{"id":2094,"date":"2022-11-09T05:19:30","date_gmt":"2022-11-09T05:19:30","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=2094"},"modified":"2022-11-09T05:19:30","modified_gmt":"2022-11-09T05:19:30","slug":"ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/","title":{"rendered":"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3"},"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 11.3 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-11-1\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 11.4<\/span><\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 11.3\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1. Find the circumference of the circle with the following radius: (Take \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<br \/>\n<\/strong><strong>(a) 14 cm\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (b) 28 cm\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(c) 21 cm<\/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>(a) 14 cm<br \/>\n<\/strong>Given, radius of circle = 14 cm<br \/>\nCircumference of the circle = 2\u03c0r<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 14<br \/>\n= 2 \u00d7 22 \u00d7 2<br \/>\n= 88 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(b) 28 cm<br \/>\n<\/strong>Given, radius of circle = 28 cm<br \/>\nCircumference of the circle = 2\u03c0r<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 28<br \/>\n= 2 \u00d7 22 \u00d7 4<br \/>\n= 176 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(c) 21 cm<br \/>\n<\/strong>Given, radius of circle = 21 cm<br \/>\nCircumference of the circle = 2\u03c0r<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 21<br \/>\n= 2 \u00d7 22 \u00d7 3<br \/>\n= 132 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. Find the area of the following circles, given that:<br \/>\n<\/strong><strong>(a) Radius = 14 mm (Take \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<br \/>\n(b) Diameter = 49 m<br \/>\n(c) Radius = 5 cm<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>(a) Radius = 14 mm (Take \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><br \/>\nGiven, radius of circle = 14 mm<br \/>\nThen,<br \/>\nArea of the circle = \u03c0r<sup>2<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 14<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 196<br \/>\n= 22 \u00d7 28<br \/>\n= 616 mm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(b) Diameter = 49 m<br \/>\n<\/strong>Given, diameter of circle (d) = 49 m<br \/>\nWe know that, radius (r) = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{d}{2}\" alt=\"\\frac{d}{2}\" align=\"absmiddle\" \/><br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{49}{2}\" alt=\"\\frac{49}{2}\" align=\"absmiddle\" \/> = 24.5 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Then,<br \/>\nArea of the circle = \u03c0r<sup>2<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 (24.5)<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 600.25<br \/>\n= 22 \u00d7 85.75<br \/>\n= 1886.5 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(c) Radius = 5 cm<br \/>\n<\/strong>Given, radius of circle = 5 cm<br \/>\nThen,<br \/>\nArea of the circle = \u03c0r<sup>2<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 25<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{550}{7}\" alt=\"\\frac{550}{7}\" align=\"absmiddle\" \/><br \/>\n= 78.57 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3.<\/strong>\u00a0<strong>If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>) <\/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 \/>\nCircumference of the circle = 154 m<br \/>\nThen,<br \/>\nWe know that, Circumference of the circle = 2\u03c0r<br \/>\n154 = 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 r<br \/>\n154 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{44}{7}\" alt=\"\\frac{44}{7}\" align=\"absmiddle\" \/> \u00d7 r<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">r = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{154\\times&amp;space;7}{44}\" alt=\"\\frac{154\\times 7}{44}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">r = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{14\\times&amp;space;7}{4}\" alt=\"\\frac{14\\times 7}{4}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">r = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7\\times&amp;space;7}{2}\" alt=\"\\frac{7\\times 7}{2}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">r = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{49}{2}\" alt=\"\\frac{49}{2}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">r = 24.5 m<br \/>\nNow,<br \/>\nArea of the circle = \u03c0r<sup>2<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 (24.5)<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 600.25<br \/>\n= 22 \u00d7 85.75<br \/>\n= 1886.5 m<sup>2<br \/>\n<\/sup>So, the radius of circle is 24.5 and area of circle is 1886.5\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs \u20b9 4 per meter. (Take \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-2111 alignnone\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-11.3-Q4.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"117\" height=\"114\" \/><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 \/>\nDiameter of the circular garden = 21 m<br \/>\nWe know that, radius (r) = d\/2<br \/>\n= 21\/2<br \/>\n= 10.5 m<br \/>\nThen,<br \/>\nCircumference of the circle = 2\u03c0r<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 10.5 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{462}{7}\" alt=\"\\frac{462}{7}\" align=\"absmiddle\" \/><br \/>\n= 66 m<br \/>\nSo, the length of rope required = 2 \u00d7 66 = 132 m<br \/>\nCost of 1 m rope = \u20b9 4<br \/>\nCost of 132 m rope = \u20b9 4 \u00d7 132<br \/>\n= \u20b9 528<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take \u03c0 = 3.14) <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>Give that,<br \/>\nRadius of circular sheet R = 4 cm<br \/>\nA circle of radius to be removed r = 3 cm<br \/>\nThen,<br \/>\nThe area of the remaining sheet = \u03c0R<sup>2\u00a0<\/sup>\u2013 \u03c0r<sup>2<br \/>\n<\/sup>= \u03c0 (R<sup>2<\/sup>\u00a0\u2013 r<sup>2<\/sup>)<br \/>\n= 3.14 (4<sup>2<\/sup>\u00a0\u2013 3<sup>2<\/sup>)<br \/>\n= 3.14 (16 \u2013 9)<br \/>\n= 3.14 \u00d7 7<br \/>\n= 21.98 cm<sup>2<br \/>\n<\/sup>So, the area of the remaining sheet is 21.98 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>6. Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs \u20b9 15. (Take \u03c0 = 3.14)<\/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 \/>\nDiameter of the circular table cover = 1.5 m<br \/>\nWe know that, radius (r) = d\/2<br \/>\n= 1.5\/2<br \/>\n= 0.75 m<br \/>\nThen,<br \/>\nCircumference of the circular table cover = 2\u03c0r<br \/>\n= 2 \u00d7 3.14 \u00d7 0.75<br \/>\n= 4.71 m<br \/>\nSo, the length of lace = 4.71 m<br \/>\nCost of 1 m lace = \u20b9 15<br \/>\nCost of 4.71 m lace = \u20b9 15 \u00d7 4.71<br \/>\n= \u20b9 70.65<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2112\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-11.3-Q7.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"237\" height=\"144\" \/><\/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 \/>\nDiameter of semi-circle = 10 cm<br \/>\nWe know that, radius (r) = d\/2<br \/>\n= 10\/2<br \/>\n= 5 cm<br \/>\nThen,<br \/>\nCircumference of the semi-circle = \u03c0r<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{110}{7}\" alt=\"\\frac{110}{7}\" align=\"absmiddle\" \/><br \/>\n= 15.71 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Now,<br \/>\nPerimeter of the given figure = Circumference of the semi-circle + semi-circle diameter<br \/>\n= 15.71 + 10<br \/>\n= 25.71 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is \u20b915\/m<sup>2<\/sup>. (Take \u03c0 = 3.14) <\/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 \/>\nDiameter of the circular table-top = 1.6 m<br \/>\nWe know that, radius (r) = d\/2<br \/>\n= 1.6\/2<br \/>\n= 0.8 m<br \/>\nThen,<br \/>\nArea of the circular table-top = \u03c0r<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 0.8<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 0.8 \u00d70.8<br \/>\n= 2.0096 m<sup>2<br \/>\n<\/sup>Cost of polishing 1 m<sup>2<\/sup> area = \u20b9 15<br \/>\nCost of polishing 2.0096 m<sup>2<\/sup> area = \u20b9 15 \u00d7 2.0096<br \/>\n= \u20b9 30.144<br \/>\nHence, the cost of polishing 2.0096 m<sup>2<\/sup>\u00a0area is \u20b9 30.144.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/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 \/>\nLength of wire that Shazli took =44 cm<br \/>\nThen,<br \/>\nIf the wire is bent into a circle,<br \/>\nWe know that, circumference of the circle = 2\u03c0r<br \/>\n44 = 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 r <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">44 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{44}{7}\" alt=\"\\frac{44}{7}\" align=\"absmiddle\" \/> \u00d7 r<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{44\\times&amp;space;7}{44}\" alt=\"\\frac{44\\times 7}{44}\" align=\"absmiddle\" \/>\u00a0 = r<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">r = 7 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Area of the circle = \u03c0r<sup>2<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7 \u00d77<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 22 \u00d7 7<br \/>\n= 154 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Now,<br \/>\nIf the wire is bent into a square,<br \/>\nLength of wire = perimeter of square<br \/>\n44 = 4 x side<br \/>\n44 = 4s<br \/>\ns = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{44}{4}\" alt=\"\\frac{44}{4}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">s = 11cm<br \/>\nArea of square = (side)<sup>2<\/sup>\u00a0=\u00a0\u00a011<sup>2<br \/>\n<\/sup>= 121 cm<sup>2<br \/>\n<\/sup>Therefore circle has more area than square<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2113\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-11.3-Q10.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"151\" height=\"148\" \/><\/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 \/>\nRadius of the circular card sheet = 14 cm<br \/>\nRadius of the two small circles = 3.5 cm<br \/>\nLength of the rectangle = 3 cm<br \/>\nBreadth of the rectangle = 1 cm<br \/>\nFirst we have to find out the area of circular card sheet, two circles and rectangle to find out the remaining area.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Now,<br \/>\nArea of the circular card sheet = \u03c0r<sup>2<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 14<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/>\u00d7 14 \u00d7 14<br \/>\n= 22 \u00d7 2 \u00d7 14<br \/>\n= 616 cm<sup>2<br \/>\n<\/sup>Area of the 2 small circles = 2 \u00d7 \u03c0r<sup>2<br \/>\n<\/sup>= 2 \u00d7 (<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 3.5<sup>2<\/sup>)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 2 \u00d7 (<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 3.5 \u00d7 3.5)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 2 \u00d7 (22\u00a0\u00d7 0.5 \u00d7 3.5)<br \/>\n= 2 \u00d7 38.5<br \/>\n= 77 cm<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Area of the rectangle = Length \u00d7 Breadth<br \/>\n= 3 \u00d7 1<br \/>\n= 3 cm<sup>2<br \/>\n<\/sup>Now,<br \/>\nArea of the remaining sheet = Area of circular card sheet \u2013 (Area of two small circles + Area of the rectangle)<br \/>\n= 616 \u2013 (77 + 3)<br \/>\n= 616 \u2013 80<br \/>\n= 536 cm<sup>2<br \/>\n<\/sup>Hence, the area of the remaining sheet is 536 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>11. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side <\/strong><strong>6 cm. What is the area of the left over aluminium sheet? (Take \u03c0 = 3.14)<\/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 circle = 2 cm<br \/>\nSquare sheet side = 6 cm<br \/>\nFirst we have to find out the area of square aluminium sheet and circle to find out the remaining area.<br \/>\nNow,<br \/>\nArea of the square = side<sup>2<br \/>\n<\/sup>Hence, the area of the square aluminium sheet = 6<sup>2<\/sup>\u00a0= 36 cm<sup>2<br \/>\n<\/sup>Area of the circle = \u03c0r<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 2<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 2 \u00d7 2 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 3.14 \u00d7 4<br \/>\n= 12.56 cm<sup>2<br \/>\n<\/sup>Now,<br \/>\nArea of aluminium sheet left = Area of the square aluminum sheet \u2013 Area of the circle<br \/>\n= 36 \u2013 12.56<br \/>\n= 23.44 cm<sup>2<br \/>\n<\/sup>Hence, the area of the aluminium sheet left is 23.44 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take \u03c0 = 3.14)<\/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 \/>\nCircumference of a circle = 31.4 cm<br \/>\nCircumference of a circle = 2\u03c0r<br \/>\n31.4 = 2 \u00d7 3.14 \u00d7 r<br \/>\n31.4 = 6.28 \u00d7 r<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{31.4}{6.28}\" alt=\"\\frac{31.4}{6.28}\" align=\"absmiddle\" \/> = r<br \/>\nr = 5 cm<br \/>\nThen,<br \/>\nArea of the circle = \u03c0r<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 (5cm)<sup>2<br \/>\n<\/sup>= 3. 14 \u00d7 25 cm<sup>2<br \/>\n<\/sup>= 78.5 cm<sup>2<br \/>\n<\/sup>Therefore, radius of the circle is 5 cm and area of the circle is 78.5 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (\u03c0 = 3.14)<\/strong><\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2114\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-11.3-Q13.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"129\" height=\"128\" \/><\/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 \/>\nDiameter of the flower bed = 66 m<br \/>\nThen,<br \/>\nRadius of the flower bed = d\/2<br \/>\n= 66\/2<br \/>\n= 33 m<br \/>\nArea of flower bed = \u03c0r<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 33<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 1089<br \/>\n= 3419.46 m<br \/>\nNow we have to find area of the flower bed and path together<br \/>\nSo, radius of flower bed and path together = 33 + 4 = 37 m<br \/>\nArea of the flower bed and path together = \u03c0r<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 37<sup>2<br \/>\n<\/sup>= 3.14 \u00d7 1369<br \/>\n= 4298.66 m<br \/>\nArea of the path = Area of the flower bed and path together \u2013 Area of flower bed<br \/>\n= 4298.66 \u2013 3419.46<br \/>\n= 879.2 m<sup>2<br \/>\n<\/sup>Hence, the area of the path is 879.2 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>14. A circular flower garden has an area of 314 m<sup>2<\/sup>. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take \u03c0 = 3.14)<\/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 \/>\nArea of the circular flower garden = 314 m<sup>2<br \/>\n<\/sup>Sprinkler at the centre of the garden can cover an area that has a radius = 12 m<br \/>\nArea of the circular flower garden = \u03c0r<sup>2<br \/>\n<\/sup>314 = 3.14 \u00d7 r<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{314}{3.14}\" alt=\"\\frac{314}{3.14}\" align=\"absmiddle\" \/> = r<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">r<sup>2<\/sup> = 100<br \/>\nr = \u221a100<br \/>\nr = 10 m<br \/>\n\u2234 Radius of the circular flower garden is 10 m.<br \/>\nSince, the sprinkler can cover an area of radius 12 m<br \/>\nHence, the sprinkler will water the whole garden.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take \u03c0 = 3.14)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2115\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-11.3-Q15.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"140\" height=\"133\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>Radius of inner circle = outer circle radius \u2013 10<br \/>\n= 19 \u2013 10<br \/>\n= 9 m<br \/>\nCircumference of the inner circle = 2\u03c0r<br \/>\n= 2 \u00d7 3.14 \u00d7 9<br \/>\n= 56.52 m<br \/>\nThen,<br \/>\nRadius of outer circle = 19 m<br \/>\nCircumference of the outer circle = 2\u03c0r<br \/>\n= 2 \u00d7 3.14 \u00d7 19<br \/>\n= 119.32 m<br \/>\nTherefore, the circumference of the inner circle is 56.52 m and the circumference of the outer circle is 119.32 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/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 \/>\nRadius of the wheel = 28 cm<br \/>\nTotal distance = 352 m = 35200 cm<br \/>\nCircumference of the wheel = 2\u03c0r<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 28<br \/>\n= 2 \u00d7 22 \u00d7 4<br \/>\n= 176 cm<br \/>\nNow we have to find the number of rotations of the wheel,<br \/>\nNumber of times the wheel should rotate = Total distance covered by wheel \/ Circumference of the wheel<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{352m}{176cm}\" alt=\"\\frac{352m}{176cm}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{35200cm}{176cm}\" alt=\"\\frac{35200cm}{176cm}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">= 200<br \/>\nHence, wheel rotates 200 times<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>17. The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take \u03c0 = 3.14)<\/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 \/>\nLength of the minute hand of the circular clock = 15 cm<br \/>\nThen,<br \/>\nDistance travelled by the tip of minute hand in 1 hour = circumference of the clock<br \/>\n= 2\u03c0r<br \/>\n= 2 \u00d7 3.14 \u00d7 15<br \/>\n= 94.2 cm<br \/>\nTherefore, the minute hand moves 94.2 cm in 1 hour<\/span><\/p>\n<p>&nbsp;<\/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 11.3 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-2094","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) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 7 Mathematics\u00a0Chapter - 11 (Perimeter and Area)The NCERT Solutions in English Language for Class 7 Mathematics Chapter - 11 Perimeter and Area Exercise 11.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.3\u00a0\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3\" \/>\n<meta property=\"og:description\" content=\"NCERT Solutions Class 7 Mathematics\u00a0Chapter - 11 (Perimeter and Area)The NCERT Solutions in English Language for Class 7 Mathematics Chapter - 11 Perimeter and Area Exercise 11.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.3\u00a0\" \/>\n<meta property=\"og:url\" content=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/\" \/>\n<meta property=\"og:site_name\" content=\"TheExamPillar NCERT\" \/>\n<meta property=\"article:published_time\" content=\"2022-11-09T05:19:30+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/latex.codecogs.com\/gif.latex?mathbffrac227\" \/>\n<meta name=\"author\" content=\"Admin\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@exampillar\" \/>\n<meta name=\"twitter:site\" content=\"@exampillar\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Admin\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"20 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/\"},\"author\":{\"name\":\"Admin\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"headline\":\"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3\",\"datePublished\":\"2022-11-09T05:19:30+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/\"},\"wordCount\":1439,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/latex.codecogs.com\\\/gif.latex?\\\\mathbf{\\\\frac{22}{7}}\",\"keywords\":[\"NCERT Class 7 Maths Chapter 11 Perimeter and Area in English\",\"NCERT Solutions Class 7 Maths Chapter 11 in english\",\"NCERT Solutions Class 7 Maths in English\",\"NCERT Solutions in English\"],\"articleSection\":[\"Class 7 Maths\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/\",\"name\":\"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 | TheExamPillar NCERT\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/#primaryimage\"},\"image\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/latex.codecogs.com\\\/gif.latex?\\\\mathbf{\\\\frac{22}{7}}\",\"datePublished\":\"2022-11-09T05:19:30+00:00\",\"description\":\"NCERT Solutions Class 7 Mathematics\u00a0Chapter - 11 (Perimeter and Area)The NCERT Solutions in English Language for Class 7 Mathematics Chapter - 11 Perimeter and Area Exercise 11.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.3\u00a0\",\"breadcrumb\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/#primaryimage\",\"url\":\"https:\\\/\\\/latex.codecogs.com\\\/gif.latex?\\\\mathbf{\\\\frac{22}{7}}\",\"contentUrl\":\"https:\\\/\\\/latex.codecogs.com\\\/gif.latex?\\\\mathbf{\\\\frac{22}{7}}\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\\\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#website\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/\",\"name\":\"TheExamPillar NCERT\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\"},\"alternateName\":\"NCERT Solution\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":[\"Person\",\"Organization\"],\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/#\\\/schema\\\/person\\\/521fbdbd2eb8621382a3096b5e3ecaf1\",\"name\":\"Admin\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"contentUrl\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\",\"width\":512,\"height\":512,\"caption\":\"Admin\"},\"logo\":{\"@id\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/wp-content\\\/uploads\\\/2022\\\/07\\\/cropped-ExamPillar-PNG.png\"},\"sameAs\":[\"https:\\\/\\\/theexampillar.com\\\/ncert\"],\"url\":\"https:\\\/\\\/theexampillar.com\\\/ncert\\\/author\\\/ncert_eng_vikram\\\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 | TheExamPillar NCERT","description":"NCERT Solutions Class 7 Mathematics\u00a0Chapter - 11 (Perimeter and Area)The NCERT Solutions in English Language for Class 7 Mathematics Chapter - 11 Perimeter and Area Exercise 11.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.3\u00a0","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3","og_description":"NCERT Solutions Class 7 Mathematics\u00a0Chapter - 11 (Perimeter and Area)The NCERT Solutions in English Language for Class 7 Mathematics Chapter - 11 Perimeter and Area Exercise 11.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.3\u00a0","og_url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/","og_site_name":"TheExamPillar NCERT","article_published_time":"2022-11-09T05:19:30+00:00","og_image":[{"url":"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}","type":"","width":"","height":""}],"author":"Admin","twitter_card":"summary_large_image","twitter_creator":"@exampillar","twitter_site":"@exampillar","twitter_misc":{"Written by":"Admin","Est. reading time":"20 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/#article","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/"},"author":{"name":"Admin","@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"headline":"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3","datePublished":"2022-11-09T05:19:30+00:00","mainEntityOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/"},"wordCount":1439,"commentCount":0,"publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/#primaryimage"},"thumbnailUrl":"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}","keywords":["NCERT Class 7 Maths Chapter 11 Perimeter and Area in English","NCERT Solutions Class 7 Maths Chapter 11 in english","NCERT Solutions Class 7 Maths in English","NCERT Solutions in English"],"articleSection":["Class 7 Maths"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/","url":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/","name":"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 | TheExamPillar NCERT","isPartOf":{"@id":"https:\/\/theexampillar.com\/ncert\/#website"},"primaryImageOfPage":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/#primaryimage"},"image":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/#primaryimage"},"thumbnailUrl":"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}","datePublished":"2022-11-09T05:19:30+00:00","description":"NCERT Solutions Class 7 Mathematics\u00a0Chapter - 11 (Perimeter and Area)The NCERT Solutions in English Language for Class 7 Mathematics Chapter - 11 Perimeter and Area Exercise 11.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Exercise - 11.3\u00a0","breadcrumb":{"@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/#primaryimage","url":"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}","contentUrl":"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}"},{"@type":"BreadcrumbList","@id":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-11-perimeter-and-area-ex-11-3\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/theexampillar.com\/ncert\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3"}]},{"@type":"WebSite","@id":"https:\/\/theexampillar.com\/ncert\/#website","url":"https:\/\/theexampillar.com\/ncert\/","name":"TheExamPillar NCERT","description":"","publisher":{"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1"},"alternateName":"NCERT Solution","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/theexampillar.com\/ncert\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":["Person","Organization"],"@id":"https:\/\/theexampillar.com\/ncert\/#\/schema\/person\/521fbdbd2eb8621382a3096b5e3ecaf1","name":"Admin","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","url":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","contentUrl":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png","width":512,"height":512,"caption":"Admin"},"logo":{"@id":"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/cropped-ExamPillar-PNG.png"},"sameAs":["https:\/\/theexampillar.com\/ncert"],"url":"https:\/\/theexampillar.com\/ncert\/author\/ncert_eng_vikram\/"}]}},"_links":{"self":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/2094","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/comments?post=2094"}],"version-history":[{"count":3,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/2094\/revisions"}],"predecessor-version":[{"id":2509,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/posts\/2094\/revisions\/2509"}],"wp:attachment":[{"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/media?parent=2094"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/categories?post=2094"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/theexampillar.com\/ncert\/wp-json\/wp\/v2\/tags?post=2094"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}