{"id":11492,"date":"2021-04-23T10:54:06","date_gmt":"2021-04-23T05:24:06","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=11492"},"modified":"2021-09-15T11:33:04","modified_gmt":"2021-09-15T06:03:04","slug":"ap-board-6th-class-maths-solutions-chapter-11-intext-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-6th-class-maths-solutions-chapter-11-intext-questions\/","title":{"rendered":"AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 6th Class Maths Solutions<\/a> Chapter 11 Perimeter and Area InText Questions and Answers.<\/p>\n

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

\"AP<\/p>\n

Check Your Progress<\/span> (Pg.No. 147)<\/p>\n

Question 1.
\nFind the perimeters of the given figures.
\n\"AP
\nIn the above figures (I) and (ii), find the perimeter of \u0394KLM, \u0394KMN and \u25a1 KLMN.
\n(a) Compare the perimeters of \u0394KLM and \u25a1 KLMN
\nCompare the perimeters of \u0394KMN and \u25a1 KLMN.
\nWhat can you say? –
\nSolution:
\ni) In \u0394KLM, given KL = 2cm; LM = 2.6 cm; MK = 3.8 cm
\nPerimeter of \u0394KLM = KL + LM + MK
\n= 2 + 2.6 + 3.8
\n= 8.4 cm<\/p>\n

\"AP<\/p>\n

In \u0394KMN, given KM = 3.8 cm; MN = 2cm; KN = 2.6 cm
\nPerimeter of \u0394KMN = KM + MN + KN
\n= 3.8 + 2 + 2.6 = 8.4 cm<\/p>\n

ii) In \u25a1 KLMN, given KL = 2cm, LM = 2.6cm, MN = 2cm and KN = 2.6 cm
\nPerimeter of \u25a1 KLMN = KL + LM + MN + KN = 2 + 2.6 + 2 + 2.6 = 9.2 cm<\/p>\n

a) By comparing the perimeter of \u0394KLM and \u25a1 KLMN
\n8.4 cm < 9.2 cm
\n\u2234 Perimeter of \u0394KLM < Perimeter of \u25a1 KLMN
\nBy comparing the perimeter of \u0394KMN and \u25a1 KLMN
\n8.4 cm < 9.2 cm
\n\u2234 Perimeter of \u0394KMN < Perimeter of \u25a1 KLMN.
\nBy comparing the perimeters of triangle with quadrilateral ‘
\nPerimeter of the triangle is always less than the perimeter of quadrilateral.<\/p>\n

Let’s Explore<\/span> (Pg.No. 150)<\/p>\n

Question 1.
\nIf the radius of a circle is doubled, then, what is the change in its circumference?
\nSolution:
\nIf the radius is r, then its circumference of the circle is 2\u03c0r.
\nIf radius is doubled means r = 2r (new)
\nWrite r = 2r in 2\u03c0r.
\nThen, the circumference = 2\u03c0 x 2r
\n= 4\u03c0r = 2.27\u03c0r
\nCircumference is doubled.
\nSo, if the radius of a circle is doubled, then the circumference is also doubled.<\/p>\n

Question 2.
\nIf the radius is halved, then, what is the change in its circumference?
\nSolution:
\nIf the radius is r, then its circumference of the circle is 2\u03c0r.
\nIf radius is halved means r = \\(\\frac{\\mathbf{r}}{2}\\) (new)
\nWrite r = \\(\\frac{\\mathbf{r}}{2}\\) in 2\u03c0r.
\nThen, the circumference = 2\u03c0 x \\(\\frac{r}{2}=\\frac{1}{2}\\). 2\u03c0r
\nCircumference is halved.
\nSo if the radius of a circle is halved, then the circumference is also halved.<\/p>\n

\"AP<\/p>\n

Check Your Progress<\/span> (Page No. 151)<\/p>\n

Question 1.
\nFind the area of a square with side 16 cm.
\nSolution:
\nGiven side of the square (s) = 16 cm
\nArea of the square A = s x s = 16 x 16 = 256 sq.cm<\/p>\n

Question 2.
\nLength and breadth of a rectangle are 16 cm and 12 cm respectively. Find its area.
\nSolution:
\nGiven length of the rectangle l = 16 cm
\nbreadth of the rectangle b = 12 cm
\nArea of the rectangle A = l x b = 16 x 12 = 192 sq.cm<\/p>\n

Let’s Think<\/span> (Page No. 151)<\/p>\n

Question 1.
\nFind the perimeter and area of a square of side 4 cm. Are these same? Give some examples to support your answer.
\nSolution:
\n\"AP<\/p>\n

\"AP<\/p>\n

From the above table,
\nIf side is 4 cm, then only its perimeter and areas are equal.
\nIn other cases perimeter and areas are not same.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions and Answers. AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area InText Questions Check Your Progress (Pg.No. 147) Question 1. Find the perimeters of the given figures. In the above figures (I) and (ii), find the perimeter …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[10],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/11492"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=11492"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/11492\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=11492"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=11492"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=11492"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}