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 $A'$ . If $AB = BB' = 6$ , $BC = CC' = 7$ , $CD = DD' = 8$ , and $DA = AA' = 9$ , and the area of $ABCD$ is 10, determine the area of quadrilateral $A'B'C'D'$ .

$$ (Error compiling LaTeX. Unknown error_msg) [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] $\[\]$ [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] $$ (Error compiling LaTeX. Unknown error_msg)

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).

Page

Toolbox

Search

1978 AHSME Problems/Problem 29

Problem

Solution