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

**mathwizard** · June 19th 2008, 04:01 PM

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.

**masters** · June 20th 2008, 02:26 PM

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 $b_1$ and the base of triangle B be $b_2$

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

Now, the areas of the two triangles:

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

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

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

Substituting, we get,

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

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

That means:

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

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

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

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

Since we know from the * above,

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

$T=A+B+h_1b_2$

Now,

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

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

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

$4AB=(h_1b_2)^2$

$2\sqrt{AB}=h_1b_2$

Substituting back into our last T equation we finally get:

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

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