1978 AHSME Problems/Problem 29
Problem
Sides , , , and , respectively of convex quadrilateral are extended past , , , and to points , , , and . If , , , and , and the area of is 10, determine the area of quadrilateral .
$$ (Error compiling LaTeX. Unknown error_msg) [asy] unitsize(1 cm);
pair[] A, B, C, D;
A[0] = (0,0); B[0] = (0.6,1.2); C[0] = (-0.3,2.5); D[0] = (-1.5,0.7); B[1] = interp(A[0],B[0],2); C[1] = interp(B[0],C[0],2); D[1] = interp(C[0],D[0],2); A[1] = interp(D[0],A[0],2);
draw(A[1]--B[1]--C[1]--D[1]--cycle); draw(A[0]--B[1]); draw(B[0]--C[1]); draw(C[0]--D[1]); draw(D[0]--A[1]);
label("", A[0], SW); label("", B[0], SE); label("", C[0], NE); label("", D[0], NW); label("", A[1], SE); label("", B[1], NE); label("", C[1], N); label("", D[1], SW);
[\asy] [asy] unitsize(1 cm);
pair[] A, B, C, D;
A[0] = (0,0); B[0] = (0.6,1.2); C[0] = (-0.3,2.5); D[0] = (-1.5,0.7); B[1] = interp(A[0],B[0],2); C[1] = interp(B[0],C[0],2); D[1] = interp(C[0],D[0],2); A[1] = interp(D[0],A[0],2);
draw(A[1]--B[1]--C[1]--D[1]--cycle); draw(A[0]--B[1]); draw(B[0]--C[1]); draw(C[0]--D[1]); draw(D[0]--A[1]);
label("", A[0], SW); label("", B[0], SE); label("", C[0], NE); label("", D[0], NW); label("", A[1], SE); label("", B[1], NE); label("", C[1], N); label("", D[1], SW);
[\asy] [asy] unitsize(1 cm);
pair[] A, B, C, D;
A[0] = (0,0); B[0] = (0.6,1.2); C[0] = (-0.3,2.5); D[0] = (-1.5,0.7); B[1] = interp(A[0],B[0],2); C[1] = interp(B[0],C[0],2); D[1] = interp(C[0],D[0],2); A[1] = interp(D[0],A[0],2);
draw(A[1]--B[1]--C[1]--D[1]--cycle); draw(A[0]--B[1]); draw(B[0]--C[1]); draw(C[0]--D[1]); draw(D[0]--A[1]);
label("", A[0], SW); label("", B[0], SE); label("", C[0], NE); label("", D[0], NW); label("", A[1], SE); label("", B[1], NE); label("", C[1], N); label("", D[1], SW);
[\asy] $$ (Error compiling LaTeX. Unknown error_msg)
Solution
Notice that the area of is the same as that of (same base, same height). Thus, the area of is twice that (same height, twice the base). Similarly, [ ] = 2 [ ], and so on.
Adding all of these, we see that the area the four triangles around is twice [ ] + [ ] + [ ] + [ ], which is itself twice the area of the quadrilateral . Finally, [] = [] + 4 [] = 5 [] = .
~ Mathavi
Note: Anyone with a diagram would be of great help (still new to LaTex).