{"id":8942,"date":"2022-01-25T17:41:22","date_gmt":"2022-01-25T12:11:22","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=8942"},"modified":"2025-05-28T16:44:17","modified_gmt":"2025-05-28T11:14:17","slug":"ncert-solutions-for-class-7-maths-chapter-11-perimeter-and-area-ex-11-4","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-material\/maths\/ncert-solutions-for-class-7\/chapter-11-perimeter-and-area-ex-11-4\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4"},"content":{"rendered":"

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4<\/h2>\n

Infinity Learn provides detailed solutions to all the subjects, chapter-wise.<\/p>\n

 <\/p>\n

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.4<\/strong>
\nEx 11.4 Class 7 Maths 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. Find the area of the path. Also find the area of the garden in hectare.
\nSolution:
\n\"NCERT
\nGiven: Length = 90 m
\nBreadth = 75 m
\nArea of the garden = l \u00d7 b
\n= 90 m \u00d7 75 m = 6750 m2<\/sup>
\nLength of the garden including path
\n= 90m + 5m + 5m = 100 m
\nBreadth of the garden including path
\n= 75m + 5m + 5m = 85m
\nArea of the garden including path
\n= l \u00d7 b
\n= 100 m \u00d7 85 m = 8500 m2<\/sup>
\nArea of the path = 8500 m2<\/sup> \u2013 6750 m2<\/sup> = 1750 m2<\/sup>
\nHence, required area of path = 1750 m2<\/sup> and area of the garden = 6750 m2<\/sup> = 0.675 ha<\/p>\n

Ex 11.4 Class 7 Maths 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 of the park = 125 m
\nBreadth of the park = 65 m
\nArea of the park = l \u00d7 b
\n= 125 m \u00d7 65 m = 8125 m2<\/sup>
\n\"NCERT
\nLength of the park including path
\n= 125 m + 3m + 3m = 131 m
\nBreadth of the park including path
\n= 65m + 3m + 3m = 71m
\nArea of the park including path
\n= 131 m \u00d7 71 m = 9301 m2<\/sup>
\n\u2234 Area of the path
\n= 9301 m2<\/sup> \u2013 8125 m2<\/sup> = 1176 m2<\/sup>
\nHence, the required area = 1176 m2<\/sup>.<\/p>\n

Ex 11.4 Class 7 Maths 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\"NCERT
\nLength = 8 cm, breadth = 5 cm
\nArea of the cardboard = l \u00d7 b
\n= 8 cm \u00d7 5 cm = 40 cm2<\/sup>
\nWidth of the margin = 1.5 cm
\nLength of the inner cardboard
\n= 8 cm \u2013 1.5 \u00d7 2 cm
\n= 8 cm \u2013 3 cm = 5 cm
\nBreadth of the inner cardboard
\n= 5 cm \u2013 1.5 \u00d7 2 cm
\n= 5 cm \u2013 3 cm = 2 cm
\nArea of the inner rectangle = l \u00d7 b
\n= 5 cm \u00d7 2 cm = 10 cm2<\/sup> Area of the margin
\n= 40 cm2<\/sup> \u2013 10 cm2<\/sup> = 30 cm2<\/sup>
\nHence, the required area = 30 cm2<\/sup>.<\/p>\n

Ex 11.4 Class 7 Maths 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:
\n(i) the area of the verandah.
\n(ii) the cost of cementing the floor of the verandah at the rate of \u20b9 200 per m2<\/sup>.
\nSolution:
\n\"NCERT
\nLength of the room = 5.5 m
\nBreadth of the room = 4 m
\n\u2234 Area of the room = l \u00d7 b = 5.5 m \u00d7 4 m = 22 m2<\/sup>
\nWidth of the verandah = 2.25 m
\nLength of the room including verandah
\n= 5.5 m + 2 \u00d7 2.25 m = 10 m
\nBreadth of the room including verandah
\n= 4 m + 2 \u00d7 2.25 m = 8.50 m2<\/sup>
\nArea of the room including verandah = l \u00d7 b
\n= 10 m \u00d7 8.50 m = 85 m2<\/sup>
\n(i) Area of the verandah = 85 m2<\/sup> \u2013 22 m2<\/sup>
\n= 63 m2<\/sup>
\n(ii) Cost of cementing the floor of the verandah = \u20b9 63 \u00d7 200 = \u20b912600<\/p>\n

Ex 11.4 Class 7 Maths Question 5.
\nA path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
\n(i) the area of the path.
\n(ii) the cost of planting grass in the remaining portion of the garden at the rate of \u20b9 40 per m2<\/sup>.
\nSolution:
\n\"NCERT
\nArea of the square garden = (Side)2<\/sup>
\n= 30 m \u00d7 30 m = 900 m2<\/sup>
\nLength of the garden excluding the path = 30 m \u2013 2 \u00d7 1 m = 28 m
\n\u2234 Area of the garden excluding the path = 28 m \u00d7 28 m = 784 m2<\/sup>
\n(i) Area of the path = 900 m2<\/sup> \u2013 784 m2<\/sup>
\n= 116 m2<\/sup>
\n(ii) Cost of the planting the remaining portion at the rate of \u20b9 40 per m2<\/sup>
\n= \u20b9 40 \u00d7 784 = \u20b9 31,360<\/p>\n

Ex 11.4 Class 7 Maths Question 6.
\nTwo cross roads, each of width 10 m, cut a 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\"NCERT
\nLength of the road parallel to the length of the park = 700 m
\nWidth of the road = 10 m
\n\u2234 Area of the road = l \u00d7 b = 700 m \u00d7 10 m = 7000 m2<\/sup>
\nLength of the road parallel to the breadth of the park = 300 m
\nWidth of the road = 10 m Area of this road = l \u00d7 b = 300 m \u00d7 10 = 3000 m2<\/sup>
\nArea of the both roads
\n= 7000 m2<\/sup> + 3000 m2<\/sup> \u2013 Area of the common portion
\n= 10,000 m2<\/sup> \u2013 10 m \u00d7 10 m
\n= 10,000 m2<\/sup> \u2013 100 m2<\/sup>
\n= 9900 m2<\/sup> = 0.99 ha
\nArea of the park = l \u00d7 b
\n= 700 m \u00d7 300 m = 210000 m2<\/sup>
\nArea of the park excluding the roads
\n= 210000 m2<\/sup> \u2013 9900 m2<\/sup>
\n= 200100 m2<\/sup> = 20.01 ha<\/p>\n

Ex 11.4 Class 7 Maths Question 7.
\nThrough a rectangular field 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 m, find
\n(i) the area covered by the roads.
\n(ii) the cost of constructing the roads at of the rate of \u20b9 110 per m2<\/sup>.
\nSolution:
\n\"NCERT
\nLength of the road along the length of the field = 90 m
\nBreadth = 3 m
\n\u2234 Area of this road = l \u00d7 b
\n= 90 m \u00d7 3 m = 270 m2<\/sup>
\nSimilarly, the area of the road parallel to the breadth of the field = l \u00d7 b
\n= 60 m \u00d7 3 m = 180 m2<\/sup> Area of the common portion
\n= 3m \u00d7 3m = 9m2<\/sup>
\n(i) Area of the two roads
\n= 270 m2<\/sup> + 180 m2<\/sup> \u2013 9 m2<\/sup>
\n= 450 m2<\/sup> \u2013 9 m2<\/sup> = 441 m2<\/sup>
\n(ii) Cost of constructing the roads
\n= \u20b9 110 \u00d7 441 = \u20b9 48,510<\/p>\n

Ex 11.4 Class 7 Maths Question 8.
\nPragya wrapped a card around a circular pipe of radius 4 cm and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (\u03c0 = 3.14)
\n\"NCERT
\nSolution:
\nLength of the cord = Circumference of the circular pipe
\n= 2\u03c0r = 2 \u00d7 3.14 \u00d7 4 = 25.12 cm
\nPerimeter of the square box
\n= 4 \u00d7 side = 4 \u00d7 4 cm = 16 cm
\nLength of the cord left
\n= 25.12 cm \u2013 16 cm = 9.12 cm
\nYes, 9.12 cm cord is left.<\/p>\n

Ex 11.4 Class 7 Maths Question 9.
\nThe given figure represents a rectangular lawn with a circular flower bed in the middle. Find:
\n(i) the area of the whole land.
\n(ii) the area of the flower bed.
\n(iii) the area of the lawn excluding the area of the flower bed.
\n(iv) the circumference of the flower bed.
\n\"NCERT
\nSolution:
\n(i) Length of the lawn = 10 m
\nBreadth of the lawn = 5 m
\nArea of the lawn = l \u00d7 b
\n= 10 m \u00d7 5 m = 50 m2<\/sup>
\n(ii) Area of the circular flower bed = \u03c0r2<\/sup>
\n\\(=\\frac{22}{7} \\times 2 \\times 2=\\frac{88}{7} \\mathrm{m}^{2}=12.57 \\mathrm{m}^{2}\\)
\n(iii) Area of the lawn excluding the area of the flower bed
\n\"NCERT<\/p>\n

Ex 11.4 Class 7 Maths Question 10.
\nIn the following figures, find the area of the shaded portion.
\n\"NCERT
\nSolution:
\n(i) Area of the rectangle = l \u00d7 b
\n= 18 cm \u00d7 (6 cm + 4 cm)
\n= 18 cm \u00d7 10 cm = 180 cm2<\/sup>
\nArea of right triangle
\n\\(=\\frac{1}{2} \\times b \\times h=\\frac{1}{2} \\times 6 \\times 10=30 \\mathrm{cm}^{2}\\)
\nArea of right \u2206BCE = \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h
\n= \\(\\frac{1}{2}\\) \u00d7 8 \u00d7 10 =40 cm2<\/sup>
\nArea of the two right triangles
\n= 30 cm2<\/sup> + 40 cm2<\/sup> = 70 cm2<\/sup>
\nArea of the shaded portion
\n= 180 cm2<\/sup> \u2013 70 cm2<\/sup> = 110 cm2<\/sup>
\n(ii) Area of the square PQRS = (Side)2<\/sup>
\n= (20)2<\/sup> = 400 cm2<\/sup>
\n\"NCERT
\nArea of the three triangles
\n= 50 cm2<\/sup> + 100 cm2<\/sup> + 100 cm2<\/sup> = 250 cm2<\/sup>
\nArea of the shaded portion
\n= 400 cm2<\/sup> \u2013 250 cm2<\/sup> = 150 cm2<\/sup><\/p>\n

Ex 11.4 Class 7 Maths Question 11.
\nFind the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM \u22a5 AC, DN \u22a5 AC.
\n\"NCERT
\nSolution:
\n\"NCERT
\nArea of the quadrilateral ABCD
\n= Area of \u2206ABC + Area of \u2206ADC
\n= 33 cm2<\/sup> + 33 cm2<\/sup> = 66 cm2<\/sup>
\nHence, the required area = 66 cm2<\/sup>.<\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

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