{"id":15673,"date":"2020-09-10T10:08:24","date_gmt":"2020-09-10T04:38:24","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=15673"},"modified":"2021-07-02T15:01:53","modified_gmt":"2021-07-02T09:31:53","slug":"kseeb-solutions-for-class-7-maths-chapter-11-ex-11-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-11-ex-11-2\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2"},"content":{"rendered":"

Students can Download Chapter 11 Perimeter and Area Ex 11.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2<\/h2>\n

Question 1.
\nFind the area of each of the following parallelograms :
\na) \"KSEEB
\nSolution:
\nBase = B = 7cm
\nHeight = h = 4 cm
\n\u2234 Area of the parallelogram = B \u00d7 h
\n= 7 \u00d7 4 = 28 sq.cms<\/p>\n

b) Base = B = 5 cm
\nHeight = h = 3 cm.
\n\"KSEEB
\nArea of the parallelogram = Base \u00d7 height
\n= B \u00d7 h = 5 \u00d7 3
\n= 15 sq.cms<\/p>\n

c) Base = B = 2.5 cms
\nHeight = h = 3.5 cms
\n\"KSEEB
\nArea of the parallelogram
\nBase \u00d7 height = 2.5 \u00d7 3.5 = 8.75 sq.cms<\/p>\n

\"KSEEB<\/p>\n

d) Base = B = 5 cms
\nHeight = h = 4.8 cms
\n\"KSEEB
\nArea of the parallelogram = Base \u00d7 height
\n= 5 \u00d7 4.8 = 24 sq.cms<\/p>\n

e) Base = B = 2 cms
\n\"KSEEB
\nHeight = h = 4.4 cms
\n\u2234 Area of the parallelogram = Base \u00d7 height
\n2 \u00d7 4.4 = 8.8 sq.cms<\/p>\n

Question 2.
\nFind the area of each of the following triangles :
\nSolution:
\na) Base = B = 4 cms
\nHeight = h = 3 cms
\n\"KSEEB
\n\"KSEEB<\/p>\n

b) Base = B = 5cms
\nHeight = h = 3.2 cms
\n\"KSEEB
\n\u2234 Area of the triangle
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

c) Base = B = 3 cms
\nHeight = h = 4 cms
\nArea of the triangle
\n\"KSEEB
\n\"KSEEB<\/p>\n

d) Base = B = 3 cms
\nHeight = h = 2 cms
\nArea of the parallelogram
\n\"KSEEB<\/p>\n

Question 3.
\nFind the missing values :
\n\"KSEEB
\nSolution:
\na) Area of the parallelogram = Base \u00d7 height
\nBase = 20 cm
\nArea = 246 sq cm
\nHeight = ?
\n246 = 20 \u00d7 h
\n\"KSEEB
\nHeight = 12.3 cms<\/p>\n

b) Height = 15 cm
\nArea = 154.5 cm2<\/sup>
\nBase = ?
\nArea of the parallelogram = Base \u00d7 height
\n154.5 = B \u00d7 15
\n\"KSEEB
\n\u2234 Base = 10.3 cms<\/p>\n

\"KSEEB<\/p>\n

c) Height = 8.4 cm
\nArea = 48.72 cm2<\/sup>
\nBase = ?
\nArea of the parallelogram = Base \u00d7 height
\n48.72 = B \u00d7 8.4
\n\"KSEEB
\n\u2234 Base = 5.8 cms<\/p>\n

d) Base = 15.6 cm
\nArea = 16.38 sq cm
\nHeight = ?
\nArea of the parallelogram = Base \u00d7 height
\n16.38 = 15.6 \u00d7 h
\n\"KSEEB
\n\u2234 Height = 1.05 cms
\n\"KSEEB<\/p>\n

Question 4.
\nFind the missing values :
\n\"KSEEB
\nSolution:
\ni) Base = 15 cm Area = 87 cm2<\/sup> Height = ?
\n\"KSEEB<\/p>\n

ii) Height = 31.4 mm
\nArea = 1256 mm2<\/sup>
\nBase = ?
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

iii) Base = 22 cm Area = 170.5 cm2<\/sup>
\nHeight = ?
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nPQRS is a parallelogram (Fig.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find :
\nSolution:
\n\"KSEEB
\na) the area of the parallelogram PQRS
\nArea of the parallelogram PQRS = Base \u00d7 height
\nBase = SR =12 cm
\nHeight = QM = 7.6 cm
\n= 12 \u00d7 7.6 = 91.2 sq.cms
\n\u2234 Area = 91.2 cm2<\/sup><\/p>\n

b) QN, if PS = 8 cm
\nBase = PS = 8 cm Height = QN = ?
\nArea = 91.2 cm2<\/sup>
\nArea of the parallelogram = Base \u00d7 height
\n91.2 = 8 \u00d7 QN
\n\"KSEEB
\n\u2234 QN = 11.4 cms.<\/p>\n

Question 6.
\nDL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm2<\/sup>, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
\nSolution:
\n\"KSEEB
\nArea of the parallelogram = Base \u00d7 height
\ni) Base = AB = 35 cm
\nHeight = DL = ?
\nArea = 1470 cm2<\/sup>
\n1470 = 35 \u00d7 DL
\n\"KSEEB<\/p>\n

ii) Base = AD = 49 cm
\nHeight = BM = ?
\nArea = 1470 cm2<\/sup>
\nArea of the parallelogram = Base \u00d7 height
\n1470 = 49 \u00d7 BM
\n\"KSEEB
\n\u2234 BM = 30 cms.<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\n\u2206 ABC is right-angled at A (Fig 11.25). AD Is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of \u2206 ABC. Also, find the length of AD.
\nSolution:
\n\"KSEEB<\/p>\n

Question 8.
\n\u2206 ABC is isosceles with AB AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC is 6 cm. Find the area is 6 cm. Find the area of \u2206ABC. What will be the height from C to AB i.e., CE?
\nSolution:
\n\"KSEEB
\n\"KSEEB
\n\"KSEEB
\n\u2234 The height from C to AB = CE = 7.2 cms<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 11 Perimeter and Area Ex 11.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 Question 1. Find the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15673"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=15673"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15673\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=15673"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=15673"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=15673"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}