{"id":79699,"date":"2021-10-18T17:00:02","date_gmt":"2021-10-18T11:30:02","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=79699"},"modified":"2021-10-21T12:39:04","modified_gmt":"2021-10-21T07:09:04","slug":"squares-and-square-roots-ncert-extra-questions-for-class-8-maths-chapter-6","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/squares-and-square-roots-ncert-extra-questions-for-class-8-maths-chapter-6\/","title":{"rendered":"Squares and Square Roots Class 8 Extra Questions Maths Chapter 6"},"content":{"rendered":"
Extra Questions for Class 8 Maths Chapter 6 Squares and Square Roots<\/strong><\/p>\n Squares And Square Roots Class 8 Worksheet Question 1.<\/strong> Square And Square Roots Class 8 Extra Questions Question 2.<\/strong> Square Root Questions For Class 8 Question 3.<\/strong> Squares And Square Roots Class 8 Extra Questions With Answers Question 4.<\/strong> Square And Square Roots Class 8 Worksheet Question 5.<\/strong> Square And Square Roots Class 8 Test With Answers Question 6.<\/strong> Class 8 Square And Square Roots Extra Questions Question 7.<\/strong> Class 8 Maths Chapter 6 Extra Questions Question 8.<\/strong> Square And Square Roots For Class 8 Worksheets With Answers Question 9.<\/strong> Class 8 Square And Square Roots Worksheet Question 10.<\/strong> Class 8 Maths Square And Square Roots Extra Questions Question 11.<\/strong> Square And Square Roots For Class 8 Worksheets With Answers Pdf Question 12.<\/strong> Class 8 Maths Chapter 6 Test Paper Question 13.<\/strong> Question 14. Question 15. Question 16. Question 17. Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. Question 24. Squares and Square Roots Class 8 Extra Questions Maths Chapter 6 Extra Questions for Class 8 Maths Chapter 6 Squares and Square Roots Squares and Square Roots Class 8 Extra Questions Very Short Answer Type Squares And Square Roots Class 8 Worksheet Question 1. Find the perfect square numbers between 40 and 50. Solution: Perfect …<\/p>\n","protected":false},"author":27,"featured_media":159551,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nSquares and Square Roots Class 8 Extra Questions Very Short Answer Type<\/h3>\n
\nFind the perfect square numbers between 40 and 50.
\nSolution:
\nPerfect square numbers between 40 and 50 = 49.<\/p>\n
\nWhich of the following 242<\/sup>, 492<\/sup>, 772<\/sup>, 1312<\/sup> or 1892<\/sup> end with digit 1?
\nSolution:
\nOnly 492<\/sup>, 1312<\/sup> and 1892<\/sup> end with digit 1.<\/p>\n
\nFind the value of each of the following without calculating squares.
\n(i) 272<\/sup> – 262<\/sup>
\n(ii) 1182<\/sup> – 1172<\/sup>
\nSolution:
\n(i) 272<\/sup> – 262<\/sup> = 27 + 26 = 53
\n(ii) 1182<\/sup> – 1172<\/sup> = 118 + 117 = 235<\/p>\n
\nWrite each of the following numbers as difference of the square of two consecutive natural numbers.
\n(i) 49
\n(ii) 75
\n(iii) 125
\nSolution:
\n(i) 49 = 2 \u00d7 24 + 1
\n49 = 252<\/sup> – 242<\/sup>
\n(ii) 75 = 2 \u00d7 37 + 1
\n75 = 382<\/sup> – 372<\/sup>
\n(iii) 125 = 2 \u00d7 62 + 1
\n125 = 632<\/sup> – 622<\/sup><\/p>\n
\nWrite down the following as sum of odd numbers.
\n(i) 72<\/sup>
\n(ii) 92<\/sup>
\nSolution:
\n(i) 72<\/sup> = Sum of first 7 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13
\n(ii) 92<\/sup> = Sum of first 9 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17<\/p>\n
\nExpress the following as the sum of two consecutive integers.
\n(i) 152<\/sup>
\n(ii) 192<\/sup>
\nSolution:
\n<\/p>\n
\nFind the product of the following:
\n(i) 23 \u00d7 25
\n(ii) 41 \u00d7 43
\nSolution:
\n(i) 23 \u00d7 25 = (24 – 1) (24 + 1) = 242<\/sup> – 1 = 576 – 1 = 575
\n(ii) 41 \u00d7 43 = (42 – 1) (42 + 1) = 422<\/sup> – 1 = 1764 – 1 = 1763<\/p>\n
\nFind the squares of:
\n(i) \\(\\frac { -3 }{ 7 }\\)
\n(ii) \\(\\frac { -9 }{ 17 }\\)
\nSolution:
\n<\/p>\n
\nCheck whether (6, 8, 10) is a Pythagorean triplet.
\nSolution:
\n2m, m2<\/sup> – 1 and m2<\/sup> + 1 represent the Pythagorean triplet.
\nLet 2m = 6 \u21d2 m = 3
\nm2<\/sup> – 1 = (3)2<\/sup> – 1 = 9 – 1 = 8
\nand m2<\/sup> + 1 = (3)2<\/sup> + 1 = 9 + 1 = 10
\nHence (6, 8, 10) is a Pythagorean triplet.
\nAlternative Method:
\n(6)2<\/sup> + (8)2<\/sup> = 36 + 64 = 100 = (10)2<\/sup>
\n\u21d2 (6, 8, 10) is a Pythagorean triplet.<\/p>\n
\nUsing property, find the value of the following:
\n(i) 192<\/sup> – 182<\/sup>
\n(ii) 232<\/sup> – 222<\/sup>
\nSolution:
\n(i) 192<\/sup> – 182<\/sup> = 19 + 18 = 37
\n(ii) 232<\/sup> – 222<\/sup> = 23 + 22 = 45<\/p>\nSquares and Square Roots Class 8 Extra Questions Short Answer Type<\/h3>\n
\nUsing the prime factorisation method, find which of the following numbers are not perfect squares.
\n(i) 768
\n(ii) 1296
\nSolution:
\n
\n768 = 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 3
\nHere, 3 is not in pair.
\n768 is not a perfect square.
\n
\n1296 = 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 3 \u00d7 3<\/span> \u00d7 3 \u00d7 3<\/span>
\nHere, there is no number left to make a pair.
\n1296 is a perfect square.<\/p>\n
\nWhich of the following triplets are Pythagorean?
\n(i) (14, 48, 50)
\n(ii) (18, 79, 82)
\nSolution:
\nWe know that 2m, m2<\/sup> – 1 and m2<\/sup> + 1 make Pythagorean triplets.
\n(i) For (14, 48, 50),
\nPut 2m =14 \u21d2 m = 7
\nm2<\/sup> – 1 = (7)2<\/sup> – 1 = 49 – 1 = 48
\nm2<\/sup> + 1 = (7)2<\/sup> + 1 = 49 + 1 = 50
\nHence (14, 48, 50) is a Pythagorean triplet.
\n(ii) For (18, 79, 82)
\nPut 2m = 18 \u21d2 m = 9
\nm2<\/sup> – 1 = (9)2<\/sup> – 1 = 81 – 1 = 80
\nm2 <\/sup>+ 1 = (9)2<\/sup> + 1 = 81 + 1 = 82
\nHence (18, 79, 82) is not a Pythagorean triplet.<\/p>\n
\nFind the square root of the following using successive subtraction of odd numbers starting from 1.
\n(i) 169
\n(ii) 81
\n(iii) 225
\nSolution:
\n(i) 169 – 1 = 168, 168 – 3 = 165, 165 – 5 = 160, 160 – 7 = 153, 153 – 9 = 144, 144 – 11 = 133, 133 – 13 = 120, 120 – 15 = 105, 105 – 17 = 88, 88 – 19 = 69,
\n69 – 21 = 48, 48 – 23 = 25, 25 – 25 = 0
\nWe have subtracted odd numbers 13 times to get 0.
\n\u221a169 = 13
\n(ii) 81 – 1 = 80, 80 – 3 = 77, 77 – 5 = 72, 72 – 7 = 65, 65 – 9 = 56, 56 – 11 = 45, 45 – 13 = 32, 32 – 15 = 17, 17 – 17 = 0
\nWe have subtracted 9 times to get 0.
\n\u221a81 = 9
\n(iii) 225 – 1 = 224, 224 – 3 = 221, 221 – 5 = 216, 216 – 7 = 209, 209 – 9 = 200, 200 – 11 = 189, 189 – 13 = 176, 176 – 15 = 161, 161 – 17 = 144, 144 – 19 = 125,
\n125 – 21 = 104, 104 – 23 = 81, 81 – 25 = 56, 56 – 27 = 29, 29 – 29 = 0
\nWe have subtracted 15 times to get 0.
\n\u221a225 = 15<\/p>\n
\nFind the square rootofthe following using prime factorisation
\n(i) 441
\n(ii) 2025
\n(iii) 7056
\n(iv) 4096
\nSolution:
\n(i) 441 = 3 \u00d7 3<\/span> \u00d7 7 \u00d7 7<\/span>
\n\u221a441 = 3 \u00d7 7 = 21
\n
\n(ii) 2025 = 3 \u00d7 3<\/span> \u00d7 3 \u00d7 3<\/span> \u00d7 5 \u00d7 5<\/span>
\n\u221a2025 = 3 \u00d7 3 \u00d7 5 = 45
\n
\n(iii) 7056 = 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 3 \u00d7 3<\/span> \u00d7 7 \u00d7 7<\/span>
\n\u221a7056 = 2 \u00d7 2 \u00d7 3 \u00d7 7 = 84
\n
\n(iv) 4096 = 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span> \u00d7 2 \u00d7 2<\/span>
\n\u221a4096 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 = 64
\n<\/p>\n
\nFind the least square number which is divisible by each of the number 4, 8 and 12.
\nSolution:
\nLCM of 4, 8, 12 is the least number divisible by each of them.
\nLCM of 4, 8 and 12 = 24
\n24 = 2 \u00d7 2<\/span> \u00d7 2 \u00d7 3
\nTo make it perfect square multiply 24 by the product of unpaired numbers, i.e., 2 \u00d7 3 = 6
\nRequired number = 24 \u00d7 6 = 144
\n<\/p>\n
\nFind the square roots of the following decimal numbers
\n(i) 1056.25
\n(ii) 10020.01
\nSolution:
\n
\n<\/p>\n
\nWhat is the least number that must be subtracted from 3793 so as to get a perfect square? Also, find the square root of the number so obtained.
\nSolution:
\nFirst, we find the square root of 3793 by division method.
\n
\nHere, we get a remainder 72
\n612<\/sup> < 3793
\nRequired perfect square number = 3793 – 72 = 3721 and \u221a3721 = 61<\/p>\n
\nFill in the blanks:
\n(\u0430) The perfect square number between 60 and 70 is …………
\n(b) The square root of 361 ends with digit …………..
\n(c) The sum of first n odd numbers is …………
\n(d) The number of digits in the square root of 4096 is ………..
\n(e) If (-3)2<\/sup> = 9, then the square root of 9 is ……….
\n(f) Number of digits in the square root of 1002001 is …………
\n(g) Square root of \\(\\frac { 36 }{ 625 }\\) is ………..
\n(h) The value of \u221a(63 \u00d7 28) = …………
\nSolution:
\n(a) 64
\n(b) 9
\n(c) n2<\/sup>
\n(d) 2
\n(e) \u00b13
\n(f) 4
\n(g) \\(\\frac { 6 }{ 25 }\\)
\n(h) 42<\/p>\n
\nSimplify: \u221a900 + \u221a0.09 + \u221a0.000009
\nSolution:
\nWe know that \u221a(ab) = \u221aa \u00d7 \u221ab
\n\u221a900 = \u221a(9 \u00d7 100) = \u221a9 \u00d7 \u221a100 = 3 \u00d7 10 = 30
\n\u221a0.09 = \u221a(0.3 \u00d7 0.3) = 0.3
\n\u221a0.000009 = \u221a(0.003 \u00d7 0.003) = 0.003
\n\u221a900 + \u221a0.09 + \u221a0.000009 = 30 + 0.3 + 0.003 = 30.303<\/p>\nSquares and Square Roots Class 8 Extra Questions Higher Order Thinking Skills (HOTS)<\/h3>\n
\nFind the value of x if
\n\\(\\sqrt { 1369 } +\\sqrt { 0.0615+x } =37.25\\)
\nSolution:
\n<\/p>\n
\nSimplify:
\n
\nSolution:
\n<\/p>\n
\nA ladder 10 m long rests against a vertical wall. If the foot of the ladder is 6 m away from the wall and the ladder just reaches the top of the wall, how high is the wall? (NCERT Exemplar)
\n
\nSolution:
\nLet AC be the ladder.
\nTherefore, AC = 10 m
\nLet BC be the distance between the foot of the ladder and the wall.
\nTherefore, BC = 6 m
\n\u2206ABC forms a right-angled triangle, right angled at B.
\nBy Pythagoras theorem,
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n102 <\/sup>= AB2<\/sup> + 62<\/sup>
\nor AB2<\/sup> = 102<\/sup> – 62<\/sup> = 100 – 36 = 64
\nor AB = \u221a64 = 8m
\nHence, the wall is 8 m high.<\/p>\n
\nFind the length of a diagonal of a rectangle with dimensions 20 m by 15 m. (NCERT Exemplar)
\nSolution:
\nUsing Pythagoras theorem, we have Length of diagonal of the rectangle = \\(\\sqrt { { l }^{ 2 }+{ b }^{ 2 } }\\) units
\n
\nHence, the length of the diagonal is 25 m.<\/p>\n
\nThe area of a rectangular field whose length is twice its breadth is 2450 m2. Find the perimeter of the field.
\nSolution:
\nLet the breadth of the field be x metres. The length of the field 2x metres.
\nTherefore, area of the rectangular field = length \u00d7 breadth = (2x)(x) = (2x2<\/sup>) m2<\/sup>
\nGiven that area is 2450 m2<\/sup>.
\nTherefore, 2x2<\/sup> = 2450
\n\u21d2 x2<\/sup> = 1225
\n\u21d2 x = \u221a1225 or x = 35 m
\nHence, breadth = 35 m
\nand length = 35 \u00d7 2 = 70 m
\nPerimeter of the field = 2 (l + b ) = 2(70 + 35) m = 2 \u00d7 105 m = 210 m.<\/p>\n
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\n<\/p>\nExtra Questions for Class 8 Maths<\/a><\/h4>\n
NCERT Solutions for Class 8 Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"