{"id":3347,"date":"2022-12-11T11:31:38","date_gmt":"2022-12-11T11:31:38","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=3347"},"modified":"2023-01-31T06:51:18","modified_gmt":"2023-01-31T06:51:18","slug":"ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-11-mensuration-ex-11-2\/","title":{"rendered":"NCERT Solutions Class 8 Maths Chapter 11 Mensuration Ex 11.2"},"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 11.2 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-11-3\" 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; 11.2\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1.The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3367\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-1.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"409\" height=\"257\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-1.png 409w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-1-300x189.png 300w\" sizes=\"auto, (max-width: 409px) 100vw, 409px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>One parallel side of the trapezium\u00a0(a)\u00a0= 1 m<br \/>\nAnd second side\u00a0(b)\u00a0= 1.2 m and<br \/>\nheight\u00a0 (h) = 0.8 m<br \/>\nArea of top surface of the table=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 (a + b)h<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 (1 + 1.2)0.8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 2.2 \u00d7 0.8 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 0.88<br \/>\nArea of top surface of the table is 0.88 m<sup>2<\/sup>\u00a0.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. The area of a trapezium is 34\u00a0 cm<sup>2<\/sup>and the length of one of the parallel sides is 10 cm and its height is 4 cm Find the length of the other parallel side.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3366\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-2.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"296\" height=\"130\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<\/strong> Let the length of the other parallel side be\u00a0b.<br \/>\nLength of one parallel side, a = 10 cm<br \/>\nheight, (h) = 4 cm and<br \/>\nArea of a trapezium is 34 cm<sup>2<br \/>\n<\/sup>Formula for, Area of trapezium = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d7(a + b)h<br \/>\n34 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (10 + b)\u00d74 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">34 = 2\u00d7(10+b)<br \/>\n17 = (10+b)<br \/>\nb = 17 &#8211; 10<br \/>\nb = 7<br \/>\nHence another required parallel side is 7 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3368\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-3.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"296\" height=\"137\" \/><br \/>\n<\/strong>BC = 48 m,<br \/>\nCD = 17 m,<br \/>\nAD = 40 m<br \/>\nPerimeter = 120 m<br \/>\n\u2235 Perimeter of trapezium ABCD<br \/>\n= AB+BC+CD+DA<br \/>\n120 = AB + 48 + 17 + 40<br \/>\n120 = AB + 105<br \/>\nAB = 120 \u2013 105<br \/>\nAB = 15 m<br \/>\nNow, Area of the field=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7(BC+AD)\u00d7AB<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7(48 + 40) \u00d7 15<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 88 \u00d7 15<br \/>\n= 660<br \/>\nHence, area of the field ABCD is 660m<sup>2\u00a0<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3363\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-4.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"257\" height=\"179\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Answ-4.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"300\" height=\"200\" \/><br \/>\n<\/strong>From the diagram, the area for\u00a0quadrilateral will be the sum of the area of two triangles. <\/span><br \/>\n<span style=\"color: #000000;\">Area of quadrilateral ABCD =\u00a0Area of \u0394ABC + Area of \u0394ADC <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 (AC \u00d7 BE) + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d7 (AC \u00d7 FD)<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 AC (BE + FD)<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 24 m (13 m + 8 m)<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 24 m \u00d7 21 m = 252 m\u00b2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Given:<br \/>\nd1 = 7.5 cm<br \/>\nd2 = 12 cm<br \/>\nWe know that the area of a rhombus =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 d1 \u00d7 d2<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 7.5 \u00d7 12 = 45<br \/>\nTherefore, the area of the rhombus is 45 cm<sup>2<\/sup>\u00a0.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>Since a rhombus is also a kind of parallelogram,<br \/>\nThe formula for Area of rhombus = Base \u00d7 Altitude<br \/>\nPutting values, we have<br \/>\nArea of rhombus = 6 \u00d7 4 = 24<br \/>\nArea of rhombus is 24 cm<sup>2<br \/>\n<\/sup>Also, Formula for Area of rhombus = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 d<sub>1 <\/sub>d<sub>2<br \/>\n<\/sub>After substituting the values, we get<br \/>\n24 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 8 \u00d7 d<sub>2<br \/>\n<\/sub>d<sub>2<\/sub> = 6<br \/>\nHence, the length of the other diagonal is 6 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per\u00a0m<sup>2<\/sup>is Rs. 4.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Length of one diagonal,<br \/>\nd<sub>1<\/sub> = 45 cm<br \/>\nd<sub>2<\/sub>= 30 cm<br \/>\n\u2235 Area of one tile =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> d<sub>1 <\/sub>d<sub>2<br \/>\n<\/sub>=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 45 \u00d7 30<br \/>\n= 675<br \/>\nArea of one tile is 675\u00a0cm<sup>2<br \/>\n<\/sup>Area of 3000 tiles is<br \/>\n= 675 \u00d7 3000<br \/>\n= 2025000 cm<sup>2<br \/>\n<\/sup>= 2025000\/10000<br \/>\n= 202.50\u00a0m<sup>2\u00a0<\/sup>[\u2235 1m<sup>2<\/sup>\u00a0= 10000 cm<sup>2<\/sup>]<br \/>\n\u2235 Cost of polishing the floor per sq. meter = 4<br \/>\nCost of polishing the floor per 202.50 sq. meter = 4 \u00d7 202.50 = 810<br \/>\nHence the total cost of polishing the floor is Rs. 810.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m<sup>2<\/sup>\u00a0and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3364\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-8.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"346\" height=\"219\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-8.png 346w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-8-300x190.png 300w\" sizes=\"auto, (max-width: 346px) 100vw, 346px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Perpendicular distance\u00a0 (h) = 100 m<br \/>\nArea of the trapezium-shaped field = 10500 m<sup>2<\/sup><br \/>\nLet the side along the road be \u2018x\u2019 m and the side along the river =\u00a0 2x m<br \/>\nArea of the trapezium field =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 (a+b) \u00d7 h<br \/>\n10500 =\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 (x + 2x) \u00d7 100<br \/>\n10500 = 3x \u00d7 50<br \/>\n150x = 10500<br \/>\nx = 70<br \/>\nwhich means the side along the river is 70 m<br \/>\nHence, the side along the river =\u00a0 2x = 2(70) = 140 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3365\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-9.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"185\" height=\"153\" \/> <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Area of rectangular surface + 2 (Area of trapeziums)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 11 \u00d7 5 + 2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{(5+11)\\times&amp;space;4}{2}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\frac{(5+11)\\times 4}{2} \\right )\" align=\"absmiddle\" \/> m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 55 + 16 \u00d7 4 m<sup>2<br \/>\n<\/sup>= 55 + 64 m<sup>2<\/sup><br \/>\n= 119 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. There is a pentagonal shaped park as shown in the figure. <\/strong><strong>For finding its area Jyoti and Kavita divided it in two different ways.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3362\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-10.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"712\" height=\"191\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-10.png 712w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-10-300x80.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-10-480x129.png 480w\" sizes=\"auto, (max-width: 712px) 100vw, 712px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Find the area of this park using both ways. Can you suggest some other way of finding its area?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><strong>Jyoti\u2019s diagram,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3371\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-10-A.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"163\" height=\"200\" \/><br \/>\n<\/strong>Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP<br \/>\n=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (AP+BC)\u00d7CP + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d7(ED+AP)\u00d7DP<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>(30+15)\u00d7CP + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d7(15+30)\u00d7DP<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 (30+15)\u00d7(CP+DP)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d745\u00d7CD<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d745\u00d715<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 337.5 m<sup>2 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Area of pentagon is 337.5 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Kavita\u2019s diagram<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3372\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-10-B.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"163\" height=\"200\" \/><br \/>\n<\/strong>Here, a perpendicular AM is drawn to BE.<br \/>\nAM = 30 \u2013 15 = 15 m<br \/>\nArea of pentagon = Area of\u00a0 triangle ABE + Area of square BCDE (from above figure)<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 15 \u00d7 15 + (15\u00d715)<br \/>\n= 112.5+225.0<br \/>\n= 337.5<br \/>\nHence, the total area of pentagon-shaped park = 337.5 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11. Diagram of the adjacent picture frame has outer dimensions = 24 cm\u00d728 cm and inner dimensions 16 cm\u00d720 cm. Find the area of each section of the frame, if the width of each section is same.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3361\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Q-11.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"214\" height=\"229\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-3373\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Ans-11.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"296\" height=\"279\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Ans-11.png 429w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-11.2-Ans-11-300x283.png 300w\" sizes=\"auto, (max-width: 296px) 100vw, 296px\" \/><br \/>\n<\/strong> Given that the width of each section is same. <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, <\/span><br \/>\n<span style=\"color: #000000;\">IB = BJ = CK = CL = DM = DN = AP = AO\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;(1)<\/span><br \/>\n<span style=\"color: #000000;\">IL = IB + BC + CL <\/span><br \/>\n<span style=\"color: #000000;\">28 = IB + 20 + CL <\/span><br \/>\n<span style=\"color: #000000;\">28 \u2212 20 = IB + CL <\/span><br \/>\n<span style=\"color: #000000;\">8 = IB + CL <\/span><br \/>\n<span style=\"color: #000000;\">But, <\/span><br \/>\n<span style=\"color: #000000;\">IB = CL [From equation (1)]<\/span><br \/>\n<span style=\"color: #000000;\">Thus, 2IB = 8<\/span><br \/>\n<span style=\"color: #000000;\">IB = 4 cm <\/span><br \/>\n<span style=\"color: #000000;\">Hence IB = BJ = CK = CL = DM = DN = AP = AO = 4 cm <\/span><br \/>\n<span style=\"color: #000000;\">We know that, ABEH, CDGF, BEFC and ADGH are all trapezium<\/span><br \/>\n<span style=\"color: #000000;\">Area of trapezium ABEH = Area of trapezium CDGF\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d7 (AB + HE) \u00d7 IB<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> (16 + 24) \u00d7 4 = 80 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Area of trapezium BEFC = Area of trapezium ADGH [By symmetry]<\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d7 (BC + EF) \u00d7 BJ<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00d7 (20 + 28) \u00d7 4 = 96 cm\u00b2<\/span><\/p>\n<p><span style=\"color: #000000;\">Area of rectangle ABCD = BC \u00d7 DC <\/span><br \/>\n<span style=\"color: #000000;\">= 20 cm \u00d7 16 cm = 320 m\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the area of section ABEH and CDGF is 80 m\u00b2, the area of section BEFC and ADGH is 96 m\u00b2 and the area of section ABCD is 320 m\u00b2.<\/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 11.2 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-3347","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 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