Maths ncert solutions class 9 chapter 12 heron’s

9 class maths,Maths NCERT Solutions Class 9 : Takshila Learning aims to provide the best subject material for Class 9 Maths along with the 9th class maths solution designed under the guidance of our expert team. We focus on designing the entire syllabus for CBSE Class 9 in such a manner that is easy to understand and offers effective learning. Now, we will proceed and discuss the Chapter 12 ‘Heron’s Formula’ of 9th Class Maths.

In Class 8, we had studied the triangle and the area of a triangle. Now, we will learn the area of a right triangle, equilateral triangle and isosceles triangle.

Area of a triangle=1/2*base*height Area of a triangle – by Heron’s formula The Heron was born in Alexandria in Egypt in 10 AD. He has derived the famous formula for the area of a triangle in terms of its three sides.

Area of a triangle=√s(s-a)(s-b)(s-c) Where a, b, and c are the sides of the triangle and s is a semi perimeter (half the perimeter of the triangle= a +b + c/2)

To understand this formula better let us go through Class 9 Maths solutions

Exercise 12.1 Question 1.A traffic signal board, indicating ‘SCHOOL AHEAD’ is an equilateral triangle with side Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board? MATHS Question Solution:-Perimeter=sum of all sides = 180=3a =side of a triangle 180/3=60cm =s=180/2=90 = using herons formula =√s (s-a)(s-b)(s-c) =√90(90-60)(90-60)(90-60) =√90*30*30*30 =√3*3*10*3*10*3*10*3*10 =3*3*10*10√3 =900√3cm² Question - Sides of a triangle are in the ratio of 12:17:25 and its perimeter is 540 cm. Find its area?

Solution- Let the ratio b x Perimeter= sum of all sides 540 =12x+17x+25x 540=54x X=540/54=10 Side a =12*10=120 Side b=17*10=170 Side c=25*10=250 S=a+b+c/2=540/2=270 cm

Area of a triangle using herons formula =√s(s-a)(s-b)(s-c) =√270 (270-120)(270-170)(270-250) =√270*150*100*20 =√3*3*3*10*3*5*10*10*10*10*2 =√3*3*3*3*2*5*10*1010*10*10*2*5 =√3*3*3*3*2*2*5*5*10*10*10*10 =3*3*2*5*10*10 =9000 cm² Application of Herons Formula in finding an area in quadrilaterals Using the Heron’s formula, we can calculate an area of a quadrilateral whose sides and one diagonal is given by dividing the quadrilateral into two triangles.

To understand better, go through online classes for CBSE Class 9 Maths Exercise 12.2 Question 4 -A triangles and a parallelogram have the same base and the same area. If the sides of the triangle are 26cm, 28 cm, and 30 cm; the parallelogram stands on the base 28 cm. Find the height of the parallelogram.

Solution- s=a+ b +c/2 = (26+28+30)/2 = 84/2 =42 cm Area= √s(s-a)(s-b)(s-c) =√42(42-26)(42-28)(42-30) =√42*16*14*12 =√7*2*3*2*2*2*2*7*2*2*2* =7*2*2*2*3

=336 cm² Area of a parallelogram=b*h 336 cm² =28 cm *h H=336/28 H=12 cm Practice question Question- An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm.Find the area of a triangle.

Answer- 9√15 cm² Question- Find the area of a trapezium whose parallel sides are 25 cm,13 cm, and other sides are 15 cm and 15 cm Answer- 82.8 cm² Question- Find the area of an equilateral triangle with side 4√5 cm Answer- 20 √3 cm²

For more solutions and last year papers, click on Class 9 NCERT solutions or CBSE question paper for Class 9. It will enable you to review the complete subject matter for Class 9 designed by the experts.

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