{"id":3348,"date":"2022-12-11T11:31:56","date_gmt":"2022-12-11T11:31:56","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=3348"},"modified":"2023-01-31T06:51:13","modified_gmt":"2023-01-31T06:51:13","slug":"ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-3\/","title":{"rendered":"NCERT Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.3"},"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 11.3 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-11-1\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 8 Maths Ex &#8211; 11.1<\/a><\/li>\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-11-4\" 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; 11.3\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3378\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-1.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"402\" height=\"158\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-1.png 402w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-1-300x118.png 300w\" sizes=\"auto, (max-width: 402px) 100vw, 402px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(a)<\/strong> Length of cuboidal box\u00a0 (l)\u00a0= 60 cm<br \/>\nBreadth of cuboidal box\u00a0 (b) \u00a0= 40 cm<br \/>\nHeight of cuboidal box\u00a0 (h) \u00a0= 50 cm<br \/>\nTotal surface area of cuboidal box =\u00a0 2 \u00d7 (lb + bh + hl)<br \/>\n= 2\u00d7(60\u00d740+40\u00d750+50\u00d760)<br \/>\n= 2\u00d7(2400+2000+3000)<br \/>\n= 14800\u00a0cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(b)\u00a0<\/strong> Length of cubical box (l) = 50 cm<br \/>\nBreadth of cubicalbox (b) = 50 cm<br \/>\nHeight of cubicalbox (h) = 50 cm<br \/>\nTotal surface area of cubical box = 6(side)<sup>2<br \/>\n<\/sup>= 6(50\u00d750)<br \/>\n= 6\u00d72500<br \/>\n= 15000<br \/>\nSurface area of the cubical box is 15000 cm<sup>2<br \/>\n<\/sup>From the result of (a) and (b), cuboidal box\u00a0requires the lesser amount of material to make.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. A suitcase with measures 80 cm\u00a0x\u00a048 cm\u00a0x\u00a024 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Length of suitcase box, l = 80 cm,<br \/>\nBreadth of suitcase box, b= 48 cm<br \/>\nAnd Height of cuboidal box , h = 24 cm<br \/>\nTotal surface area of suitcase box = 2(lb+bh+hl)<br \/>\n= 2(80\u00d748+48\u00d724+24\u00d780)<br \/>\n= 2 (3840+1152+1920)<br \/>\n= 2\u00d76912<br \/>\n= 13824<br \/>\nTotal surface area of suitcase box is 13824\u00a0cm<sup>2<br \/>\n<\/sup>Area of Tarpaulin cloth = Surface area of suitcase<br \/>\nl\u00d7b = 13824<br \/>\nl \u00d796 = 13824<br \/>\nl = 144<br \/>\nRequired tarpaulin for 100 suitcases = 144\u00d7100 = 14400 cm = 144 m<br \/>\nHence tarpaulin cloth required to cover 100 suitcases is 144 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Find the side of a cube whose surface area is 600 cm<sup>2<\/sup> .<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Surface area of cube = 600 cm<sup>2<\/sup>\u00a0(Given)<br \/>\nFormula for surface area of a cube = 6(side)<sup>2<br \/>\n<\/sup>Substituting the values, we get<br \/>\n6(side)<sup>2<\/sup> = 600<br \/>\n(side)<sup>2<\/sup> = 100<br \/>\nOr side = \u00b110<br \/>\nSince side cannot be negative, the measure of each side of a cube is 10 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Rukshar painted the outside of the cabinet of measure 1 m \u00d7 2 m \u00d7 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3379\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-4.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"266\" height=\"143\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Length of cabinet, l\u00a0= 2 m,<br \/>\nBreadth of cabinet, b = 1 m<br \/>\nHeight of cabinet, h \u00a0= 1.5 m<br \/>\nSurface area of cabinet =\u00a0 lb+2(bh+hl )<br \/>\n= 2\u00d71+2(1\u00d71.5+1.5\u00d72)<br \/>\n= 2+2(1.5+3.0)<br \/>\n= 2+9.0<br \/>\n= 11<br \/>\nRequired surface area of cabinet is 11m<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Daniel is paining the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100\u00a0m^2of area is painted. How many cans of paint will she need to paint the room?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Length of wall, l \u00a0= 15 m,<br \/>\nBreadth of wall, b = 10 m<br \/>\nHeight of wall, h \u00a0= 7 m<br \/>\nTotal Surface area of classroom = lb+2(bh+hl )<br \/>\n= 15\u00d710+2(10\u00d77+7\u00d715)<br \/>\n= 150+2(70+105)<br \/>\n= 150+350<br \/>\n= 500<br \/>\nNow, Required number of cans =\u00a0 Area of hall\/Area of one can<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{500}{100}\" alt=\"\\frac{500}{100}\" align=\"absmiddle\" \/><br \/>\n= 5<br \/>\nTherefore, 5 cans are required to paint the room.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Describe how the two figures below are alike and how they are different. Which box has larger lateral surface areas?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3376\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-6.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"325\" height=\"158\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-6.png 325w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-6-300x146.png 300w\" sizes=\"auto, (max-width: 325px) 100vw, 325px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Diameter of cylinder = 7 cm<br \/>\nRadius of cylinder, r \u00a0=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/> cm<br \/>\nHeight of cylinder, h = 7 cm<br \/>\nLateral surface area of cylinder = 2\u03c0rh<br \/>\n=\u00a0 2\u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/> \u00d7 7<br \/>\n= 154<br \/>\nSo, Lateral surface area of cylinder is 154 cm<sup>2<br \/>\n<\/sup>Now, lateral surface area of cube =\u00a0 4 (side)<sup>2<br \/>\n<\/sup>= 4\u00d77<sup>2<\/sup><br \/>\n= 4\u00d749<br \/>\n= 196<br \/>\nLateral surface area of cube is 196 cm<sup>2<br \/>\n<\/sup>Hence, the cube has a larger lateral surface area.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>Radius of cylindrical tank, r \u00a0= 7 m<br \/>\nHeight of cylindrical tank\u00a0, h\u00a0= 3 m<br \/>\nTotal surface area of cylindrical tank = 2\u03c0r(h+r)<br \/>\n=\u00a0 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7 (3 + 7)<br \/>\n= 44 \u00d7 10<br \/>\n= 440<br \/>\nTherefore, 440\u00a0 m<sup>2<\/sup>\u00a0metal sheet is required.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. The lateral surface area of a hollow cylinder is 4224cm<sup>2<\/sup>. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Lateral surface area of a hollow cylinder = 4224 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Width of rectangular sheet = 33 cm<\/span><br \/>\n<span style=\"color: #000000;\">Let length of the rectangular sheet = l <\/span><br \/>\n<span style=\"color: #000000;\">Lateral\u00a0surface area of cylinder\u00a0=\u00a0Area of rectangular\u00a0sheet<\/span><br \/>\n<span style=\"color: #000000;\">4224\u00a0cm\u00b2 = b\u00a0\u00d7 l<\/span><br \/>\n<span style=\"color: #000000;\">4224 cm\u00b2 = 33 cm \u00d7 l <\/span><br \/>\n<span style=\"color: #000000;\">l = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4224}{33}\" alt=\"\\frac{4224}{33}\" align=\"absmiddle\" \/> cm = 128 cm <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the length of the rectangular sheet be 128 cm<\/span><\/p>\n<p><span style=\"color: #000000;\">Perimeter of the rectangular\u00a0sheet = 2 \u00d7 (l\u00a0+ b)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 (128 cm +\u00a033 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 161 cm = 322 cm <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the perimeter of the rectangular sheet is 322 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length 1 m.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3377\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-9.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"250\" height=\"220\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Diameter of road roller, d = 84 cm<br \/>\nRadius of road roller, r = d\/2 = 84\/2 = 42 cm<br \/>\nLength of road roller, h\u00a0= 1 m = 100 cm<br \/>\nFormula for Curved surface area of road roller = 2\u03c0rh<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 42 \u00d7 100<br \/>\n= 26400<br \/>\nCurved surface area of the road roller is 26400 cm<sup>2<br \/>\n<\/sup>Again, Area covered by the road roller in 750 revolutions = 26400\u00d7750cm<sup>2<br \/>\n<\/sup>= 1,98,00,000cm<sup>2<br \/>\n<\/sup>= 1980 m<sup>2<\/sup>\u00a0\u00a0[\u2235\u00a01\u00a0m<sup>2<\/sup>= 10,000 cm<sup>2<\/sup>]<br \/>\nHence the area of the road is 1980 m<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in figure). If the label is placed 2 cm from top and bottom, what is the area of the label?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3375\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.3-Q-10.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"237\" height=\"191\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Diameter of the cylindrical container , d = 14 cm<br \/>\nRadius of cylindrical container, r = d\/2 = 14\/2 \u00a0= 7 cm<br \/>\nHeight of cylindrical container = 20 cm<br \/>\nHeight of the label, say h = 20\u20132\u20132 (from the figure) = 16 cm<br \/>\nCurved surface area of label = 2\u03c0rh<br \/>\n=\u00a0 2\u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7 \u00d7 16<br \/>\n= 704<br \/>\nHence, the area of the label is 704 cm<sup>2<\/sup>.<\/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: 14pt;\"><strong>NCERT Class 8<sup>th\u00a0<\/sup>Solution\u00a0<\/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-english\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 English\u201d (Edit)\">NCERT Solutions Class 8 English<\/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-hindi\/\" aria-label=\"\u201cNCERT Solutions Class 6 Maths\u201d (Edit)\">NCERT Solutions Class 8 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 11.3 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; 11.1 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-3348","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 11.3 | 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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