Question

Explain :

Area of a Triangle - by Heron's Formula

Application of Heron's Formula in finding Areas of Quadrilaterals

Class 9​

Answer 1

Area of a Triangle - by Heron's Formula

We know that we can use the (below) mentioned formula to find area of right angled triangle:

${\small{\underline{\boxed{\pmb{\sf{\star \: Area \: of \: \triangle \: = \dfrac{1}{2} \times Base \times Height}}}}}}$

But what about the other triangles that are isosceles triangle, scalene triangle and equilateral triangle. Then the Heron's formula will be use an Heron's formula is also known as Hero's formula. That formula is mentioned below:

${\small{\underline{\boxed{\pmb{\sf{\star \: \sqrt{s \: (s-a) \: (s-b) \: (s-c)}}}}}}}$

Where, a denotes first side of triangle, b denotes the second side of triangle, denotes c third side of triangle and s denotes semi - perimeter.

How to find semi-perimeter!?

${\small{\underline{\boxed{\pmb{\sf{\star \: s \: = \dfrac{Perimeter}{2}}}}}}}$

How to find perimeter!?

${\small{\underline{\boxed{\pmb{\sf{\star \: Perimeter \: = \dfrac{a + b + c}{2}}}}}}}$

~ We can appy Heron's formula where we haven't to find the height.

Let us take an example!

Let us take an example!Question: Find out the area of triangle whose sides are given as 5 cm, 8 cm and 5 cm.

~ Firstly finding semi-perimeter by using suitable formula!

$:\implies \sf s \: = \dfrac{a + b + c}{2} \\ \\ :\implies \sf s \: = \dfrac{5 + 8 + 5}{2} \\ \\ :\implies \sf s \: = \dfrac{10 + 8}{2} \\ \\ :\implies \sf s \: = \dfrac{18}{2} \\ \\ :\implies \sf s \: = 9 \: cm \\ \\ :\implies \sf Semi - \: perimeter \: = 9 \: cm$

~ Finding area by using Heron's formula!

$:\implies \sf \sqrt{s \: (s-a) \: (s-b) \: (s-c)} \\ \\ :\implies \sf \sqrt{9 \times (9-5) \times (9-5) \times (9-8)} \\ \\ :\implies \sf \sqrt{9 \times 4 \times 4 \times 1} \\ \\ :\implies \sf \sqrt{3 \times 3 \times 2 \times 2 \times 2 \times 2 \times 1} \\ \\ :\implies \sf 3 \times 2 \times 2 \sqrt{1} \\ \\ :\implies \sf 3 \times 4\\ \\ :\implies \sf Area \: = 12 \: cm^{2}$

* Here we can't use ½ × B × H formula!

Application of Heron's Formula in finding areas of Quadrilaterals -

~ We can use Heron's formula in finding areas of Quadrilaterals by dividing the quadrilateral into triangles. Then have to use Heron's formula according to the condition.

Let us take an example!

Question: Find the area of attached quadrilateral whose diagonal is 160 m and all the sides are 100 m. Perimeter is given as 400 m.

~ Now here let us consider two triangles.

One ∆ADB and other as ∆BCD!

Firstly calculating the semi-perimeter!

$:\implies \sf s \: = \dfrac{a + b + c}{2} \\ \\ :\implies \sf s \: = \dfrac{100 + 100 + 160}{2} \\ \\ :\implies \sf s \: = \dfrac{200 + 160}{2} \\ \\ :\implies \sf s \: = \dfrac{360}{2} \\ \\ :\implies \sf s \: = 180 \: m \\ \\ :\implies \sf Semi - \: perimeter \: = 180 \: m$

~ Finding area of triangle ADB by using Heron's formula!

$:\implies \sf \sqrt{s \: (s-a) \: (s-b) \: (s-c)} \\ \\ :\implies \sf \sqrt{180 \times (180-100) \times (180-100) \times (180-160)} \\ \\ :\implies \sf \sqrt{180 \times 80 \times 80 \times 20} \\ \\ :\implies \sf \sqrt{2 \times 2 \times 3 \times 3 \times 5 \times 2 \times 2 \times 2 \times 2 \times 5\times 2 \times 2 \times 2 \times 2 \times 5 \times 2 \times 2 \times 5} \\ \\ :\implies \sf \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5 \times 5} \\ \\ :\implies \sf 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 5 \\ \\ :\implies \sf 4 \times 4 \times 4 \times 25 \times 3 \\ \\ :\implies \sf 16 \times 100 \times 3 \\ \\ :\implies \sf 16 \times 300 \\ \\ :\implies \sf 4800 \: m^2$

Second triangle have same area.

Therefore, total area will be 4800 + 4800 = 9600 m sq.