{"id":11724,"date":"2024-02-04T13:26:46","date_gmt":"2024-02-04T07:56:46","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=11724"},"modified":"2024-02-05T09:10:36","modified_gmt":"2024-02-05T03:40:36","slug":"ap-board-6th-class-maths-solutions-chapter-11-ex-11-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-board-6th-class-maths-solutions-chapter-11-ex-11-2\/","title":{"rendered":"AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2"},"content":{"rendered":"

AP State Syllabus AP Board 6th Class Maths Solutions<\/a> Chapter 11 Perimeter and Area Ex 11.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Ex 11.2<\/h2>\n

\"AP<\/p>\n

Question 1.
\nFind the circumferences of the circles with the radius given below.
\nA) 7 cm B) 3.5 cm C) 14 cm
\nSolution:
\nA) Given radius (r) = 7 cm
\nCircumference of a circle C = 2\u03c0r (\u2235 \u03c0 = \\(\\frac{22}{7}\\))
\n= 2 x \\(\\frac{22}{7}\\) x 7 = 44 cm
\n\u2234 Circumference of a circle = 44 cm ‘<\/p>\n

B) Given radius (r) = 3.5 cm
\nCircumference of a circle C = 2\u03c0r (\u2235 \u03c0 = \\(\\frac{22}{7}\\))
\n= 2 x \\(\\frac{22}{7}\\) x 3.5
\n\u2234 Circumference of a circle = 22 cm<\/p>\n

C) Given radius (r) = 14 cm
\nCircumference of a circle C = 2\u03c0r (\u2235 \u03c0 = \\(\\frac{22}{7}\\))
\n= 2 x \\(\\frac{22}{7}\\) x 14
\n\u2234 Circumference of a circle = 88 cm<\/p>\n

\"AP<\/p>\n

Question 2.
\nGiven below are the circumferences of different circles. Find the radius of each circle.
\nA) 4.4 m B) 176 cm C) 1.54 cm
\nSolution:
\nA) Given circumference of a circle = 2\u03c0r = 4.4 m
\n2 x \\(\\frac{22}{7}\\) x r = 4.4
\nDivide with 2 x \\(\\frac{22}{7}\\) on both sides.
\n\"AP
\n\u2234 Radius of the circle (r) = 0.7 m = 70 cm.<\/p>\n

B) Given circumference of a circle = 2\u03c0r =176 cm
\n2 x \\(\\frac{22}{7}\\) x r = 176
\nDivide with 2 x \\(\\frac{22}{7}\\) on both sides.
\n\"AP
\n\u2234 Radius of the circle (r) = 28 cm.<\/p>\n

C) Given circumference of a circle is C = 2\u03c0r = 1.54 cm
\n2 x \\(\\frac{22}{7}\\) x r = 1.54
\nDivide with 2 x \\(\\frac{22}{7}\\) on both sides.
\n\"AP
\n\u2234 Radius of the circle (r) = 0.245 cm<\/p>\n

Question 3.
\nA gold smith has 8.8m of gold wire with him. He has to make gold rings of 2cm radius. How many such rings he can make with it?
\nSolution:
\nGiven the radius of the gold ring r = 2 cm
\nLength of the gold wire = 8.8 m (or) = 880 cm
\nTo find the? number of rings we have to divide the length of the gold wire by circumference of the gold ring.
\nNow, circumference of the gold ring C = 2\u03c0r
\n\"AP
\nNumber of rings made by gold smith = 70<\/p>\n

\"AP<\/p>\n

Question 4.
\nA wire was bent in the shape of a circle with radius 7cm. If the same wire was again used to make a square, then find its side.
\nSolution:
\nGiven radius of the circle r = 7 cm
\nCircumference of the circle C = 2\u03c0r
\n= 2 x \\(\\frac{22}{7}\\) x 7 = 44 cm
\nGiven length of the wire same.
\nSo, perimeter of square = circumference of the circle
\n4 x side = 44 cm
\nDivide with 4 on both sides = \\(\\frac{4 \\times \\text { side }}{4}=\\frac{44}{4}\\)
\nSide = 11 cm
\n\u2234 Side of square = 11cm.<\/p>\n

Question 5.
\nIn a chemical factory two wheels of different radius were connected with a belt. Radius of the bigger wheel is 21cm and radius of the smaller wheel is 7cm. If the bigger wheel rotates completely 100 times, find out the number of times that the smaller wheel rotates. .
\nSolution:
\nGiven the radius of the bigger wheel R = 21 cm
\nCircumference of the bigger wheel C = 2\u03c0R = 2 x \\(\\frac{22}{7}\\) x 21 = 132 cm
\nIf the bigger wheel completes 100 rotations,
\nDistance covered by the bigger wheel = Number of rotates x circumference
\n= 100 x 132 = 13,200 cm
\nNow, radius of smaller wheel r = 7 cm.
\nCircumference of the smaller wheel C = 2\u03c0r = 2 x \\(\\frac{22}{7}\\) x 7 = 44cm
\nIf the smaller wheel completes ‘n’ rotations,
\nDistance covered by the smaller wheel = number of rotations x circumference
\n= n x 44 = 44n cm
\nDistance covered by the small wheel = Distance covered by the bigger wheel
\n44n = 13200 cm
\nDivide with 44 as both sides,
\n\\(\\frac{44 n}{44}=\\frac{13200}{44}\\) = 300
\nn = 300
\n\u2234 Number of rotations made by the smaller wheel = 300.<\/p>\n

Question 6.
\nMohan is playing with a ring of diameter 14 cm, which is made up of metallic wire. When his brother asked, Mohan stretched the wire and made it as two equal parts. With those parts, he made another two small rings. Find the radius of smaller ring. Sol. Given the diameter of the bigger ring
\nSolution:
\nGiven the diameter of the bigger ring = 14 cm
\nRadius of the bigger ring R = \\(\\frac{\\text { diameter }}{2}\\)
\nR = \\(\\frac{14}{2}\\) = 7 cm
\nLength of the wire = Circumference of the bigger ring
\n= 2\u03c0R = 2 x \\(\\frac{22}{7}\\) x 7 = 44 cm
\nIf Mohan stretched the wire into’ halves.
\nThen the length of half of the wire = circumference of the bigger ring \u00f7 2
\n= \\(\\frac{44}{2}\\) = 22 cm
\nCircumference of the smaller ring = 2\u03c0r = 22
\n2 x \\(\\frac{22}{7}\\) x r = 22
\nDivide with 2 x \\(\\frac{22}{7}\\) on both sides
\n\"AP
\n\u2234 Radius of the smaller ring = 3.5 cm<\/p>\n

\"AP<\/p>\n

Question 7.
\nIn designing an iron gril a black smith needed 70 rings with radius of 7cm each. Find how much length of the rod he required, if the wastage is 20 cm.
\nSolution:
\nGiven radius of the ring r = 7 cm
\nCircumference of the ring C = 2\u03c0r = 2 x \\(\\frac{22}{7}\\) x 7 = 44 cm
\nLength of rod required to make one ring = 44 cm
\nLength of rod required to make 70 rings = 44 x 70 = 3080 cm
\nWastage = 20 cm
\nRequired length of rod to make 70 rings = 3080 + 20 = 3100 c<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 Textbook Questions and Answers. AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Ex 11.2 Question 1. Find the circumferences of the circles with the radius given below. A) 7 cm B) 3.5 cm C) 14 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":38650,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[10],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/04\/Chapter-11-Perimeter-and-Area-Ex-11.2.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/11724"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=11724"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/11724\/revisions"}],"predecessor-version":[{"id":38651,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/11724\/revisions\/38651"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/38650"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=11724"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=11724"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=11724"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}