{"id":15818,"date":"2020-09-10T14:50:01","date_gmt":"2020-09-10T09:20:01","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=15818"},"modified":"2021-07-02T15:01:53","modified_gmt":"2021-07-02T09:31:53","slug":"kseeb-solutions-for-class-7-maths-chapter-11-ex-11-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-11-ex-11-4\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4"},"content":{"rendered":"

Students can Download Chapter 11 Perimeter and Area Ex 11.4, 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 11.4<\/h2>\n

Question 1.
\nA garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Found the area of the path. Also find the area of the garden in hectare.
\nSolution:
\n\"KSEEB
\nArea of the garden = l \u00d7 b
\nLength = l = 90 m
\nBreath = b = 75 m
\n\u2234 Area of the garden EFG ||)} = 90 \u00d7 75
\n= 6750 sq mtrs
\n1 Hectare = 10,000 sqmts.
\n\"KSEEB
\nArea of the outer skirt of the garden = l \u00d7 b
\nlength = 90 + 5 + = 100 m
\nbreadth = 75 + 5 + 5 = 85 m
\n\u2234 Area = 100 \u00d7 85 = 8,500sqm
\n\"KSEEB
\nArea of the path = 0.850 – 0.675 = 0.175
\n0.175 hectare or 1750 sq mts<\/p>\n

Question 2.
\nA 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
\nSolution:
\nlength = l = 125 breadth = b = 65 m
\nwidth of a path = 3 m
\nArea of a rectangular park = l \u00d7 b = 125 \u00d7 65
\n= 8125 sq mts
\nArea of ABCD = l \u00d7 b
\nl = 125 + 3 + 3 = 131 b = 65 + 3 + 3 = 71
\n\u2234 Area = 131 \u00d7 71 = 9301 sq mts
\nArea of the path = Area of ABCD \u2013 Area
\nof park 9301 \u2013 8125 = 1176 sq mts
\n\"KSEEB<\/p>\n

Question 3.
\nA picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
\nSolution:
\n\"KSEEB
\nArea of cardboard = l \u00d7 b
\nl = 8 cm b = 5 cm Margin = 1.5 cm
\nArea of picture = l \u00d7 b
\nl = 8 = (1.5 + 1.5) = 5 cms b = 5 = (1.5 + 1.5) = 2 cms
\nArea = l \u00d7 b = 5 \u00d7 2
\nArea of margin = Area of Cardboard \u2013 Area of picture
\n= 40 \u2013 10 = 30 sq cms<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nA verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find the area of the verandah: the cost of cementing the floor of the verandah at the rate of \u20b9 200 per m2. The area of the verandah with the room. (Fig KLMN)
\nSolution:
\nl = length = 5.5 + 2.25 + 2.25 = 10 m
\nb = breadth = 4 + 2.25 + 2.25 = 8.5 m
\n\u2234 Area 10 \u00d7 8.5
\n= 85 sq cms.
\n\"KSEEB
\nArea of room = l \u00d7 b
\nl = 5.5 m = 5.5 \u00d7 4 N
\n\u2234 Area of verandah = Area of KLMN – Area of room
\n= 85 – 22 = 63 sq mts
\nCost of cementing the floor = Rs 200 \/ sq m
\nTotal cost of cementing = 63 \u00d7 200 = Rs 12,600<\/p>\n

Question 5.
\nA path 1 m wide is built along the border and inside a square garden of side 30m. Find :
\ni) the area of the path
\nii) the cost of planting grass in the remaining portion of the garden at the rate of \u20b9 40 per m2<\/sup>.
\nSolution:
\n\"KSEEB
\nArea of KLMN = side \u00d7 side = 30 \u00d7 30
\nArea of ABCD = 30 – (1 + 1) = 28
\n= 28 \u00d7 28
\n= 784 sq mts
\nArea of the path = Area of KLMN – Area of ABCD
\nCost of planting grass = Rs 40 \/ sq m
\n\u2234 Total cost of planting = 784 \u00d7 40
\n= Rs 31.360<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nTwo cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
\nSolution:
\n\"KSEEB
\nArea of PQRS = l x b = 700 \u00d7 300
\n= 2,10,000 sq m
\nArea of cross Roads = Area of KLMN abcd – 1, 2, 3, 4
\n= 300 \u00d7 10 + 700 \u00d7 10 – 10 \u00d7 10
\n= 3000 + 7000 – 100
\n= 10,000 – 100 = 9,900 sq mts
\nArea of the park excluding the cross roads = 2,10,000 – 9,900
\n= 2,00,100 sq mts.
\n\"KSEEB<\/p>\n

Question 7.
\nThrough a rectangular filed of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 in. Find :
\ni) The area covered by the roads.
\n\"KSEEB
\nArea covered by the roads = Area of klmn + Area of abcd – Area of 1, 2, 3, 4
\nArea of abcd = 90 \u00d7 3 = 270 sq mtrs
\nArea of klmn = 60 \u00d7 3 = 180 sq mts.
\n\"KSEEB
\n= 180 + 270 – 9 = 441 sq mts.<\/p>\n

ii) the cost of constructing the roads at the rate of \u20b9 110 per m2<\/sup>.
\nCost of constructing the roads = Rs 110\/sq m
\n\u2234 Total cost of constructing the roads
\n= 441 \u00d7 110 = Rs. 48.510<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nPragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure)
\nand cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown).
\nDid she have any cord left? (r = 3.14)
\nSolution:
\n\"KSEEB
\nRadius of the circular pipe} r = 4 cm
\nCirumference of the circular pipe } 2\u03c0r
\n=2 \u00d7 3.14 \u00d7 4
\n= 25.12 cms
\nPerimeter of the square box = 4 \u00d7 side
\n= 4 \u00d7 4
\n= 16 cms
\nyes, the length of the cord left = 25.12 – 16
\n= 9.12 cms<\/p>\n

Question 9.
\nThe adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find
\nSolution:
\n\"KSEEB
\ni) the area of the whole land
\nArea of the whole land = l \u00d7 b
\nl = 10 mts = 10 \u00d7 5
\nb = 5 mts = 50 sq mts<\/p>\n

ii) the area of the flower bed
\nArea of the flower bed = \u03c0r2<\/sup>
\nr = 2m = 3.14 \u00d7 2 \u00d7 2
\n= 12.56 sq mts<\/p>\n

\"KSEEB<\/p>\n

iii) the area of the lawn excluding the area of the flower bed
\nSolution:
\nThe area of the lawn excluding the circular flower bed } = 50 – 12.56
\n= 37.44 sq m<\/p>\n

iv) the circumference of the flower bed.
\nThe circumference of the circular flower bed} = 2\u03c0r
\n= 2 \u00d7 3.14 \u00d7 2
\n= 12.56 mts.<\/p>\n

Question 10.
\nIn the following figures, find the area of the shaded portions :
\n\"KSEEB
\nSolution:
\n1) Area of the rectangle ABCD = l \u00d7 b
\nl = 18 cms = 18 \u00d7 10
\nb = 10 cms = 180 sq cms
\n\"KSEEB
\n= 30 sq cms
\n\"KSEEB
\n= 40 sq cms
\nArea of the shaded position = Area of rectangle ABCD – (AreaofAEF+AreaofBCE)
\n= 180 – (30 + 40)
\n= 180 – 70 = 1110 sq cms<\/p>\n

ii) Area of the square PQRS = side \u00d7 side
\n= 20 \u00d7 20
\n= 400 sq cms.
\n\"KSEEB
\n= 50 sq cm
\n\"KSEEB
\n= 100 sq cms
\n\"KSEEB
\n= 100 sq cm
\n\u2234 Area of shaded region = 400 – (50 + 100 – 100)
\n= 400 – 250
\n= 150 sq cms.<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nFind the area of the quadrilateral ABCD.
\nHere, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM \u22a5 AC, DN \u22a5 AC
\n\"KSEEB
\nSolution:
\nArea of the quadrilateral ABCD =
\nArea of the \u2206 ABC + Area of the \u2206 ACD
\n\"KSEEB
\n\u2234 Area of quadrilateral ABCD = 33 + 33
\n= 66 sq cms<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 11 Perimeter and Area Ex 11.4, 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 11.4 Question 1. A garden …<\/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\/15818"}],"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=15818"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15818\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=15818"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=15818"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=15818"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}