SOLUTION: find the area of an isosceles triangle with side lengths 15,15, and 18.

Hi, there--

The Problem--
Find the area of an isosceles triangle with side lengths 15,15, and 18.

The Solution:

The formula for the area of a triangle is 
A = (1/2)*B*H

where B is the base and H is the height. The base of this triangle is 18. 

Now we need the height.

An isosceles triangle can be split into two congruent right triangles by drawing a line segment
 from the vertex of the angle across from the base to the base so that the line segment is 
perpendicular to the base (90 degree angle). The segment cuts the base exactly in half. The 
length of this segment is the height of the big triangle.

Now you have two right triangles each with hypotenuse of 15 and base 9. (The base of the 
smaller triangles is 9 because 9 is half of 18.)

Since they are right triangles, we can use the Pythagorean Equation to find the height

a^2 + b^2 = c^2

Let a be the base, a=9.
Let b be the height. That's we are trying to find.
Let c be the hypotenuse, c=15.

9^2 + b^2 = 15^2
81 + b^2 = 225
b^2 = 144
b = 12

The small triangles have a height of 12. They each have an area of

A = (1/2)*(12)*(9)
A = 54

The two small triangles combine to give the area of the large triangle so its area is 
54+54 = 108.

Best,

Mrs.Figgy