{"id":4577,"date":"2023-02-19T05:45:03","date_gmt":"2023-02-19T05:45:03","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4577"},"modified":"2023-02-19T05:46:03","modified_gmt":"2023-02-19T05:46:03","slug":"ncert-solutions-class-9-maths-chapter-12-herons-formula-ex-12-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-12-herons-formula-ex-12-1\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 12 Heron\u2019s Formula Ex 12.1"},"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 12.1 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-12-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 12.2<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 12.1<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. A traffic signal board, indicating \u2018SCHOOL AHEAD\u2019, is an equilateral triangle with side \u2018a\u2019. Find the area of the signal board, using Heron\u2019s formula. If its perimeter is 180 cm, what will be the area of the signal board?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong> <strong>Given :<\/strong> Dimensions of the traffic signal board and its perimeter.<br \/>\nSide of the signal board = a<br \/>\nPerimeter of the signal board = 3a = 180 cm<br \/>\n\u2234 a = 180\/3 cm = 60 cm<br \/>\nSemi Perimeter,\u00a0s = (a + b + c)\/2 = (a + a + a)\/2 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3a}{2}\" alt=\"\\frac{3a}{2}\" align=\"absmiddle\" \/><br \/>\nBy using Heron\u2019s formula,<br \/>\n\u221a[s(s &#8211; a)(s &#8211; b)(s &#8211; c)]<br \/>\n= \u221a[s(s &#8211; a)(s &#8211; a)(s &#8211; a)]<br \/>\n= (s &#8211; a) \u221a[s(s &#8211; a)]<br \/>\nWe know, s = 3a\/2,<br \/>\nArea = (3a\/2 &#8211; a)\u221a[3a\/2(3a\/2 &#8211; a)]<br \/>\n=\u00a0(a\/2) \u221a[3a\/2(a\/2)]<\/span><br \/>\n<span style=\"color: #000000;\">= a\/2 \u00d7 a\/2 \u00d7 \u221a3<\/span><br \/>\n<span style=\"color: #000000;\">= (\u221a3\/4)a<sup>2<br \/>\n<\/sup>Area of the signal board = (\u221a3\/4)a<sup>2<\/sup>\u00a0sq. units<br \/>\nSubstituting the value of &#8216;a&#8217;. <\/span><br \/>\n<span style=\"color: #000000;\">Area of the signal board = (\u221a3\/4)(60)<sup>2<br \/>\n<\/sup>=\u00a0(\u221a3\/4)(3600)<\/span><br \/>\n<span style=\"color: #000000;\">=\u00a0900\u221a3<\/span><br \/>\n<span style=\"color: #000000;\">Area of the signal board = 900\u221a3 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of \u20b95000 per m<sup>2<\/sup>\u00a0per year. A company hired one of its walls for 3 months. How much rent did it pay?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4631\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.1-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"581\" height=\"296\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.1-Q2.png 581w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.1-Q2-300x153.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.1-Q2-480x245.png 480w\" sizes=\"auto, (max-width: 581px) 100vw, 581px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.<br \/>\nNow, the perimeter will be (122 + 22 + 120) = 264 m<br \/>\nSemi\u00a0Perimeter,\u00a0s = (a + b + c)\/2<br \/>\n= (122 + 22 + 120)\/2<\/span><br \/>\n<span style=\"color: #000000;\">= 264\/2<\/span><br \/>\n<span style=\"color: #000000;\">= 132 m <\/span><br \/>\n<span style=\"color: #000000;\">Using Heron\u2019s formula,<br \/>\nArea of the triangle = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c)<br \/>\n= \u221a132(132 &#8211; 122) (132 &#8211; 22) (132 &#8211; 120)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a132 \u00d7 10 \u00d7 110 \u00d7 12<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a11 \u00d7 12 \u00d7 10 \u00d7 10 \u00d7 11 \u00d712<\/span><br \/>\n<span style=\"color: #000000;\">= 11 \u00d7 12 \u00d710<\/span><br \/>\n<span style=\"color: #000000;\">= 1320 m<sup>2<br \/>\n<\/sup>We know that the rent of advertising per year = \u20b9 5000 per m<sup>2<br \/>\n<\/sup>\u2234 The rent of one wall for 3 months = Rs. (1320 \u00d7 5000 \u00d7 3)\/12 = Rs. 16,50,000<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. There is a slide in a park. One of its side walls has been painted in some colour with a message \u201cKEEP THE PARK GREEN AND CLEAN\u201d (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4632\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.1-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"482\" height=\"214\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.1-Q3.png 721w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.1-Q3-300x133.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-12.1-Q3-480x213.png 480w\" sizes=\"auto, (max-width: 482px) 100vw, 482px\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>It is given that the sides of the wall as 15 m, 11 m and 6 m.<br \/>\nSo, the semi perimeter of triangular wall (s) = (15 + 11 + 6)\/2 m = 16 m<br \/>\nUsing Heron\u2019s formula,<br \/>\nArea of the message = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c)<br \/>\n= \u221a[16(16 &#8211; 15)(16 &#8211; 11)(16 &#8211; 6)] m<sup>2<br \/>\n<\/sup>= \u221a[16 \u00d7 1 \u00d7 5 \u00d7 10] m<sup>2<br \/>\n<\/sup>= \u221a800 m<sup>2<br \/>\n<\/sup>= 20\u221a2 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>The sides of triangle given: a =18 cm, b = 10 cm<br \/>\nPerimeter of the triangle = (a + b + c)<br \/>\n42 = 18 + 10 + c<\/span><br \/>\n<span style=\"color: #000000;\">42 = 28 + c<\/span><br \/>\n<span style=\"color: #000000;\">c = 42 &#8211; 28<\/span><br \/>\n<span style=\"color: #000000;\">c = 14 cm<\/span><br \/>\n<span style=\"color: #000000;\">Semi Perimeter <\/span><br \/>\n<span style=\"color: #000000;\">s = (a + b + c) = 42\/2 = 21 cm<\/span><br \/>\n<span style=\"color: #000000;\">Using Heron\u2019s formula,<br \/>\nArea of the triangle = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c)<br \/>\n= \u221a[21(21 &#8211; 18)(21 &#8211; 10)(21 &#8211; 14)] cm<sup>2<br \/>\n<\/sup>= \u221a[21 \u00d7 3 \u00d7 11 \u00d7 7] m<sup>2<br \/>\n<\/sup>= 21\u221a11 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>The ratio of the sides of the triangle are given as 12 : 17 : 25<br \/>\nSo, we can assume the length of the sides of the triangle as\u00a012x\u00a0cm, 17x cm, and 25x cm.<br \/>\nIt is also given that the perimeter of the triangle = 540 cm<br \/>\n12x + 17x + 25x = 540 cm<br \/>\n54x = 540cm<br \/>\nx = 10<br \/>\nTherefore, the sides of the triangle: <\/span><br \/>\n<span style=\"color: #000000;\">12x = 12 \u00d7 10 = 120 cm, <\/span><br \/>\n<span style=\"color: #000000;\">17x = 17 \u00d7 10 = 170 cm, <\/span><br \/>\n<span style=\"color: #000000;\">25x = 25 \u00d7 10 = 250 cm<\/span><br \/>\n<span style=\"color: #000000;\">a = 120cm, b = 170 cm, c = 250 cm<\/span><br \/>\n<span style=\"color: #000000;\">Semi-perimeter(s) = (120 + 170 + 250)\/2 = 540\/2 = 270 cm<\/span><br \/>\n<span style=\"color: #000000;\">By using Heron\u2019s formula,<\/span><br \/>\n<span style=\"color: #000000;\">Area of a triangle = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a270(270 &#8211; 120)(270 &#8211; 170)(270 &#8211; 250)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a270 \u00d7 150 \u00d7 100 \u00d7 20<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a81000000<\/span><br \/>\n<span style=\"color: #000000;\">= 9000 cm<sup>2<br \/>\n<\/sup>Area of the triangle = 9000 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>First, let the third side be x.<br \/>\nIt is given that the length of the equal sides is 12 cm and its perimeter is 30 cm.<br \/>\n30 = 12 + 12 + x<br \/>\n\u2234 The length of the third side = 6 cm<br \/>\nSemi Perimeter (s) = (a + b + c)\/2 =<br \/>\ns = (12 + 12 + 6)\/2 = 30\/2 = 15 cm<\/span><br \/>\n<span style=\"color: #000000;\">Using Heron\u2019s formula,<br \/>\nArea of the triangle = \u221as(s &#8211; a)(s &#8211; b)(s &#8211; c)<br \/>\n= \u221a[15(15 &#8211; 12)(15 &#8211; 12)(15 &#8211; 6)] cm<sup>2<br \/>\n<\/sup>= \u221a[15 \u00d7 3 \u00d7 3 \u00d7 9] cm<sup>2<br \/>\n<\/sup>= 9\u221a15 cm<sup>2<\/sup><\/span><\/p>\n<p>&nbsp;<\/p>\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 12.1 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; 12.2 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-4577","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 v27.4) - 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