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. Find the area of $A'B'C'D'$ .

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).

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1978 AHSME Problems/Problem 29

Problem

Solution