{"id":10571,"date":"2024-01-17T11:06:33","date_gmt":"2024-01-17T05:36:33","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=10571"},"modified":"2024-01-18T09:41:03","modified_gmt":"2024-01-18T04:11:03","slug":"ap-board-7th-class-maths-solutions-chapter-13-ex-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-board-7th-class-maths-solutions-chapter-13-ex-2\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 2"},"content":{"rendered":"

AP State Syllabus AP Board 7th Class Maths Solutions<\/a> Chapter 13 Area and Perimeter Ex 2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 2<\/h2>\n

\"AP<\/p>\n

Question 1.
\nFind the area of the each of the following parallelograms.
\n\"AP
\nSolution:<\/p>\n\n\n\n\n\n\n\n
Fig<\/td>\nBase<\/td>\nHeight<\/td>\nA = Base x Height<\/td>\n<\/tr>\n
(i)<\/td>\n7<\/td>\n4<\/td>\n7 x 4 = 28cm2<\/sup><\/td>\n<\/tr>\n
(ii)<\/td>\n5<\/td>\n3<\/td>\n5 x 3\u00a0 = 15 cm2<\/sup><\/td>\n<\/tr>\n
(iii)<\/td>\n5.1<\/td>\n7.6<\/td>\n5.1 x 7.6 = 38.76 cm2<\/sup><\/td>\n<\/tr>\n
(iv)<\/td>\n4<\/td>\n6<\/td>\n4 x 6 = 24cm2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"AP<\/p>\n

Question 2.
\nPQRS is a parallelogram. PM is the height from P to \\(\\overline{\\mathrm{SR}}\\) and PN is the height from P to \\(\\overline{\\mathrm{QR}}\\). If SR = 12cm and PM = 7.6 cm.
\n(i) Find the area of the parallelogram PQRS
\n(ii) Find PN, if QR = 8 cm.
\nSolution:
\n\"AP
\ni) Given:SR = 12cm
\nPM = 7.6cm
\n\u2234 Area = base \u00d7 height
\n= 12 \u00d7 7.6 = 91.2 cm2<\/sup><\/p>\n

ii) Also area = base \u00d7 height
\n91.2 = QR \u00d7 PN
\n91.2 = 8 \u00d7 PN
\n\u2234 PN = \\(\\frac{91.2}{8}\\) = 11.4cm<\/p>\n

Question 3.
\nDF and BE are the height on sides AB and AD respectively in parallelogram ABCD. Ifthe area of the parallelogram is 1470 cm2<\/sup>, AB = 35 cm and AD = 49 cm, find the length of BE and DF.
\n\"AP
\nSolution:
\nGiven : AB = 35 cm, AD = 49 cm
\narea of \ud83d\uddbeABCD = 1470 cm2<\/sup>
\nArea of a parallelogram = base \u00d7 height
\nAB \u00d7 DF = area of \ud83d\uddbeABCD
\n35 \u00d7 DF = 1470
\nDF = \\(\\frac{1470}{35}\\) = 42cm
\nAlso AD \u00d7 BE = 1470
\n49 \u00d7 BE = 1470
\n\u2234 BE= \\(\\frac{1470}{49}\\) = 30cm<\/p>\n

\"AP<\/p>\n

Question 4.
\nThe height of a parallelogram is one third of its base. If the area of the parallelogram is 192cm2<\/sup>, find its height and base.
\nSolution:
\nArea of the parallelogram = 192 cm2<\/sup>
\nArea of parallelogram = base \u00d7 height = 192 cm2<\/sup> (given)
\nLet the base of the parallelogram be x.
\nheight = \\(\\frac{x}{3}\\)
\nArea = base \u00d7 height = x x \\(\\frac{x}{3}=\\frac{x^{2}}{3}\\) = 192
\n\u21d2 x2<\/sup> = 192 \u00d7 3 = 576
\nx = \\(\\sqrt{576}\\) = 24
\n\u2234 base = 24 cm
\nheight = \\(\\frac{x}{3}=\\frac{24}{3}\\) = 8cm.<\/p>\n

\"AP<\/p>\n

Question 5.
\nIn a parallelogram the base and height are is in the ratio of 5:2. If the area of the parallelogram is 360m2<\/sup>, find its base and height.
\nSolution:
\nLet the base of the parallelogram be 5x and height of the parallelogram be 2x.
\nArea = height x base = 5x \u00d7 2x = 10x2<\/sup> = 360 cm2<\/sup> (given)
\n10x2<\/sup> = 360
\nx2<\/sup> = \\(\\frac{360}{10}\\) = 36
\nx = \u221a36 = 6
\nbase = 5x = 5(6) = 30 cm
\nheight = 2x = 2(6) = 12 cm<\/p>\n

Question 6.
\nA square and a parallelogram have the same area. If a side of the square is 40m and the height of the parallelogram is 20m, find the base of the parallelogram.
\nSolution:
\nGiven: Side of a square = 40m
\nHeight of a parallelogram = 20m
\nArea of parallelogram = Area of the square
\nbase \u00d7 height side \u00d7 side
\nbase \u00d7 20 40 \u00d7 40
\n\u2234 base = \\(\\frac{40 \\times 40}{20}\\) = 80 m.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 2 Textbook Questions and Answers. AP State Syllabus 7th Class Maths Solutions 13th Lesson Area and Perimeter Exercise 2 Question 1. Find the area of the each of the following parallelograms. Solution: Fig Base Height A = Base x Height … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":37869,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/04\/Chapter-13-Area-and-Perimeter-Ex-2.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/10571"}],"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=10571"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/10571\/revisions"}],"predecessor-version":[{"id":37870,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/10571\/revisions\/37870"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/37869"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=10571"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=10571"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=10571"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}