Question

Find the area of the triangle whose lengths of sides are 15m, 17m, 21m (use Heron’s Formula) and verify your answer by using the formula A = 1/2 bh .

Answer 1

Dear Student,

Answer:126.18 sq-m

Solution:

Heron's Formula:

Area of triangle =
$\sqrt{s(s - a)(s - b)(s - c)}$
here S is semi-perimeter of the triangle

here a,b and c are the sides of triangle

$s = \frac{a + b + c}{2}$
a = 15 m

b = 17 m

c = 21 m

$s \: = \frac{15 + 17 + 21}{2} \\ \\ s = 26.5 \: m$
Area
$= \sqrt{26.5(26.5 - 15)(26.5 - 17)(26.5 - 21)} \\ \\ = \sqrt{26.5 \times 11.5 \times 9.5 \times 5.5} \\ \\ = \sqrt{15923.1875} \\ \\ = 126.18 \: {m}^{2}$
Area of triangle by heron's Formula = 126.18 sq-m

Since the given triangle is not right angle triangle so we can't apply 1/2 (base)(height ) formula.
and no additional information given about altitude of triangle.

Hope it helps you.

Answer 2

Hi ,

********************************************

Heron's formula :

Area of the triangle whose lengths

of the sides are a , b , c is A .

A = √[ s( s - a )( s - b )( s - c )]

Where , s = ( a + b + c )/2

*********************************************

a = 15 , b = 17 , c = 21 ,

s = ( 15 + 17 + 21 )/2

s = 53/2 = 26.5

A=√[26.5(26.5 - 15)(26.5 - 17)(26.5-21)]

A = √( 26.5 × 11.5 × 9.5 × 5.5 )

A = 126.19 sq m
___________________________

Here ,. a = 15 m , b = 17 m , c = 21 m

not a Phythogarian triplet .

Therefore ,

It is not possible to find area using

A = ( bh )/2 formula

I hope this helps you.

: )