NCERT Solutions for Class 9 Maths Heron’s Formula Chapter 12

NCERT Solutions for Class 9 Maths Heron’s Formula Chapter 12

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 Chapter 12 ‘Heron’s Formula’ of 9 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)

NCERT Solutions for Class 9 Maths Heron’s Formula

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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