe remainder is 41
\nThe required least number to be subtracted from the given number is 41.
\n825 – 41 = 784
\n\u2234 \u221a784 = 28
\n\"NCERT<\/p>\n

(v) 4000
\nThe remainder is 31
\nThe required least number to be subtracted from the given number is 31.
\n4000 – 31 = 3969
\n\u2234 \u221a3969 = 63
\n\"NCERT<\/p>\n

Question 5.
\nFind the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square, so obtained.
\n(i) 525
\n(ii) 1750
\n(iii) 252
\n(iv) 1825
\n(v) 6412
\nSolution:
\n(i) 525
\nThe remainder is 41.
\ni.e. 525 > 222<\/sup> and the next perfect square number is 232<\/sup> = 529
\nThe required number to be added = 232<\/sup> – 525
\n= 529 – 525
\n= 4
\nHence, 4 should be added to 525 to get a perfect square
\nnumber.
\n525 + 4 = 529
\n\u2234 \u221a529 = 23
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(ii) 1750
\nThe remainder is 69.
\n1750 > 412<\/sup> and the next perfect square number is 422<\/sup> = 1764
\nThe required number to be added = 422<\/sup> – 1750
\n= 1764 – 1750
\n= 14
\nThe next perfect square number = 1750 + 14 = 1764
\n\u2234 \u221a1764 = 42
\n\"NCERT<\/p>\n

(iii) 252
\nThe remainder is 27.
\nThis shows that 152<\/sup> < 252
\nThe next perfect square number is 162<\/sup> = 256.
\nThe required number to be added = 162<\/sup> – 252
\n= 256 – 252
\n= 4
\nThe perfect square number = 252 + 4 = 256
\n\u2234 \u221a256 = 16
\n\"NCERT<\/p>\n

(iv) 1825
\nThe remainder is 61.
\nThis shows that 422<\/sup> < 1825
\nThe next perfect square is 432<\/sup> = 1849
\nHence, the required number to be added = 432<\/sup> – 1825
\n= 1849 – 1825
\n= 24
\nThe perfect square number so obtained 1825 + 24 = 1849
\nHence, \u221a1849 = 43
\n\"NCERT<\/p>\n

(v) 6412
\nThe remainder is 12.
\nThis shows that 802<\/sup> < 6412
\nThe next perfect square number is 812<\/sup> = 6561
\nThe required number to be added = 6561 – 6412 = 149
\nThe perfect square number = 6412 + 149 = 6561
\nHence, \u221a6561 = 81
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nFind the length of the side of a square where area is 441m2<\/sup>.
\nSolution:
\nLet x be the side of a square.
\n\u2234 Area of square = x2<\/sup>
\nGiven, Area of square = 441
\nx2<\/sup> = 441
\n\u21d2 x = \u221a441 = 21
\nSo, length of the side of the square = 21 m.
\n\"NCERT<\/p>\n

Question 7.
\nIn a right triangle ABC, \u2220B = 90\u00b0.
\n(a) If AB = 6 cm, BC = 8 cm, find AC.
\n(b) If AC = 13 cm, BC = 5 cm, find AB.
\nSolution:
\n(a) In the right triangle ABC, \u2220B = 90\u00b0
\n\"NCERT<\/p>\n

\"NCERT
\nBy Pythagoras Theorem,
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n\u21d2 AC2<\/sup> = 62<\/sup> + 82<\/sup>
\n\u21d2 AC2<\/sup> = 36 + 64
\n\u21d2 AC2<\/sup> = 100
\n\u21d2 AC = \u221a100 = 10
\n\u2234 Length of AC = 10 cm<\/p>\n

\"NCERT<\/p>\n

(b) In the right triangle ABC, \u2220B = 90\u00b0
\n\"NCERT
\nBy Pythagoras theorem,
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n\u21d2 AB2<\/sup> = AC2<\/sup> – BC2<\/sup>
\n\u21d2 AB2<\/sup> = 132<\/sup> – 52<\/sup>
\n\u21d2 AB2<\/sup> = 169 – 25
\n\u21d2 AB2<\/sup> = 144
\n\u21d2 AB = \u221a144 = 12
\n\u2234 Length of AB = 12 cm<\/p>\n

Question 8.
\nA gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
\nSolution:
\nLet the number of rows be \u2018x\u2019
\nThe number of columns is x
\nThe number of plants = x \u00d7 x = x2<\/sup>
\nwhich is a perfect square
\nLet us find out the square root of 1000 by division method
\nThe remainder is 39
\nThis shows that 312<\/sup> < 1000
\nThe next perfect square number is 322<\/sup> = 1024
\nMinimum number of plants he needs besides 100 plants = 322<\/sup> – 1000
\n= 1024 – 1000
\n= 24
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nThere are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.
\nSolution:
\nLet the number of rows be x and the number of columns is x
\n\u2234 Number of children = x \u00d7 x = x2<\/sup>
\nwhich is a perfect square
\n\u2234 500 = x2<\/sup>
\nThe remainder is 16.
\nNumber of children left out in this arrangement = 16.
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.4 Question 1. Find the square root of each of the following numbers by division method. …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-6-ex-6-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.4 Question 1. 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