{"id":4578,"date":"2023-02-19T05:45:10","date_gmt":"2023-02-19T05:45:10","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4578"},"modified":"2023-02-19T05:46:14","modified_gmt":"2023-02-19T05:46:14","slug":"ncert-solutions-class-9-maths-chapter-12-herons-formula-ex-12-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-12-herons-formula-ex-12-2\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 12 Heron&#8217;s Formula Ex 12.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 12 (Heron\u2019s Formula)\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 9 Mathematics\u00a0<strong>Chapter &#8211; 12 Heron\u2019s Formula <\/strong>Exercise 12.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 12: Heron\u2019s Formula<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-12-herons-formula-ex-12-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 12.1<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 12.2<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. A park, in the shape of a quadrilateral ABCD, has C = 90\u00b0, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>First, construct a quadrilateral ABCD and join BD.<br \/>\nWe know that<br \/>\nC = 90\u00b0, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m<br \/>\nThe diagram is:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4634\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"326\" height=\"267\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans1.png 359w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans1-300x246.png 300w\" sizes=\"auto, (max-width: 326px) 100vw, 326px\" \/><br \/>\nNow, apply Pythagoras theorem in \u0394BCD<br \/>\nBD<sup>2<\/sup>\u00a0= BC<sup>2\u00a0<\/sup>+CD<sup>2<br \/>\n<\/sup>\u21d2 BD<sup>2<\/sup>\u00a0= 12<sup>2<\/sup>+5<sup>2<br \/>\n<\/sup>\u21d2 BD<sup>2<\/sup>\u00a0= 169<br \/>\n\u21d2 BD = 13 m<br \/>\nNow, the area of \u0394BCD = (\u00bd \u00d7 12 \u00d7 5) = 30 m<sup>2<br \/>\n<\/sup>The semi perimeter of \u0394ABD<br \/>\n(s) = (perimeter\/2)<br \/>\n= (8 + 9 + 13)\/2 m<br \/>\n= 30\/2 m<br \/>\n= 15 m<br \/>\nUsing Heron\u2019s formula,<br \/>\nArea of \u0394ABD = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a15(15 &#8211; 9)(15 &#8211; 8)(15 &#8211; 13)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a15 \u00d7 6 \u00d7 7 \u00d7 2<\/span><br \/>\n<span style=\"color: #000000;\">= 6\u221a35<\/span><br \/>\n<span style=\"color: #000000;\">=\u00a035.5 m<sup>2<\/sup>\u00a0(approx.)<\/span><br \/>\n<span style=\"color: #000000;\">Area of \u0394ABD = 35.5 m<sup>2<br \/>\n<\/sup>\u2234 The area of quadrilateral ABCD = Area of \u0394BCD+Area of \u0394ABD<br \/>\n= 30 m<sup>2<\/sup>+35.5m<sup>2<\/sup>\u00a0= 65.5 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>First, construct a diagram with the given parameter.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4635\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"335\" height=\"329\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans2.png 335w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans2-300x295.png 300w\" sizes=\"auto, (max-width: 335px) 100vw, 335px\" \/><br \/>\nFor \u2206ABC, consider<\/span><br \/>\n<span style=\"color: #000000;\">AC<sup>2 <\/sup>= AB<sup>2<\/sup>\u00a0+ BC<sup>2<\/sup><\/span><br \/>\n<span style=\"color: #000000;\">AC<sup>2 <\/sup>= 3<sup>2<\/sup>\u00a0+\u00a04<sup>2\u00a0<\/sup>= 25<\/span><br \/>\n<span style=\"color: #000000;\">AC = 5<\/span><br \/>\n<span style=\"color: #000000;\">Thus, it can be concluded that \u0394ABC\u00a0is a right angled at B.<br \/>\nSo, area of \u0394ABC\u00a0= (\u00bd \u00d7 3 \u00d7 4) = 6 cm<sup>2<br \/>\n<\/sup>Now, In \u2206ADC<br \/>\nThe semi perimeter of \u0394ADC (s) = (perimeter\/2) = (5 + 5 + 4)\/2 cm = 14\/2 cm = 7 m<br \/>\nNow, using Heron\u2019s formula,<br \/>\nArea of \u0394ADC = \u221a[s(s &#8211; a)(s &#8211; b)(s &#8211; c)]<br \/>\n= \u221a[7(7 &#8211; 5)(7 &#8211; 4)(7 &#8211; 5)]<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a[7 \u00d7 2 \u00d7 3 \u00d7 2]<\/span><br \/>\n<span style=\"color: #000000;\">= 2\u221a21\u00a0cm<sup>2<br \/>\n<\/sup>Area of \u0394ADC = 9.2 cm<sup>2<\/sup>\u00a0(approx.)<\/span><br \/>\n<span style=\"color: #000000;\">Area of the quadrilateral ABCD = Area of \u0394ADC + Area of \u0394ABC<\/span><br \/>\n<span style=\"color: #000000;\">= 9.2 cm<sup>2<\/sup>\u00a0+ 6 cm<sup>2<br \/>\n<\/sup>= 15.2 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4636\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"448\" height=\"473\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q3.png 609w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q3-284x300.png 284w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q3-480x507.png 480w\" sizes=\"auto, (max-width: 448px) 100vw, 448px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong><strong>For For the triangle marked as I<\/strong><br \/>\nIt is an isosceles triangle, therefore\u00a0a = 5 cm, b = 5 cm, c = 1 cm<\/span><br \/>\n<span style=\"color: #000000;\">Semi Perimeter: (s) = (Perimeter\/2)<\/span><br \/>\n<span style=\"color: #000000;\">s = (a + b + c)\/2<\/span><br \/>\n<span style=\"color: #000000;\">= (5 + 5 + 1)\/2<\/span><br \/>\n<span style=\"color: #000000;\">= 11\/2<\/span><br \/>\n<span style=\"color: #000000;\">= 5.5 cm<\/span><br \/>\n<span style=\"color: #000000;\">By using\u00a0Heron\u2019s formula,<\/span><br \/>\n<span style=\"color: #000000;\">Area of triangle\u00a0marked I = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a5.5 (5.5 &#8211; 5) (5.5 &#8211; 5) (5.5 &#8211; 1)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a5.5 \u00d7 0.5 \u00d7 0.5 \u00d7 4.5<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a6.1875 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 2.5 cm<sup>2<\/sup>\u00a0(approx.)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>For the rectangle marked as II:<\/strong><\/span><br \/>\n<span style=\"color: #000000;\">The measures of the sides are 6.5 cm, 1 cm,\u00a0 6.5 cm, and 1 cm.<\/span><br \/>\n<span style=\"color: #000000;\">Area of rectangle = length \u00d7 breadth<\/span><br \/>\n<span style=\"color: #000000;\">= 6.5 cm \u00d7 1 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 6.5 cm<sup>2<\/sup><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>For the trapezium marked as III<br \/>\n<\/strong>It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the adjoining figure. Its height is given by<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4638\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"214\" height=\"151\" \/><\/span><br \/>\n<span style=\"color: #000000;\">h = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{1^2&amp;space;-\\left&amp;space;(&amp;space;\\frac{1}{2}&amp;space;\\right&amp;space;)^2}\" alt=\"\\sqrt{1^2 -\\left ( \\frac{1}{2} \\right )^2}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{1-\\frac{1}{4}}\" alt=\"\\sqrt{1-\\frac{1}{4}}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\frac{3}{4}}&amp;space;=&amp;space;\\frac{\\sqrt{3}}{2}\" alt=\"\\sqrt{\\frac{3}{4}} = \\frac{\\sqrt{3}}{2}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><b><br \/>\nNote:<\/b>\u00a0The perpendicular distance between the parallel sides is called the height of the trapezium.<\/span><br \/>\n<span style=\"color: #000000;\">Area of trapezium = 1\/2 \u00d7 sum of parallel sides \u00d7 distance between them<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/2 \u00d7 (2 + 1) \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{\\sqrt{3}}{2}\" alt=\"\\frac{\\sqrt{3}}{2}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">= 1\/2 \u00d7 3 \u00d7 0.9<\/span><br \/>\n<span style=\"color: #000000;\">= 1.4 cm<sup>2<\/sup>\u00a0(approx.)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>For the triangle marked as IV and V<br \/>\n<\/strong>Triangles IV and V are congruent right-angled triangles with base 6 cm and height 1.5 cm.<\/span><br \/>\n<span style=\"color: #000000;\">Area of the triangle\u00a0= 1\/2 \u00d7 base \u00d7 height<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/2 \u00d7 6 \u00d7 1.5<\/span><br \/>\n<span style=\"color: #000000;\">= 4.5 cm<sup>2<br \/>\n<\/sup>Area of two triangles (IV and\u00a0V) = 4.5 cm<sup>2<\/sup>\u00a0+ 4.5 cm<sup>2<\/sup>\u00a0= 9 cm<sup>2<br \/>\n<\/sup>Total area of the paper used = Area I + Area II + Area III + Area IV + Area V <\/span><br \/>\n<span style=\"color: #000000;\">= 2.5 cm<sup>2<\/sup>\u00a0+ 6.5 cm<sup>2<\/sup>\u00a0+ 1.4 cm<sup>2<\/sup>\u00a0+ 9 cm<sup>2<\/sup>\u00a0= 19.4 cm<sup>2<br \/>\n<\/sup>Thus, the total area of the paper used is\u00a019.4 cm<sup>2<\/sup>\u00a0(approx.).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>It is given that the parallelogram and triangle have equal areas.<br \/>\nThe sides of the triangle are given as 26 cm, 28 cm and 30 cm.<br \/>\nSo, the perimeter = 26 + 28 + 30 = 84 cm<br \/>\nits semi perimeter = 84\/2 cm = 42 cm<br \/>\nNow, by using Heron\u2019s formula, area of the triangle = = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c)<br \/>\n= \u221a[42(42 &#8211; 26)(42 &#8211; 28)(42 &#8211; 30)] cm<sup>2<br \/>\n<\/sup>= \u221a[42 \u00d7 16 \u00d7 14 \u00d7 12] cm<sup>2<br \/>\n<\/sup>= 336 cm<sup>2<br \/>\n<\/sup>Now, let the height of parallelogram be h.<br \/>\nAs the area of parallelogram = area of the triangle,<br \/>\n28 cm\u00d7 h = 336 cm<sup>2<br \/>\n<\/sup>\u2234 h = 336\/28 cm<br \/>\nSo, the height of the parallelogram is 12 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4639\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"337\" height=\"240\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans5.png 337w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans5-300x214.png 300w\" sizes=\"auto, (max-width: 337px) 100vw, 337px\" \/><br \/>\nConsider the triangle BCD,<br \/>\nIts semi-perimeter = (48 + 30 + 30)\/2 m = 54 m<br \/>\nUsing Heron\u2019s formula,<br \/>\nArea of the \u0394BCD<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\sqrt{s(s-a)(s-b)(s-c)}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\sqrt{s(s-a)(s-b)(s-c)} \\right )\" align=\"absmiddle\" \/><br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{54(54-48)(54-30)(54-30)}\" alt=\"\\sqrt{54(54-48)(54-30)(54-30)}\" align=\"absmiddle\" \/> m<sup>2<\/sup><br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{54\\times&amp;space;6\\times24\\times24}\" alt=\"\\sqrt{54\\times 6\\times24\\times24}\" align=\"absmiddle\" \/> m<sup>2<br \/>\n<\/sup>= 432 m<sup>2<br \/>\n<\/sup>\u2234 Area of field = 2 \u00d7 area of the \u0394BCD = (2 \u00d7 432) m<sup>2\u00a0<\/sup>= 864 m<sup>2<br \/>\n<\/sup>Thus, the area of the grass field that each cow will be getting = (864\/18) m<sup>2\u00a0<\/sup>= 48 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4640\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"236\" height=\"332\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q6.png 236w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q6-213x300.png 213w\" sizes=\"auto, (max-width: 236px) 100vw, 236px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>For each triangular piece, The semi perimeter will be<br \/>\ns = (50+50+20)\/2 cm = 120\/2 cm = 60cm<br \/>\nUsing Heron\u2019s formula,<br \/>\nArea of the triangular piece<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\sqrt{s(s-a)(s-b)(s-c)}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\sqrt{s(s-a)(s-b)(s-c)} \\right )\" align=\"absmiddle\" \/><br \/>\n= \u221a[60(60 &#8211; 50)(60 &#8211; 50)(60 &#8211; 20)] cm<sup>2<br \/>\n<\/sup>= \u221a[60 \u00d7 10 \u00d7 10 \u00d7 40] cm<sup>2<br \/>\n<\/sup>= 200\u221a6 cm<sup>2<br \/>\n<\/sup>\u2234 The area of all the triangular pieces = 5 \u00d7 200\u221a6 cm<sup>2\u00a0<\/sup>= 1000\u221a6 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4641\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q7.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"159\" height=\"158\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q7.png 159w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q7-150x150.png 150w\" sizes=\"auto, (max-width: 159px) 100vw, 159px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>As the kite is in the shape of a square, its area will be<br \/>\nA = (\u00bd) \u00d7 (diagonal)<sup>2<br \/>\n<\/sup>Area of the kite = (\u00bd) \u00d7 32 \u00d7 32 = 512 cm<sup>2.<br \/>\n<\/sup>The area of shade I = Area of shade II<br \/>\n512\/2 cm<sup>2\u00a0<\/sup>= 256 cm<sup>2<br \/>\n<\/sup>So, the total area of the paper that is required in each shade = 256 cm<sup>2<br \/>\n<\/sup>For the triangle section (III),<br \/>\nThe sides are given as 6 cm, 6 cm and 8 cm<br \/>\nNow, the semi perimeter of this isosceles triangle = (6 + 6 + 8)\/2 cm = 10 cm<br \/>\nBy using Heron\u2019s formula, the area of the III triangular piece will be<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\sqrt{s(s-a)(s-b)(s-c)}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\sqrt{s(s-a)(s-b)(s-c)} \\right )\" align=\"absmiddle\" \/><br \/>\n= \u221a[10(10 &#8211; 6)(10 &#8211; 6)(10 &#8211; 8)] cm<sup>2<br \/>\n<\/sup>= \u221a(10 \u00d7 4 \u00d7 4 \u00d7 2) cm<sup>2<br \/>\n<\/sup>= 8\u221a5 cm<sup>2<\/sup><br \/>\n= 17.92 cm<sup>2<\/sup>\u00a0(approx.)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm<sup>2<\/sup>.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4642\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"360\" height=\"354\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q8.png 360w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Q8-300x295.png 300w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>The semi perimeter of the each triangular shape = (28+9+35)\/2 cm = 36 cm<br \/>\nBy using Heron\u2019s formula,<br \/>\nThe area of each triangular shape will be<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\sqrt{s(s-a)(s-b)(s-c)}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\sqrt{s(s-a)(s-b)(s-c)} \\right )\" align=\"absmiddle\" \/><br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{36(36-35)(36-28)(36-9)}\" alt=\"\\sqrt{36(36-35)(36-28)(36-9)}\" align=\"absmiddle\" \/><br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{36\\times&amp;space;1\\times&amp;space;8\\times&amp;space;27}\" alt=\"\\sqrt{36\\times 1\\times 8\\times 27}\" align=\"absmiddle\" \/><br \/>\n= 36\u221a6 cm<sup>2\u00a0<\/sup>= 88.2 cm<sup>2<br \/>\n<\/sup>Now, the total area of 16 tiles = 16 \u00d7 88.2 cm<sup>2\u00a0<\/sup>= 1411.2 cm<sup>2<br \/>\n<\/sup>It is given that the polishing cost of tiles = 50 paise\/cm<sup>2<br \/>\n<\/sup>\u2234 The total polishing cost of the tiles = Rs. (1411.2\u00d70.5) = Rs. 705.6<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Given ABCD is a field. Draw CG \u22a5 AB from C on AB, and CF parallel to DA.<br \/>\n<strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4643\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans9.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"518\" height=\"340\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans9.png 653w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans9-300x197.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.2-Ans9-480x315.png 480w\" sizes=\"auto, (max-width: 518px) 100vw, 518px\" \/><br \/>\n<\/strong>DC = AF = 10 m, DA = CF = 13 m (opposite sides of parallelogram)<\/span><br \/>\n<span style=\"color: #000000;\">So, FB = 25 &#8211; 10 = 15 m<\/span><br \/>\n<span style=\"color: #000000;\">In \u2206CFB, a = 15 m, b = 14 m, c = 13 m.<\/span><br \/>\n<span style=\"color: #000000;\">Semi Perimeter(s) = (a + b + c)\/2<\/span><br \/>\n<span style=\"color: #000000;\">s = (15 + 14 + 13)\/2<\/span><br \/>\n<span style=\"color: #000000;\">s = 42\/2<\/span><br \/>\n<span style=\"color: #000000;\">s = 21 m<\/span><br \/>\n<span style=\"color: #000000;\">By using\u00a0Heron\u2019s formula, <\/span><br \/>\n<span style=\"color: #000000;\">Area of \u2206CFB = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c) <\/span><br \/>\n<span style=\"color: #000000;\">= \u221a21(21 &#8211; 15)(21 &#8211; 14)(21 &#8211; 13)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a21 \u00d7 6 \u00d7 7 \u00d7 8<\/span><br \/>\n<span style=\"color: #000000;\">= 84 m<sup>2<br \/>\n<\/sup>Area of \u2206CFB = 1\/2 \u00d7 base \u00d7 height<\/span><br \/>\n<span style=\"color: #000000;\">84 = 1\/2 \u00d7 BF \u00d7 CG<\/span><br \/>\n<span style=\"color: #000000;\">84 = 1\/2 \u00d7 15 \u00d7 CG<\/span><br \/>\n<span style=\"color: #000000;\">CG = (84 \u00d7 2)\/15<\/span><br \/>\n<span style=\"color: #000000;\">CG = 11.2 m\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Area of trapezium ABCD = 1\/2 \u00d7 sum of parallel sides \u00d7 distance between them <\/span><br \/>\n<span style=\"color: #000000;\">=\u00a01\/2 \u00d7 (AB\u00a0+ DC) \u00d7 CG<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/2 \u00d7 (25 + 10) \u00d7 11.2<\/span><br \/>\n<span style=\"color: #000000;\">= 196 m<sup>2<br \/>\n<\/sup>Hence the area of the field is 196 m<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 12 (Heron\u2019s Formula)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 12 Heron\u2019s Formula Exercise 12.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 12: Heron\u2019s Formula NCERT Solution Class 9 Maths Ex &#8211; 12.1 Exercise &#8211; [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[723,702,186,724,701,5],"class_list":["post-4578","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-12-herons-formula-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-12-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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