In the arrow-shaped polygon [see figure], the angles at vertices $A,C,D,E$ and $F$ are right angles, $BC=FG=5, CD=FE=20, DE=10$, and $AB=AG$. The area of the polygon is closest to
$\text{(A) } 288\quad \text{(B) } 291\quad \text{(C) } 294\quad \text{(D) } 297\quad \text{(E) } 300$
$\fbox{E}$
Since they tell us that AB=AG and that angle A is a right angle, triangle ABG is a 45-45-90 triangle. Thus we can find the legs of the triangle by dividing $20$ (the hypotenuse) by $\sqrt2$. Both legs have length $10\sqrt2$, so the area of the right triangle is $100$. The rectangle is simple, just $20\times10$, so the area is 200. Adding these areas, we get $300$ as the total area.