## Maths NCERT Solutions Class 9 – Chapter 12 Heron’s

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**

## No comments, be the first one to comment !