{"id":2095,"date":"2022-11-09T05:19:38","date_gmt":"2022-11-09T05:19:38","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=2095"},"modified":"2022-11-09T05:19:38","modified_gmt":"2022-11-09T05:19:38","slug":"ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-4\/","title":{"rendered":"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 7 Mathematics\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 11 (Perimeter and Area)<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 7 Mathematics <strong>Chapter &#8211; 11 Perimeter and Area <\/strong>Exercise 11.4 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Chapter : 11 Perimeter and Area<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-1\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 11.1<\/span><\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-2\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 11.2<\/span><\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 11.3<\/span><\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 11.4\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1. A 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.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2121\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q1.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"274\" height=\"183\" \/><br \/>\n<\/strong>Given that,<br \/>\nLength of the garden (L) = 90 m<br \/>\nBreadth of the garden (B) = 75 m<br \/>\nArea of the garden = length \u00d7 breadth<br \/>\n= 90 m \u00d7 75 m<br \/>\n= 6750 m<sup>2<br \/>\n<\/sup>From the figure,<br \/>\nLength of the garden including path<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 90m + 5m + 5m = 100 m<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Breadth of the garden including path<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 75m + 5m + 5m = 85m<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Area of the garden including path<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= l \u00d7 b<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 100 m \u00d7 85 m = 8500 m<sup>2<br \/>\n<\/sup>Area of the path = 8500 m<sup>2<\/sup>\u00a0\u2013 6750 m<sup>2<\/sup>\u00a0= 1750 m<sup>2<br \/>\n<\/sup>Hence, required area of path = 1750 m<sup>2<\/sup><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">1 hectare = 10000 m<sup>2<br \/>\n<\/sup>Hence, area of garden in hectare = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{6750}{10000}\" alt=\"\\frac{6750}{10000}\" align=\"absmiddle\" \/> = 0.675 hectare<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. A 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.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2122\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q2.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"292\" height=\"172\" \/><br \/>\n<\/strong>Given that,<br \/>\nLength of the park (L) = 125 m<br \/>\nBreadth of the park (B) = 65 m<br \/>\nArea of the park = length \u00d7 breadth<br \/>\n= 125 m \u00d7 65 m<br \/>\n= 8125 m<sup>2<br \/>\n<\/sup>From the figure,<br \/>\nLength of the park including path = 125 m + 3m + 3m = 131 m<br \/>\nBreadth of the park including path = 65m + 3m + 3m = 71m<br \/>\nThe area of path = New area of the park including path \u2013 Area of park.<br \/>\nArea of the park including path = 131 m \u00d7 71 m = 9301 m<sup>2<\/sup><br \/>\n\u2234 Area of the path = 9301 m<sup>2<\/sup>\u00a0\u2013 8125 m<sup>2<\/sup>\u00a0= 1176 m<sup>2<\/sup><br \/>\nHence, the required area = 1176 m<sup>2<\/sup>.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3. A 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.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2124\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q3.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"259\" height=\"151\" \/><br \/>\n<\/strong>Given that,<br \/>\nLength of the cardboard (L) = 8 cm<br \/>\nBreadth of the cardboard (B) = 5 cm<br \/>\nArea of the cardboard = length \u00d7 breadth<br \/>\n= 8 cm \u00d7 5 cm<br \/>\n= 40 cm<sup>2<br \/>\n<\/sup>From the figure,<br \/>\nWidth of the margin = 1.5 cm<br \/>\nLength of the inner cardboard = 8 cm \u2013 1.5 \u00d7 2 cm<br \/>\n= 8 cm \u2013 3 cm = 5 cm<br \/>\nBreadth of the inner cardboard = 5 cm \u2013 1.5 \u00d7 2 cm<br \/>\n= 5 cm \u2013 3 cm = 2 cm<br \/>\nArea of the inner rectangle = l \u00d7 b<br \/>\n= 5 cm \u00d7 2 cm = 10 cm<sup>2<\/sup><br \/>\nThe area of margin = Area of the cardboard when margin is including \u2013 Area of the<br \/>\n= 40 cm<sup>2<\/sup>\u00a0\u2013 10 cm<sup>2<\/sup>\u00a0= 30 cm<sup>2<\/sup><br \/>\nHence, the required area = 30 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>4. A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:<br \/>\n<\/strong><strong>(i) the area of the verandah.<br \/>\n<\/strong><strong>(ii) the cost of cementing the floor of the verandah at the rate of \u20b9 200 per m<sup>2<\/sup>.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2125\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Que4.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"226\" height=\"137\" \/><br \/>\n<\/strong>Given that,<br \/>\nLength of the room (L) = 5.5 m<br \/>\nBreadth of the room (B) = 4 m<br \/>\nArea of the room = length \u00d7 breadth<br \/>\n= 5.5 m \u00d7 4 m<br \/>\n= 22 m<sup>2<br \/>\n<\/sup>From the figure,<br \/>\nWidth of the verandah = 2.25 m<br \/>\nLength of the room including verandah = 5.5 m + 2 \u00d7 2.25 m<br \/>\n= 10 m<br \/>\nBreadth of the room including verandah = 4 m + 2 \u00d7 2.25 m<br \/>\n= 8.50 m<sup>2<br \/>\n<\/sup>Area of the room including verandah = l \u00d7 b<br \/>\n= 10 m \u00d7 8.50 m<br \/>\n= 85 m<sup>2<\/sup><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(i)<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The area of verandah = Area of the room when verandah is included \u2013 Area of the room<br \/>\n= 85 \u2013 22<br \/>\n= 63 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii)<br \/>\n<\/strong>Given, the cost of cementing the floor of the verandah at the rate of \u20b9 200 per m<sup>2<br \/>\n<\/sup>Then the cost of cementing the 63 m<sup>2<\/sup> area of floor of the verandah = 200 \u00d7 63<br \/>\n= \u20b9 12600<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>5. A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:<br \/>\n<\/strong><strong>(i) the area of the path<br \/>\n<\/strong><strong>(ii) the cost of planting grass in the remaining portion of the garden at the rate of \u20b9 40 per m<sup>2<\/sup>. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2127\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q5.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"152\" height=\"162\" \/><br \/>\n<\/strong>Given that,<br \/>\nSide of square garden (s) = 30 m<br \/>\nArea of the square garden = (Side)<sup>2<br \/>\n<\/sup>= 30 m \u00d7 30 m = 900 m<sup>2<br \/>\n<\/sup>Length of the garden excluding the path = 30 m \u2013 2 \u00d7 1 m = 28 m<br \/>\n\u2234 Area of the garden excluding the path = 28 m \u00d7 28 m = 784 m<sup>2<\/sup><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(i)<\/strong> The area of path = Area of the square garden when path is included \u2013 Area of the garden excluding the path<br \/>\n= 900 m<sup>2<\/sup>\u00a0\u2013 784 m<sup>2<\/sup><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 116 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii)<\/strong> Cost of the planting the remaining portion at the rate of \u20b9 40 per m<sup>2<\/sup><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Then the cost of planting the grass in 784 m<sup>2<\/sup> area of the garden = \u20b9 40 \u00d7 784<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">= \u20b9 31,360<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>6. Two 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.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2128\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q6.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"254\" height=\"136\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Given that,<br \/>\nLength of the road parallel to the length of the park = 700 m<br \/>\nWidth of the road = 10 m<br \/>\n\u2234 Area of the road = l \u00d7 b<br \/>\n= 700 m \u00d7 10 m<br \/>\n= 7000 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Length of the road parallel to the breadth of the park = 300 m<br \/>\nWidth of the road = 10 m<br \/>\nArea of this road = l \u00d7 b<br \/>\n= 300 m \u00d7 10 = 3000 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Area of the both roads<br \/>\n= 7000 m<sup>2<\/sup>\u00a0+ 3000 m<sup>2<\/sup> \u2013 Area of the common portion<br \/>\n= 10,000 m<sup>2<\/sup> \u2013 10 m \u00d7 10 m<br \/>\n= 10,000 m<sup>2<\/sup>\u00a0\u2013 100 m<sup>2<br \/>\n<\/sup>= 9900 m<sup>2<\/sup> = 0.99 hectare<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Area of the park = l \u00d7 b<br \/>\n= 700 m \u00d7 300 m<br \/>\n= 210000 m<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Area of the park excluding the roads<br \/>\n= 210000 m<sup>2<\/sup>\u00a0\u2013 9900 m<sup>2<br \/>\n<\/sup>= 200100 m<sup>2<\/sup> = 20.01 hectare<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>7. Through 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<br \/>\n<\/strong><strong>(i) the area covered by the roads.<br \/>\n<\/strong><strong>(ii) the cost of constructing the roads at the rate of \u20b9 110 per m<sup>2<\/sup>.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2129\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q7.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"231\" height=\"125\" \/><br \/>\n<\/strong>Given that,<br \/>\nLength of the road along the length of the field = 90 m<br \/>\nBreadth = 3 m<br \/>\nArea of the field = length \u00d7 breadth<br \/>\n= 90 m \u00d7 3 m<br \/>\n= 270 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Similarly, the area of the road parallel to the breadth of the field = l \u00d7 b<br \/>\n= 60 m \u00d7 3 m<br \/>\n= 180 m<sup>2<\/sup>\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Area of the common portion<br \/>\n= 3m \u00d7 3m = 9m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(i)<\/strong> Area of the two roads = 270 m<sup>2<\/sup>\u00a0+ 180 m<sup>2<\/sup>\u00a0\u2013 9 m<sup>2<br \/>\n<\/sup>= 450 m<sup>2<\/sup>\u00a0\u2013 9 m<sup>2<\/sup><br \/>\n= 441 m<sup>2<\/sup><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii)<\/strong> Cost of constructing the roads = \u20b9 110 \u00d7 441<br \/>\n= \u20b9 48,510<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>8. Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) 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)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2130\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q8.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"157\" height=\"232\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Given that,<br \/>\nRadius of a circular pipe = 4 cm<br \/>\nSide of a square = 4 cm<br \/>\nPerimeter of the circular pipe = 2\u03c0r<br \/>\n= 2 \u00d7 3.14 \u00d7 4<br \/>\n= 25.12 cm<br \/>\nPerimeter of the square = 4 \u00d7 side of the square<br \/>\n= 4 \u00d7 4<br \/>\n= 16 cm<br \/>\nSo, the length of cord left with Pragya = Perimeter of circular pipe \u2013 Perimeter of square<br \/>\n= 25.12 \u2013 16<br \/>\n= 9.12 cm<br \/>\nYes, 9.12 cm cord is left.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>9. The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:<br \/>\n<\/strong><strong>(i) the area of the whole land<br \/>\n(ii) the area of the flower bed<br \/>\n<\/strong><strong>(iii) the area of the lawn excluding the area of the flower bed<br \/>\n<\/strong><strong>(iv) the circumference of the flower bed.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2131\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q9.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"283\" height=\"154\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>(i) <\/strong>Length of rectangular lawn = 10 m<br \/>\nBreadth of rectangular lawn = 5 m<br \/>\nArea of the rectangular lawn = Length \u00d7 Breadth<br \/>\n= 10 m \u00d7 5 m<br \/>\n= 50 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii) <\/strong>Radius of the flower bed = 2 m<br \/>\nArea of the flower bed = \u03c0r<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 2<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 4<br \/>\n= 12.56 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(iii) <\/strong>The area of the lawn excluding the area of the flower bed = Area of rectangular lawn \u2013 Area of flower bed<br \/>\n= 50 \u2013 12.56<br \/>\n= 37.44 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(iv) <\/strong>The circumference of the flower bed = 2\u03c0r<br \/>\n= 2 \u00d7 3.14 \u00d7 2<br \/>\n= 12.56 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>10. In the following figures, find the area of the shaded portions:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2132\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q10.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"968\" height=\"317\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q10.png 968w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q10-300x98.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q10-768x252.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q10-480x157.png 480w\" sizes=\"auto, (max-width: 968px) 100vw, 968px\" \/><br \/>\n<\/strong><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(i)<\/strong> To find the area of EFDC, first we have to find the area of \u0394AEF, \u0394EBC and rectangle ABCD<br \/>\nArea of \u0394AEF = \u00bd \u00d7 Base \u00d7 Height<br \/>\n= \u00bd \u00d7 6 \u00d7 10<br \/>\n= 1 \u00d7 3 \u00d7 10<br \/>\n= 30 cm<sup>2<br \/>\n<\/sup>Area of \u0394EBC = \u00bd \u00d7 Base \u00d7 Height<br \/>\n= \u00bd \u00d7 8 \u00d7 10<br \/>\n= 1 \u00d7 4 \u00d7 10<br \/>\n= 40 cm<sup>2<br \/>\n<\/sup>Area of rectangle ABCD = length \u00d7 breadth<br \/>\n= 18 \u00d7 10<br \/>\n= 180 cm<sup>2<br \/>\n<\/sup>Then,<br \/>\nArea of EFDC = ABCD area \u2013 (\u0394AEF + \u0394EBC)<br \/>\n= 180 \u2013 (30 + 40)<br \/>\n= 180 \u2013 70<br \/>\n= 110 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii) <\/strong>To find the area of \u0394QTU, first we have to find the area of \u0394STU, \u0394TPQ, \u0394QRU and square PQRS<br \/>\nArea of \u0394STU = \u00bd \u00d7 Base \u00d7 Height<br \/>\n= \u00bd \u00d7 10 \u00d7 10<br \/>\n= 1 \u00d7 5 \u00d7 10<br \/>\n= 50 cm<sup>2<br \/>\n<\/sup>Area of \u0394TPQ = \u00bd \u00d7 Base \u00d7 Height<br \/>\n= \u00bd \u00d7 10 \u00d7 20<br \/>\n= 1 \u00d7 5 \u00d7 20<br \/>\n= 100 cm<sup>2<br \/>\n<\/sup>Area of \u0394QRU = \u00bd \u00d7 Base \u00d7 Height<br \/>\n= \u00bd \u00d7 10 \u00d7 20<br \/>\n= 1 \u00d7 5 \u00d7 20<br \/>\n= 100 cm<sup>2<br \/>\n<\/sup>Area of square PQRS = Side<sup>2<br \/>\n<\/sup>= 20 \u00d7 20<br \/>\n= 400 cm<sup>2<br \/>\n<\/sup>Then,<br \/>\nArea of \u0394QTU = PQRS area \u2013 (\u0394STU + \u0394TPQ + \u0394QRU)<br \/>\n= 400 \u2013 (50 + 100 + 100)<br \/>\n= 400 \u2013 250<br \/>\n= 150 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>11. Find the area of the quadrilateral ABCD. <\/strong><strong>Here, AC = 22 cm, BM = 3 cm, <\/strong><strong>DN = 3 cm, and BM \u22a5 AC, DN \u22a5 AC<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2133\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q11.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"335\" height=\"320\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q11.png 335w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/11\/NCERT-Solutions-Class-7-Maths-Ex-11.4-Q11-300x287.png 300w\" sizes=\"auto, (max-width: 335px) 100vw, 335px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given that,<br \/>\nAC = 22 cm, BM = 3 cm DN = 3 cm and BM \u22a5 AC, DN \u22a5 AC<br \/>\nArea of \u0394ABC = \u00bd \u00d7 Base \u00d7 Height<br \/>\n= \u00bd \u00d7 22 \u00d7 3<br \/>\n= 1 \u00d7 11 \u00d7 3<br \/>\n= 33 cm<sup>2<br \/>\n<\/sup>Area of \u0394ADC = \u00bd \u00d7 Base \u00d7 Height<br \/>\n= \u00bd \u00d7 22 \u00d7 3<br \/>\n= 1 \u00d7 11 \u00d7 3<br \/>\n= 33 cm<sup>2<br \/>\n<\/sup>Area of the quadrilateral ABCD = Area of \u2206ABC + Area of \u2206ADC<br \/>\n= 33 cm<sup>2<\/sup>\u00a0+ 33 cm<sup>2<\/sup>\u00a0= 66 cm<sup>2<\/sup><br \/>\nHence, the required area = 66 cm<sup>2<\/sup>.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 7 Mathematics\u00a0 Chapter &#8211; 11 (Perimeter and Area) The NCERT Solutions in English Language for Class 7 Mathematics Chapter &#8211; 11 Perimeter and Area Exercise 11.4 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 11 Perimeter and Area NCERT Solution Class 7 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[219],"tags":[341,342,283,5],"class_list":["post-2095","post","type-post","status-publish","format-standard","hentry","category-class-7-maths","tag-ncert-class-7-maths-chapter-11-perimeter-and-area-in-english","tag-ncert-solutions-class-7-maths-chapter-11-in-english","tag-ncert-solutions-class-7-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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