# Thread: Express the area of the trapezoid in terms of A and B.

1. ## Express the area of the trapezoid in terms of A and B.

A trapezoid is divided into four triangles by its diagonals. Suppose the top and bottom triangles have areas of A and B. Express the AREA of the trapezoid in terms of A and B.
Here is a picture of the problem:

http://i216.photobucket.com/albums/c...hchallenge.jpg

I know the solution already, but just want to see if there's a much simpler way.

2. Originally Posted by mathwizard
A trapezoid is divided into four triangles by its diagonals. Suppose the top and bottom triangles have areas of A and B. Express the AREA of the trapezoid in terms of A and B.
Here is a picture of the problem:

http://i216.photobucket.com/albums/c...hchallenge.jpg

I know the solution already, but just want to see if there's a much simpler way.
DISCLAIMER: If this is wrong, just shoot me. I spent way too much time here.

First, draw a line through the intersection of the diagonals and perpendicular to both parallel bases. This will give us the heights of the two triangles whose areas we know.

Label the base of triangle A be $\displaystyle b_1$ and the base of triangle B be $\displaystyle b_2$

The height of triangle A will be $\displaystyle h_1$ and the height of triangle B will be $\displaystyle h_2$

Now, the areas of the two triangles:

$\displaystyle A=\frac{1}{2}b_1h_1$

$\displaystyle B=\frac{1}{2}b_2h_2$

The area of the trapezoid is given as: $\displaystyle T=\frac{1}{2}h(b_1+b_2)$

Substituting, we get,

$\displaystyle T=\frac{1}{2}(h_1+h_2)(b_1+b_2)$

$\displaystyle \triangle{A}\sim\triangle {B}$ by AA postulate

That means:

$\displaystyle \frac{b_1}{b_2}=\frac{h_1}{h_2}$

*$\displaystyle \boxed{h_1b_2=h_2b_1}\Longrightarrow Remember \ \ this!$

$\displaystyle T=\frac{1}{2}h_1+h_2)(b_1+b_2)$

$\displaystyle T=\frac{1}{2}(h_1b_1+h_2b_2+h_1b_2+h_2b_1)$

Since we know from the * above,

$\displaystyle T=\frac{1}{2}h_1b_1 + \frac{1}{2}h_2b_2 +\frac{1}{2} \underbrace{(h_1b_2 + h_2b_1)}_{\text{equal to each other}}$

$\displaystyle T=A+B+h_1b_2$

Now,

$\displaystyle AB=(\frac{1}{2}b_1h_1)(\frac{1}{2}h_2b_2)$

$\displaystyle AB=\frac{1}{4} \underbrace{(b_1h_2)(b_2h_1)}_{\text{equal to each other}}$

$\displaystyle AB=\frac{1}{4}(h_1b_2)^2$

$\displaystyle 4AB=(h_1b_2)^2$

$\displaystyle 2\sqrt{AB}=h_1b_2$

Substituting back into our last T equation we finally get:

$\displaystyle T=A+B+2\sqrt{AB}$