Area of an isosceles triangle = c/4*sqrt(4a^2 - c^2).

Can someone give proof for this?

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- Jun 11th 2014, 08:00 PMhisajeshProof: Area of an isosceles triangle
Area of an isosceles triangle = c/4*sqrt(4a^2 - c^2).

Can someone give proof for this? - Jun 11th 2014, 08:09 PMProve ItRe: Proof: Area of an isosceles triangle
Are "a" and "c" the equal sides of the triangle or are they two sides with different lengths?

- Jun 11th 2014, 08:30 PMTheTutorRe: Proof: Area of an isosceles triangle
Hi hisajesh,

An isosceles triangle has two equal sides - so let's call the length of these sides "a". Also, let's call the other side of the isosceles triangle "c".

Something neat about isosceles triangles, if you create a segment (call it "b") that bisects the angle between the two equivalent sides to "c" (try to visualize this before moving on), then the segment will also bisect the length of "c" into two equal parts.

This will split your isosceles triangle into two congruent right triangles; the hypotenuse of these right triangles is "a", one leg is "c/2", and the last leg is our new segment "b".

Let's try to express "b" in terms of "a" and "c". Since we are dealing with a right triangle, we have a^2 = b^2 + (c/2)^2 (*important* remember that "a" is our hypotenuse - not "c"!!!)

This means b = sqrt (a^2 - (c/2)^2)

Let's try to visualize our isosceles triangle again - we have two equivalent sides ("a") and a third side between them ("c"). Then we created a line segment from the angle between the two equivalent sides to the middle of "side c", and we called this "b".

The area of a triangle is:

A = (1/2)*base*height = (1/2) * c * b = (1/2) * c * sqrt(a^2 - (c/2)^2) = (1/2) * c * sqrt( 1 * (a^2 - (c/2)^2)) = (1/2) * c * sqrt( (4/4) * (a^2 - (c/2)^2))

= (1/2) * c * sqrt( (1/4) * (4*a^2 - 4*(c/2)^2)) = (1/2) * c * sqrt( (1/4)) * sqrt (4*a^2 - 4 * (c/2)^2)

= (1/2) * c * 1/2 * sqrt(4*a^2 - 4*c^2/4) = (1/4) * c * sqrt(4*a^2 - c^2)

= c/4 * sqrt(4*a^2 - c^2)