Question

Find the area of isosceles triangle by using heron formula where the measure of one its equal sides is a and third side is being b?

Answer 1

Heron's Formula:√(s(s-a)(s-b)(s-c)

Here a=a,b=a&c=b

Perimeter=a+a+b=2a+b

Hence,s=2a+b/2

So,area=√(2a+b/2)([2a-b/2]-a)

([2a-b/2]-a)([2a+b/2]-b])-----1

=√(2a+b/2)(2a+b-2a/2)

(2a+b-2a/2)(2a+b-2b/2)----2

=√(2a+b/2)(b/2)(b/2)

(2a-b/2)--------------------------3

=b/2√((2a/2)^2-(b/2)^2)--------4

=b/2√((a^2)-((b/2)^2))-----------5 So, Area=b/2√(a^2)-(b^2/4)

OR

After 4,

*use the algebraic identity,

a^2-b^2=(a+b)(a-b)

So, area=b/2√(a+b/2)(a-b/2)

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