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

Diagram:

Asymptote code below [asy] unitsize(1 cm);

pair[] A, B, C, D;

A[0] = (0,0); B[0] = (0.6,1.2); C[0] = (-0.3,2.5); D[0] = (-1.5,0.7); B[1] = interp(A[0],B[0],2); C[1] = interp(B[0],C[0],2); D[1] = interp(C[0],D[0],2); A[1] = interp(D[0],A[0],2);

draw(A[1]--B[1]--C[1]--D[1]--cycle); draw(A[0]--B[1]); draw(B[0]--C[1]); draw(C[0]--D[1]); draw(D[0]--A[1]);

label(" $A$ ", A[0], SW); label(" $B$ ", B[0], SE); label(" $C$ ", C[0], NE); label(" $D$ ", D[0], NW); label(" $A'$ ", A[1], SE); label(" $B'$ ", B[1], NE); label(" $C'$ ", C[1], N); label(" $D'$ ", D[1], SW); [/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

Page

Toolbox

Search

1978 AHSME Problems/Problem 29

Problem

Solution