Question

using heron's formula find the area of an isoscles triangle whose one side is 7 m greater than its equal side and perimeter is 70 m

Answer 1

Perimeter=x+x+x +7=70
3x+7=70
3x=70-7
x= 63/3=21m
here a 21,b=21,c=28
S=a+b+c+/2= 70/2 =35
=√s(s-a)(s-b)(s-c)
=√35x14x13x7
=7x7x2√5=98√5 m²

Answer 2

Solution :

Here , we are given an isosceles triangle of perimeter 70 m ; and we need to calculate the area of the triangle using Herons formula .

Let us assume that the length of the equal side of the isosceles triangle is x m .

So, the length of the third side is ( x + 7 ) m.

Total perimeter of the triangle :

> x + x + x + 7

> 3x + 7

But , it is given as 70 m.

So,

> 3x + 7 = 70

> 3x = 63

> x = 21 m.

( x + 7) = 28 m.

Thus, the sides of the triangle are 21 m, 21 m and 28m respectively.

We need to find the area .

Semi perimeter = 70/2 = 35.

Area = $\sqrt{s(s-a)(s-b)(s-c) }$

> Square root of [ 35 * 14 * 14 * 7 ]

> 98 root 5 m^2 .

This is the required answer.

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