# 1978 AHSME Problems/Problem 29

## Problem

Sides $AB,~ BC, ~CD$ and $DA$, respectively, of convex quadrilateral $ABCD$ are extended past $B,~ C ,~ D$ and $A$ to points $B',~C',~ D'$ and $A'$. Also, $AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8$ and $DA = AA' = 9$; and the area of $ABCD$ is $10$. The area of $A 'B 'C'D'$ is

## Solution

Notice that the area of $\triangle$ $DAB$ is the same as that of $\triangle$ $A'AB$ (same base, same height). Thus, the area of $\triangle$ $A'AB$ is twice that (same height, twice the base). Similarly, [ $\triangle$ $BB'C$] = 2 $\cdot$ [ $\triangle$ $ABC$], and so on.

Adding all of these, we see that the area the four triangles around $ABCD$ is twice [ $\triangle$ $DAB$] + [ $\triangle$ $ABC$] + [ $\triangle$ $BCD$] + [ $\triangle$ $CDA$], which is itself twice the area of the quadrilateral $ABCD$. Finally, [ $A'B'C'D'$] = [ $ABCD$] + 4 $\cdot$ [ $ABCD$] = 5 $\cdot$ [ $ABCD$] = $\fbox{50}$.

~ Mathavi

Note: Anyone with a diagram would be of great help (still new to LaTex).