{"id":6691,"date":"2023-02-23T09:20:11","date_gmt":"2023-02-23T03:50:11","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=6691"},"modified":"2023-03-30T12:30:22","modified_gmt":"2023-03-30T07:00:22","slug":"ap-board-8th-class-maths-solutions-chapter-6-ex-6-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-board-8th-class-maths-solutions-chapter-6-ex-6-2\/","title":{"rendered":"AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.2"},"content":{"rendered":"

AP State Syllabus AP Board 8th Class Maths Solutions<\/a> Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.2<\/h2>\n

\"AP<\/p>\n

Question 1.
\nFind the square roots of the following numbers by Prime factorization method.
\n(1) 441
\n(ii) 784
\n(iii) 4096
\n(iv) 7056
\nSolution:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the smallest number by which 3645 must be multiplied to get a perfect square.
\nSolution:
\nThe prime factorization of 3645
\n= (3 \u00d7 3) \u00d7 (3 \u00d7 3) (3 \u00d7 3) \u00d7 5
\n\u2234 Deficiency of one \u20185\u2019 is appeared in the above product.
\n\u2234 3645 is multiplied with 5 then we will get a perfect square.
\n\"AP<\/p>\n

Question 3.
\nFind the smallest number by which 2400 is to be multiplied to get a perfect square and also find the square root of the resulting number.
\nSolution:
\nThe prime factorization of 2400
\n=(2 \u00d7 2) \u00d7 (2 \u00d7 2) \u00d7 2 \u00d7 (5 \u00d7 5) \u00d7 3
\n\u2234 2,3 are needed to form a pair
\n\u2234 2 \u00d7 3 = 6
\n\u2234 6 should be multiplied with 2400 then we will get a perfect square number.
\n\u2234 2400 \u00d7 6 = 14400
\n\u2234 \\(\\sqrt{14400}\\) = 120
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 4.
\nFind the smallest number by which 7776 is to be divided to get a perfect square.
\nSolution:
\n\"AP
\nThe prime factorization of 7776
\n=(2 \u00d7 2) \u00d7 (2 \u00d7 2) \u00d7 2 \u00d7\u00a0(3 \u00d7 3) \u00d7 (3 \u00d7 3) \u00d7 3
\n\u2234 2, 3 are needed to form a pair
\n\u2234 2 \u00d7 3 = 6
\n\u2234 7776 should be divided by 6 then we will get a perfect square number.<\/p>\n

\"AP<\/p>\n

Question 5.
\n1521 trees are planted in a garden in such a way that there are as many trees in each row as there are rows in the garden. Find the number of rows and number of trees in each row.
\nSolution:
\nLet the no. of trees planted in a garden for each row = x say.
\nNo. of rows in the garden = x
\n\u2234 Total no. of trees in the garden = x \u00d7 x = x2<\/sup>
\nAccording to the sum x2<\/sup> = 1521
\nx = \\(\\sqrt{1521}=\\sqrt{39 \\times 39}\\) = 39
\n\u2234 No. of trees for each row = 39
\nNo. of rows in the garden = 39<\/p>\n

Question 6.
\nA school collected \u20b9 2601 as fees from its students. If fee paid by each student and number students in the school were equal, how many students were there in the school?
\nSolution:
\nLet the no. of students in a school = x say
\nThe (amount) fee paid by each student = \u20b9 x
\nAmount collected by all the students
\n= x \u00d7 x = x2<\/sup>
\nAccording to the sum
\n\u2234 x2<\/sup> = 2601
\nx = \\(\\sqrt{2601}=\\sqrt{51 \\times 51}\\) = 51
\n\u2234 x = 51
\n\u2234 No. of students in the school = 51<\/p>\n

Question 7.
\nThe product of two numbers is 1296. If one number is 16 times the other, find the two numbers?
\nSolution:
\nGiven that the product of two numbers = 1296.
\nLet the second number = x say
\nThen first number = 16 \u00d7 x = 16x
\n\u2234 The product of two numbers
\n= x \u00d7 16x= 16x2<\/sup>
\nAccording to the sum
\n16x2<\/sup> = 1296
\n\u21d2 x2<\/sup> = \\(\\frac { 1296 }{ 16 }\\) = 81
\n\u21d2 x2<\/sup> = 81
\n\u21d2 x = \\(\\sqrt{8} \\overline{1}=\\sqrt{9 \\times 9}\\) = 9
\n\u21d2 x = 9
\n\u2234 The first number = 16x
\n= 16 \u00d7 9
\n=144
\nThe second number = x = 9<\/p>\n

\"AP<\/p>\n

Question 8.
\n7921 soldiers sat in an auditorium in such a way that there are as many soldiers in a row as there are rows in the auditorium. How many rows are there in the auditorium\u2019?
\nSolution:
\nLet the number of soldiers sat in an auditorium for each row = x say
\n\u2234 No. of rows in an auditorium = x
\n\u2234 Total no. of soldiers = x \u00d7 x = x2<\/sup>
\nAccording to the sum,
\nx2<\/sup> = 7921
\nx = \\(\\sqrt{7921}=\\sqrt{89 \\times 89}\\) = 89
\n\u2234 No. of rowS in an auditorium = 89<\/p>\n

Question 9.
\nThe area of a square field is 5184 m2<\/sup>. Find the area of a rectangular field, whose perimeter is equal to the perimeter of the square field and whose length is twice of its breadth.
\nSolution:
\nArea of a square field = 5184 m2<\/sup>
\nA = s2<\/sup> = 5184
\n:. s = \\(\\sqrt{5184}=\\sqrt{72 \\times 72}\\) = 72
\n\u2234 s = 72
\n\u2234 Perimeter of the square field = 4 \u00d7 s
\n= 4 \u00d7 72
\n= 288 m
\nAccording to the sum,
\nPerimeter of a rectangular field
\n= Perimeter of a square field = 288 m
\nLet the breadth of a rectangular field
\n= x m say
\n\u2234 Length = 2 \u00d7 x = 2 \u00d7 m
\n\u2234 Perimeter of the rectangular field
\n= 2 (1 + b)
\n= 2 (2x + x)
\n= 2 \u00d7 3x
\n= 6x
\n\u2234 6x = 288 .
\nx = \\(\\frac { 288 }{ 6 }\\)
\nx = 48
\n\u2234 Breadth of the rectangular field
\n= x = 48 m
\nLength of the rectangular field = 2x
\n= 2 \u00d7 48
\n=96m
\n\u2234 Area of the rectangular field
\n= l \u00d7 b
\n= 96 \u00d7 48
\n= 4608 m2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbook Questions and Answers. AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.2 Question 1. Find the square roots of the following numbers by Prime factorization method. (1) 441 (ii) … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":36677,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2023\/02\/Chapter-6-Square-Roots-and-Cube-Roots-Ex-6.2.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/6691"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=6691"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/6691\/revisions"}],"predecessor-version":[{"id":34320,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/6691\/revisions\/34320"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/36677"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=6691"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=6691"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=6691"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}