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

$\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad \textbf{(E) }60$

Solution

\usepackage{asymptote}

\begin{asy} [asy] unitsize(1 cm);

pair A, Ap, B, C, P, Q;

A = 3*dir(60); Ap = (1,0); B = (0,0); C = (3,0); P = 8/5*dir(60); Q = C + 5/4*dir(120);

draw(B--P--Q--C--cycle); draw(P--Ap--Q); draw(P--A--Q,dashed);

label(" $A$ ", A, N); label(" $A'$ ", Ap, S); label(" $B$ ", B, SW); label(" $C$ ", C, SE); label(" $P$ ", P, NW); label(" $Q$ ", Q, NE); [/asy] \end{asy}

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