1978 AHSME Problems/Problem 29

Revision as of 13:17, 23 August 2022 by Mathavi (talk | contribs) (Problem)

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).