Similar Pentagon Areas Part 2

A pentasquish consists of two $45-45-90$ triangles with an isosceles triangle between them:

The Pythagorean Theorem can be used to find the height of the isosceles triangle:

$h^2+(\frac{s}{2})^2=(s\sqrt{2})^2$

Solving this yields $h=\frac{s\sqrt{7}}{2}$

The area of this isosceles triangle is $\frac{1}{2}\times s\times\frac{s\sqrt{7}}{2}=\frac{s^2\sqrt{7}}{4}$

The two $45-45-90$ triangles together have an area of $s^2$ .

Thus, the area of the pentasquish is $\left(1+\frac{\sqrt{7}}{4}\right)s^2$

$a=1$ , $b=7$ , and $c=4$ , and so $a+b+c=\boxed{12}$ .

WLOG, we may assume that s=1. Draw segments from the top vertex of the pentasquish to the 2 vertices of the pentasquish. This gives us 2 right triangles with legs equal to 1, and an isosceles triangle both base 1. The area of each of the right triangles is 1/2, and the hypotenuse of each of these triangles is sqrt(2). Drop a perpendicular from the top vertice to the base of the pentasquish splitting the isosceles triangle into 2 equivalent right triangles with equal sides of length sqrt(2). Using the Pythagorean theorem, the height of this isosceles triangles is sqrt(7)/2. Thus the total area is 1/2+1/2+sqrt(7)/4=1+sqrt(7)/4. Thus a=1, b=7, and c=4. Hence a+b+c=12.

Why complicate, when it is already simple :) link text