{"id":100739,"date":"2021-02-09T05:36:16","date_gmt":"2021-02-09T05:36:16","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=100739"},"modified":"2023-09-18T02:54:45","modified_gmt":"2023-09-18T02:54:45","slug":"ncert-solutions-for-9th-class-maths-chapter-12-herons-formula","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-12-herons-formula\/","title":{"rendered":"NCERT Solutions for 9th class Maths : Chapter 12 Heron&#8217;s Formula"},"content":{"rendered":"\n<p>Class 9: Maths Chapter 12 solutions. Complete Class 9 Maths Chapter 12 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>NCERT Solutions for 9th class Maths : Chapter 12 Heron&#8217;s Formula<\/strong><\/h2>\n\n\n\n<p>NCERT 9th Maths Chapter 12, class 9 Maths Chapter 12 solutions<\/p>\n\n\n\n<p>Page No: 202<\/p>\n\n\n\n<p><strong>Exercise 12.1<\/strong><\/p>\n\n\n\n<p><strong>1. A traffic signal board, indicating &#8216;SCHOOL AHEAD&#8217;, is an equilateral triangle with side &#8216;a&#8217;. 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><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of the side of equilateral triangle = a<br>Perimeter of the signal board = 3a = 180 cm<br>\u2234 3a = 180 cm \u21d2 a = 60 cm<br>Semi perimeter of the signal board (s) = 3a\/2<br>Using heron&#8217;s formula,<br>Area of the signal board = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a(3a\/2) (3a\/2 &#8211; a) (3a\/2 &#8211; a) (3a\/2 &#8211; a)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a3a\/2 \u00d7 a\/2 \u00d7 a\/2 \u00d7 a\/2<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a3a<sup>4<\/sup>\/16<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a3a<sup>2<\/sup>\/4<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a3\/4 \u00d7 60 \u00d7 60 = 900\u221a3 cm<sup>2<\/sup><\/p>\n\n\n\n<p><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> per year. A company hired one of its walls for 3 months. How much rent did it pay?<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-1.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.1 Que. 2\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>The sides of the triangle are 122 m, 22 m and 120 m.<br>Perimeter of the triangle is 122&nbsp;+ 22 + 120 = 264m<br>Semi perimeter of triangle (s) = 264\/2 = 132 m<br>Using heron&#8217;s formula,<br>Area of the advertisement = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a132(132 &#8211; 122) (132 &#8211; 22) (132 &#8211; 120) m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a132 \u00d7 10 \u00d7 110 \u00d7 12 m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 1320 m<sup>2<\/sup><br>Rent of advertising per year = \u20b9 5000 per m<sup>2<\/sup><br>Rent of one wall for 3 months = \u20b9 (1320&nbsp;\u00d7 5000&nbsp;\u00d7 3)\/12 = \u20b9 1650000<\/p>\n\n\n\n<p>Page No: 203<\/p>\n\n\n\n<p>NCERT 9th Maths Chapter 12, class 9 Maths Chapter 12 solutions<\/p>\n\n\n\n<p><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.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-2.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.1 Que. 3\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Sides of the triangular wall are 15 m, 11 m and 6 m.<br>Semi perimeter of triangular wall (s) = (15 + 11 + 6)\/2 m = 16 m<br>Using heron&#8217;s formula,<br>Area of the message = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a16(16 &#8211; 15) (16 &#8211; 11) (16 &#8211; 6) m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a16 \u00d7 1 \u00d7 5 \u00d7 10 m<sup>2 <\/sup>= \u221a800 m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 20\u221a2 m<sup>2<\/sup><\/p>\n\n\n\n<p><strong>4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Two sides of the triangle = 18cm and 10cm<br>Perimeter of the triangle = 42cm<br>Third side of triangle = 42 &#8211; (18+10) cm = 14cm<br>Semi perimeter of triangle = 42\/2 = 21cm<br>Using heron&#8217;s formula,<br>Area of the triangle = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a21(21 &#8211; 18) (21 &#8211; 10) (21 &#8211; 14) cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a21 \u00d7 3 \u00d7 11 \u00d7 7 m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 21\u221a11 cm<sup>2<\/sup><\/p>\n\n\n\n<p><strong>5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Ratio of the sides of the triangle = 12 : 17 : 25<br>Let the common ratio be x then sides are 12x, 17x and 25x<br>Perimeter of the triangle = 540cm<br>12x&nbsp;+ 17x + 25x = 540 cm<br>\u21d2 54x = 540cm<br>\u21d2 x = 10<br>Sides of triangle are,<br>12x = 12&nbsp;\u00d7 10 = 120cm<br>17x = 17 \u00d7 10 = 170cm<br>25x = 25 \u00d7 10 = 250cm<br>Semi perimeter of triangle(s) = 540\/2 = 270cm<br>Using heron&#8217;s formula,<br>Area of the triangle = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a270(270 &#8211; 120) (270 &#8211; 170) (270 &#8211; 250)cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a270 \u00d7 150 \u00d7 100 \u00d7 20 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 9000 cm<sup>2<\/sup><\/p>\n\n\n\n<p><strong>6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of the equal sides = 12cm<br>Perimeter of the triangle = 30cm<br>Length of the third side = 30 &#8211; (12+12) cm = 6cm<br>Semi perimeter of the triangle(s) = 30\/2 cm = 15cm<br>Using heron&#8217;s formula,<br>Area of the triangle = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a15(15 &#8211; 12) (15 &#8211; 12) (15 &#8211; 6)cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a15 \u00d7 3 \u00d7 3 \u00d7 9 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 9\u221a15 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Page No: 206<\/p>\n\n\n\n<p>NCERT 9th Maths Chapter 12, class 9 Maths Chapter 12 solutions<\/p>\n\n\n\n<p><strong>Exercise 12.2<\/strong><\/p>\n\n\n\n<p><strong>1. A park, in the shape of a quadrilateral ABCD, has \u2220C = 90\u00ba, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>\u2220C = 90\u00ba, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m<br>BD is joined.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-2.png\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 1\"\/><\/figure>\n\n\n\n<p>In \u0394BCD,<br>By applying Pythagoras theorem,<br>BD<sup>2<\/sup> = BC<sup>2 <\/sup>+ CD<sup>2&nbsp;<\/sup><br>\u21d2 BD<sup>2<\/sup> = 12<sup>2 <\/sup>+ 5<sup>2<\/sup><br>\u21d2 BD<sup>2<\/sup> = 169<br>\u21d2 BD = 13 m<br>Area of \u0394BCD = 1\/2 \u00d7 12 \u00d7 5 = 30 m<sup>2<\/sup><br>Now,<br>Semi perimeter of \u0394ABD(s) = (8 + 9 + 13)\/2 m = 30\/2 m = 15 m<br>Using heron&#8217;s formula,<br>Area of \u0394ABD&nbsp; = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a15(15 &#8211; 13) (15 &#8211; 9) (15 &#8211; 8) m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a15 \u00d7 2 \u00d7 6 \u00d7 7 m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 6\u221a35 m<sup>2 <\/sup>= 35.5 m<sup>2<\/sup> (approx)<br>Area of quadrilateral ABCD = Area of \u0394BCD + Area of \u0394ABD = 30 m<sup>2&nbsp;<\/sup>+35.5m<sup>2<\/sup> = 65.5m<sup>2&nbsp;<\/sup><\/p>\n\n\n\n<p><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.<\/strong><br><br><strong>Answer<\/strong><\/p>\n\n\n\n<p>AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-3.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 2\"\/><\/figure>\n\n\n\n<p>In \u0394ABC,<br>By applying Pythagoras theorem,<br>AC<sup>2<\/sup> = AB<sup>2 <\/sup>+ BC<sup>2&nbsp;<\/sup><br>\u21d2 5<sup>2<\/sup> = 3<sup>2 <\/sup>+ 4<sup>2<\/sup><br>\u21d2&nbsp;25 = 25<br>Thus, \u0394ABC is a right angled at B.<br>Area of \u0394BCD = 1\/2 \u00d7 3 \u00d7 4 = 6 cm<sup>2<\/sup><br>Now,<br>Semi perimeter of \u0394ACD(s) = (5 + 5 + 4)\/2 cm = 14\/2 cm = 7 m<br>Using heron&#8217;s formula,<br>Area of \u0394ABD&nbsp; = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a7(7 &#8211; 5) (7 &#8211; 5) (7 &#8211; 4) cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a7 \u00d7 2 \u00d7 2 \u00d7 3 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 2\u221a21 cm<sup>2 <\/sup>= 9.17 cm<sup>2<\/sup> (approx)<br>Area of quadrilateral ABCD = Area of \u0394ABC + Area of \u0394ABD = 6 cm<sup>2&nbsp;<\/sup>+9.17 cm<sup>2<\/sup> = 15.17 cm<sup>2&nbsp;<\/sup><br><br><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.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-4.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 3\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Length of the sides of the triangle section I = 5cm, 1cm and 5cm<br>Perimeter of the triangle = 5 + 5 + 1 = 11cm<br>Semi perimeter = 11\/2 cm = 5.5cm<br>Using heron&#8217;s formula,<br>Area of section I&nbsp; = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a5.5(5.5 &#8211; 5) (5.5 &#8211; 5) (5.5 &#8211; 1) cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a5.5 \u00d7 0.5 \u00d7 0.5 \u00d7 4.5 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.75\u221a11 cm<sup>2 <\/sup>= 0.75 \u00d7 3.317cm<sup>2<\/sup> = 2.488cm<sup>2<\/sup> (approx)<br>Length of the sides of the rectangle of section I = 6.5cm and 1cm<br>Area of section II = 6.5 \u00d7 1 cm<sup>2<\/sup> =6.5 cm<sup>2<\/sup><br>Section III is an isosceles trapezium which is divided into 3 equilateral of side 1cm each.<br>Area of the trapezium = 3 \u00d7 \u221a3\/4 \u00d7 1<sup>2&nbsp;<\/sup>cm<sup>2&nbsp;<\/sup>= 1.3 cm<sup>2 <\/sup>(approx)<br>Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm<br>Area of region IV and V = 2 \u00d7 1\/2 \u00d7 6 \u00d7 1.5cm<sup>2&nbsp;<\/sup>= 9cm<sup>2<\/sup><br>Total area of the paper used = (2.488 + 6.5 + 1.3 + 9)cm<sup>2&nbsp;<\/sup>=19.3 cm<sup>2<\/sup><\/p>\n\n\n\n<p><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.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given,<br>Area of the parallelogram and triangle are equal.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-6.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 4\"\/><\/figure>\n\n\n\n<p>Length of the sides of the triangle are 26 cm, 28 cm and 30 cm.<br>Perimeter of the triangle = 26&nbsp;+ 28 + 30 = 84 cm<br>Semi perimeter of the triangle = 84\/2 cm = 42 cm<br>Using heron&#8217;s formula,<br>Area of the triangle = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a42(42 &#8211; 26) (46 &#8211; 28) (46 &#8211; 30) cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a46 \u00d7 16 \u00d7 14 \u00d7 16 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 336 cm<sup>2<\/sup>Let height of parallelogram be h.<br>Area of parallelogram = Area of triangle<br>28cm \u00d7 h = 336 cm<sup>2<\/sup><br>\u21d2h = 336\/28 cm<br>\u21d2h = 12 cm<br>The height of the parallelogram is 12 cm.<\/p>\n\n\n\n<p>Page No: 207<\/p>\n\n\n\n<p>NCERT 9th Maths Chapter 12, class 9 Maths Chapter 12 solutions<\/p>\n\n\n\n<p><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?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Diagonal AC divides the rhombus ABCD into two congruent triangles of equal area.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-7.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 5\"\/><\/figure>\n\n\n\n<p>Semi perimeter of \u0394ABC = (30 + 30 + 48)\/2 m = 54 m<br>Using heron&#8217;s formula,<br>Area of the \u0394ABC = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a54(54 &#8211; 30) (54 &#8211; 30) (54 &#8211; 48) m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a54 \u00d7 24 \u00d7 24 \u00d7 6 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 432 m<sup>2<\/sup><br>Area of field = 2 \u00d7 area of the \u0394ABC = (2 \u00d7 432)m<sup>2&nbsp;<\/sup>= 864 m<sup>2<\/sup><br>Thus,<br>Area of grass field which each cow will be getting = 864\/18m<sup>2&nbsp;<\/sup>=48 m<sup>2<\/sup><\/p>\n\n\n\n<p><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?<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-8.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 6\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Semi perimeter of each triangular piece = (50 + 50 + 20)\/2 cm = 120\/2 cm = 60cm<br>Using heron&#8217;s formula,<br>Area of the triangular piece = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a60(60 &#8211; 50) (60 &#8211; 50) (60 &#8211; 20) cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a60 \u00d7 10 \u00d7 10 \u00d7 40 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 200\u221a6 cm<sup>2<\/sup><br>Area of triangular piece = 5 \u00d7 200\u221a6 cm<sup>2 <\/sup>= 1000\u221a6 cm<sup>2<\/sup><\/p>\n\n\n\n<p><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?<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-9.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 7\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>We know that,<br>As the diagonals of a square bisect each other at right angle.<br>Area of given kite = 1\/2 (diagonal)<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = 1\/2 \u00d7 32 \u00d7 32 = 512 cm<sup>2<\/sup><br>Area of shade I = Area of shade II<br>\u21d2 512\/2 cm<sup>2&nbsp;<\/sup>= 256cm<sup>2<\/sup><br>So, area of paper required in each shade = 256 cm<sup>2<\/sup><br>For the III section,<br>Length of the sides of triangle = 6cm, 6cm and 8cm<br>Semi perimeter of triangle = (6 + 6 + 8)\/2 cm = 10cm<br>Using heron&#8217;s formula,<br>Area of the III triangular piece = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a10(10 &#8211; 6) (10 &#8211; 6) (10 &#8211; 8) cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a10 \u00d7 4 \u00d7 4 \u00d7 2 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 8\u221a6 cm<sup>2<\/sup><\/p>\n\n\n\n<p>NCERT 9th Maths Chapter 12, class 9 Maths Chapter 12 solutions<\/p>\n\n\n\n<p><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> .<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-10.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 8\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Semi perimeter of the each triangular shape = (28 + 9 + 35)\/2 cm = 36 cm<br>Using heron&#8217;s formula,<br>Area of the each triangular shape = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a36(36 &#8211; 28) (36 &#8211; 9) (36 &#8211; 35) cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a36 \u00d7 8 \u00d7 27 \u00d7 1 cm<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 36\u221a6 cm<sup>2&nbsp;<\/sup>= 88.2 cm<sup>2<\/sup><br>Total area of 16 tiles = 16 \u00d7 88.2 cm<sup>2 <\/sup>= 1411.2 cm<sup>2<\/sup>Cost of polishing tiles = 50p per cm<sup>2<\/sup><br>Total cost of polishing the tiles = Rs. (1411.2 \u00d7 0.5) = Rs. 705.6<\/p>\n\n\n\n<p><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.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class-9-maths-chapter-12-ncert-11.jpg\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 9\"\/><\/figure>\n\n\n\n<p>Let ABCD be the given trapezium with parallel sides AB = 25m and CD = 10mand the non-parallel sides AD = 13m and BC = 14m.<br>CM \u22a5 AB and CE || AD.<br>In \u0394BCE,<br>BC = 14m, CE = AD = 13 m and<br>BE = AB &#8211; AE = 25 &#8211; 10 = 15m<br>Semi perimeter of the \u0394BCE = (15 + 13 + 14)\/2 m = 21 m<br>Using heron&#8217;s formula,<br>Area of the \u0394BCE = \u221as (s-a) (s-b) (s-c)<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; = \u221a21(21 &#8211; 14) (21 &#8211; 13) (21 &#8211; 15) m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u221a21 \u00d7 7 \u00d7 8 \u00d7 6 m<sup>2<\/sup><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 84 m<sup>2<\/sup><br>also, area of the \u0394BCE = 1\/2 \u00d7 BE \u00d7 CM = 84 m<sup>2<\/sup><br>\u21d2 1\/2 \u00d7 15 \u00d7 CM = 84 m<sup>2<\/sup><br>\u21d2 CM = 168\/15 m<sup>2&nbsp;<\/sup><br>\u21d2 CM = 56\/5 m<sup>2&nbsp;<\/sup><br>Area of the parallelogram AECD = Base \u00d7 Altitude = AE \u00d7 CM = 10 \u00d7 84\/5 = 112 m<sup>2<\/sup><br>Area of the trapezium ABCD = Area of AECD + Area of \u0394BCE<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = (112+ 84) m<sup>2&nbsp; <\/sup>= 196 m<sup>2&nbsp;<\/sup><\/p>\n\n\n\n<p><strong>Chapter 12 Heron&#8217;s Formula NCERT Solutions is a great way through which you can find area of triangles. You know area of a triangles is 1\/2 \u00d7 base \u00d7 height.&nbsp;When it is not possible to find the height of the triangle easily and measures of all the three sides are known then we use Heron\u2019s formula.<\/strong><\/p>\n\n\n\n<p>\u2022 Area of a Triangle \u2014 by Heron\u2019s Formula:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/1.png\" alt=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron's Formula Ex. 12.2 Que. 9\"\/><\/figure>\n\n\n\n<p>where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the perimeter of the triangle = a+b+c\/2<\/p>\n\n\n\n<p>\u2022&nbsp;Application of Heron\u2019s Formula in Finding Areas of Quadrilaterals: We will extend the use of Heron&#8217;s Formula to find the area of quadrilaterals. We can divide the quadrilateral in triangular parts and then use the formula for area of the triangle.<\/p>\n\n\n\n<p>There are only two exercises in chapter 12 Heron&#8217;s Formula where you will learn to&nbsp;grasp key concepts of the chapter properly. Below you can find <strong>exercisewise NCERT Solutions of Heron&#8217;s Formula <\/strong>just by clicking on the links.<\/p>\n\n\n\n<p>Indcareer Schools  experts have prepared these <strong>Chapter 12 NCERT Solutions<\/strong> that are detailed and accurate through which a student can clear their doubts easily.&nbsp;By solving the questions from Class 9 NCERT textbook, a student will gain confidence that is going to help them in exams.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">NCERT Solutions for Class 9 Maths Chapters:<\/h4>\n\n\n\n<p><strong>FAQ on&nbsp;<\/strong><strong>Chapter&nbsp;<\/strong><strong>12 Heron&#8217;s Formula<\/strong><\/p>\n\n\n\n<h4 class=\"wp-block-heading\">What are the benefits of NCERT Solutions for Chapter 12 Heron&#8217;s Formula Class 9 NCERT Solutions?<\/h4>\n\n\n\n<p>NCERT Solutions are very important in improving problem skills and understand all the important points of the chapter. Here we have detailed every solutions step by step so you can easily know the concepts.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">How many exercises in Chapter 12 Heron&#8217;s Formula?<\/h4>\n\n\n\n<p>Only two exercises are there in the chapter 12 Heron&#8217;s Formula. In the first exercise, you need to find the area of triangles by using heron&#8217;s formula. In the second exercise, you have to find the area of quadrilaterals. For finding area of a quadrilateral we divide it into various triangles. After we need to use Heron\u2019s formula to find the area of the triangles.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Find the length of a diagonal of a square whose side is 2 cm.<\/h4>\n\n\n\n<p>The diagonal of a square = (\u221a2) a cm<\/p>\n\n\n\n<p>\u2234 Length of the diagonal = (\u221a2) x 2 cm = 2\u221a2cm.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Find the height of an equilateral triangle whose side is 2 cm.<\/h4>\n\n\n\n<p>Since height of an equilateral triangle is given by<\/p>\n\n\n\n<p>Height = (\u221a3\/2) x side<\/p>\n\n\n\n<p>\u21d2 height =(\u221a3\/2) x 2 cm = 3 cm.<\/p>\n\n\n\n<p>NCERT 9th Maths Chapter 12, class 9 Maths Chapter 12 solutions<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-for-9th-class-maths-chapter-12-nbsp-download-pdf\">NCERT Solutions for 9th class Maths : Chapter 12:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>NCERT Solutions for 9th class Maths : Chapter 12 Heron&#8217;s Formula<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/NCERT-Solutions-for-9th-class-Maths-_-Chapter-12-Herons-Formula.pdf\"><strong>Download PDF<\/strong>: NCERT Solutions for 9th class Maths : Chapter 12 Heron&#8217;s Formula PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-chapterwise-ncert-solutions-for-class-9-maths\"><strong>Chapterwise NCERT Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-1-number-systems\/\">Chapter 1 Number System<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-2-polynomials\/\">Chapter 2 Polynomials<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-3-coordinate-geometry\/\">Chapter 3 Coordinate Geometry<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-4-linear-equations-in-two-variables\/\">Chapter 4 Linear Equations in Two Variables<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-5-introduction-to-euclids-geometry\/\">Chapter 5 Introduction to Euclid&#8217;s Geometry<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-6-lines-and-angles\/\">Chapter 6 Lines and Angles<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-7-triangles\/\">Chapter 7 Triangles<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-8-quadrilaterals\/\">Chapter 8 Quadrilaterals<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-9-areas-of-parallelograms-and-triangles\/\">Chapter 9 Areas of Parallelograms and Triangles<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-10-circles\/\">Chapter 10 Circles<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-11-constructions\/\">Chapter 11 Constructions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-12-herons-formula\/\">Chapter 12 Heron\u2019s Formula<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-13-surface-areas-and-volumes\/\">Chapter 13 Surface Areas and Volumes<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-14-statistics\/\">Chapter 14 Statistics<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-15-probability\/\">Chapter 15 Probability<\/a><\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-about-ncert\">About NCERT<\/h2>\n\n\n\n<p>The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. <a href=\"https:\/\/ncert.nic.in\/\" target=\"_blank\" rel=\"noreferrer noopener\">Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n<div class=\"wp-block-buttons is-layout-flex wp-block-buttons-is-layout-flex\">\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions\/\">NCERT Solutions<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-class-ix\/\">NCERT Solutions for Class 9<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths\/\">NCERT Solutions for Class 9 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Class 9: Maths Chapter 12 solutions. Complete Class 9 Maths Chapter 12 Notes. NCERT Solutions for 9th class Maths : Chapter 12 Heron&#8217;s Formula NCERT 9th Maths Chapter 12, class 9 Maths Chapter 12 solutions Page No: 202 Exercise 12.1 1. A traffic signal board, indicating &#8216;SCHOOL AHEAD&#8217;, is an equilateral triangle with side &#8216;a&#8217;. [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":627796,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,921],"tags":[1491],"boards":[1180],"class_list":["post-100739","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-9","tag-ncert-maths-class-9","boards-ncert","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions for Class 9, Maths Chapter 12 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron&#039;s Formula | Browse all Class 9 Maths Chapters NCERT - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-12-herons-formula\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for 9th class Maths : Chapter 12 Heron&#039;s Formula\" \/>\n<meta property=\"og:description\" content=\"Class 9: Maths Chapter 12 solutions. 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