{"id":3346,"date":"2022-12-11T11:31:26","date_gmt":"2022-12-11T11:31:26","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=3346"},"modified":"2023-01-31T06:51:22","modified_gmt":"2023-01-31T06:51:22","slug":"ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-1\/","title":{"rendered":"NCERT Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 8 Mathematics\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 11 (Mensuration)\u00a0<\/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 8 Mathematics <strong>Chapter &#8211; 11 Mensuration <\/strong>Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<h4><span style=\"color: #000000;\"><strong>Chapter 11: Mensuration<\/strong><\/span><\/h4>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-2\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 8 Maths Ex &#8211; 11.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-3\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 8 Maths Ex &#8211; 11.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-4\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 8 Maths Ex &#8211; 11.4<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 11.1\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3355\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.1-Q-1.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"293\" height=\"153\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Side of a square = 60 m<br \/>\nLength of the rectangle = 80m<\/span><br \/>\n<span style=\"color: #000000;\">Perimeter of the square\u00a0= 4 \u00d7 (side of square) <\/span><br \/>\n<span style=\"color: #000000;\">= 4 \u00d7 60m <\/span><br \/>\n<span style=\"color: #000000;\">= 240m<\/span><br \/>\n<span style=\"color: #000000;\">Perimeter of rectangle\u00a0= 2 \u00d7 (length + breadth)<\/span><br \/>\n<span style=\"color: #000000;\">Perimeter of square = Perimeter of rectangle <\/span><br \/>\n<span style=\"color: #000000;\">240 = 2 (length + breadth) <\/span><br \/>\n<span style=\"color: #000000;\">240 = 2 (80 + breadth) <\/span><br \/>\n<span style=\"color: #000000;\">240 = 160 + 2 \u00d7 breadth <\/span><br \/>\n<span style=\"color: #000000;\">240 &#8211; 160 = 2 \u00d7 breadth <\/span><br \/>\n<span style=\"color: #000000;\">80 = 2 \u00d7 breadth<\/span><br \/>\n<span style=\"color: #000000;\">breadth = 40 m<\/span><br \/>\n<span style=\"color: #000000;\">Area of the square = side \u00d7 side = 60 \u00d7 60 = 3600 m<sup>2<br \/>\n<\/sup>Area of the rectangular field = length \u00d7\u00a0breadth = 80 \u00d7 40 = 3200 m<sup>2<br \/>\n<\/sup>Thus, area of square is larger than the area of rectangular field.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Mrs.Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs. 55 per m<sup>2<\/sup>.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3359\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.1-Q-2.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"281\" height=\"260\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Side of the square plot = 25 m<br \/>\nArea of the square plot =\u00a0 square of the side = (side)<sup>2<br \/>\n<\/sup>=\u00a0(25)<sup>2<\/sup><br \/>\n= 625<br \/>\nTherefore the area of the square plot is 625 m<sup>2<br \/>\n<\/sup>Length of the house = 20 m and<br \/>\nThe breadth of the house = 15 m<br \/>\nArea of the house = length\u00d7breadth<br \/>\n= 20 \u00d7 15<br \/>\n= 300 m<sup>2<br \/>\n<\/sup>Area of the garden = Area of the square plot \u2013 Area of the house<br \/>\n= 625\u2013300 = 325 m<sup>2<br \/>\n<\/sup>\u2235 The cost of developing the garden per sq. m is Rs. 55<br \/>\nThe cost of developing the garden 325 sq. m = Rs. 55 \u00d7 325<br \/>\n= Rs. 17,875<br \/>\nHence the total cost of developing a garden is around Rs. 17,875.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 \u2013 (3.5 + 3.5 meters]<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3356\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.1-Q-3.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"295\" height=\"100\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Total length = 20 m<br \/>\nDiameter of the semi-circle = 7 m<br \/>\nRadius of the semi-circle = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/> = 3.5 m<br \/>\nLength of the rectangular field = 20 &#8211; (3.5 + 3.5) = 20 &#8211; 7 = 13 m<br \/>\nBreadth of the rectangular field = 7 m<br \/>\nArea of rectangular field = l\u00d7b<br \/>\n= 13 \u00d7 7= 91m<sup>2<br \/>\n<\/sup>Area of the two semi-circles = 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 \u03c0 \u00d7 r<sup>2<br \/>\n<\/sup>= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \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<br \/>\n= 38.5 m<sup>2<br \/>\n<\/sup>Area of garden = 91 + 38.5<br \/>\n= 129.5 m<sup>2<br \/>\n<\/sup>Now, the perimeter of the two semi-circles =\u00a0 2\u03c0r = 2\u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 3.5<br \/>\n= 22 m<br \/>\nAnd the perimeter of the garden = 22 + 13 + 13 = 48 m.<br \/>\nThus, the area of the garden is 129.5 m\u00b2, and the perimeter of the garden is 48 m.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m<sup>2<\/sup>? [If required you can split the tiles in whatever way you want to fill up the corners]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Base of flooring tile = 24 cm = 0.24 m<br \/>\nCorresponding height of a flooring tile= 10 cm = 0.10 m<br \/>\nNow Area of flooring tile= Base \u00d7 Altitude<br \/>\n= 0.24 \u00d7 0.10<br \/>\n= 0.024<br \/>\nArea of flooring tile is 0.024m<sup>2<br \/>\n<\/sup>Number of tiles required to cover the floor = Area of floor\/Area of one tile <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1080}{0.024}\" alt=\"\\frac{1080}{0.024}\" align=\"absmiddle\" \/> = 45000 tiles<br \/>\nHence 45000 tiles are required to cover the floor.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression\u00a0C = 2\u03c0r ,where\u00a0 r \u00a0is the radius of the circle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3358\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.1-Q-5.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"795\" height=\"183\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.1-Q-5.png 795w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.1-Q-5-300x69.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.1-Q-5-768x177.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.1-Q-5-480x110.png 480w\" sizes=\"auto, (max-width: 795px) 100vw, 795px\" \/><\/strong><strong>Solution &#8211;<br \/>\n<\/strong><strong>(a)\u00a0<\/strong>Radius =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{Diameter}{2}\" alt=\"\\frac{Diameter}{2}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2.8}{2}\" alt=\"\\frac{2.8}{2}\" align=\"absmiddle\" \/> cm = 1.4 cm<br \/>\nCircumference of semi-circle =\u00a0 \u03c0r<br \/>\n=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 1.4 = 4.4<br \/>\nCircumference of the semi-circle is 4.4 cm<br \/>\nTotal distance covered by the ant = Circumference of semi-circle + Diameter<br \/>\n= 4.4 + 2.8<br \/>\n= 7.2 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(b)<\/strong> Diameter of semi-circle = 2.8 cm<br \/>\nRadius =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{Diameter}{2}\" alt=\"\\frac{Diameter}{2}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2.8}{2}\" alt=\"\\frac{2.8}{2}\" align=\"absmiddle\" \/> cm = 1.4 cm<br \/>\nCircumference of semi-circle = r<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 1.4 = 4.4 cm<br \/>\nTotal distance covered by the ant = 1.5 + 2.8 + 1.5 + 4.4<br \/>\n= 10.2 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(c)\u00a0<\/strong>Diameter of semi-circle = 2.8 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Radius =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{Diameter}{2}\" alt=\"\\frac{Diameter}{2}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2.8}{2}\" alt=\"\\frac{2.8}{2}\" align=\"absmiddle\" \/> cm = 1.4 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Circumference of semi-circle =\u00a0 \u03c0r<br \/>\n=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 1.4<br \/>\n= 4.4 cm<br \/>\nTotal distance covered by the ant = 2 + 2 + 4.4 = 8.4 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Thus, the ant will have to take a longer round for food piece in (b) because the perimeter of the figure given in (b) is the greatest among all<\/span><\/p>\n<p>&nbsp;<\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr style=\"height: 32px;\">\n<td style=\"width: 100%; background-color: #f2e079; text-align: center; height: 32px;\"><span style=\"color: #ff0000; font-size: 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Hindi<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 100%; text-align: left; height: 28px;\"><span style=\"font-size: 14pt;\"><strong><span style=\"font-family: georgia, palatino, serif; font-size: 12pt;\"><a class=\"row-title\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Maths\u201d (Edit)\">NCERT Solutions Class 8 Mathematics<\/a>\u00a0<\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 100%; text-align: left; height: 28px;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-sanskrit-ruchira\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-size: 14pt;\"><strong><span style=\"font-family: georgia, palatino, serif; font-size: 12pt;\">NCERT Solutions Class 8 Sanskrit<\/span><\/strong><\/span><\/a><\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 100%; text-align: left; height: 28px;\"><span style=\"font-size: 14pt;\"><strong><span style=\"font-family: georgia, palatino, serif; font-size: 12pt;\"><a class=\"row-title\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-science\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Social Science\u201d (Edit)\">NCERT Solutions Class 8 Science<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 100%; text-align: left; height: 28px;\"><span style=\"font-size: 14pt;\"><strong><span style=\"font-family: georgia, palatino, serif; font-size: 12pt;\"><a class=\"row-title\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-social-science\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Social Science\u201d (Edit)\">NCERT Solutions Class 8 Social Science<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 8 Mathematics\u00a0 Chapter &#8211; 11 (Mensuration)\u00a0 The NCERT Solutions in English Language for Class 8 Mathematics Chapter &#8211; 11 Mensuration Exercise 11.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 11: Mensuration NCERT Solution Class 8 Maths Ex &#8211; 11.2 NCERT Solution Class 8 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[442],"tags":[464,465,445,5],"class_list":["post-3346","post","type-post","status-publish","format-standard","hentry","category-class-8-maths","tag-ncert-class-8-maths-chapter-11-mensuration-in-english","tag-ncert-solutions-class-8-maths-chapter-11-in-english","tag-ncert-solutions-class-8-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.1 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 8 Mathematics\u00a0Chapter - 11 (Mensuration)\u00a0The NCERT Solutions in English Language for Class 8 Mathematics Chapter - 11 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