What is the area of an isosceles triangle with a perimeter of 36cm and an altitude of 12cm?

Answer 1

Let x be the length measure of the base, so the length of one side is (36 - x)/2.

Let's look at one of the right triangles that are formed when we draw the altitude of the given isosceles triangle. There we have:

hypotenuse = (36 - x)/2

base = x/2

height = 12

By the Pythagorean theorem we have:

hypotenuse^2 - base^2 = height^2

Substitute what we know, and we have:

[(36 - x)/2]^2 - (x/2)^2 = 12^2

(36^2 - (2)(36)(x) + x^2]/4 - (x^2)/4 = 144

(1,296 - 72x + x^2)/4 -( x^2)/4 = 144 multiply by 4 both sides;

1,296 - 72x + x^2 - x^2 = 576 add both x^2 and -x^2 and subtract 1,296 to both sides;

-72x = -720 divide by -72 to both sides;

x = 10

So the base of the isosceles triangle is 10 ft, and the height is 12 ft. now we can find the area of the isosceles triangle which is:

A = (bh)/2 = (10 x 12)/2 = 60 ft^2