1993 AJHSME Problems/Problem 18

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Problem

The rectangle shown has length $AC=32$, width $AE=20$, and $B$ and $F$ are midpoints of $\overline{AC}$ and $\overline{AE}$, respectively. The area of quadrilateral $ABDF$ is

[asy] pair A,B,C,D,EE,F; A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10); draw(A--C--D--EE--cycle); draw(B--D--F); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,W); [/asy]

$\text{(A)}\ 320 \qquad \text{(B)}\ 325 \qquad \text{(C)}\ 330 \qquad \text{(D)}\ 335 \qquad \text{(E)}\ 340$

Solution

The area of the quadrilateral $ABDF$ is equal to the areas of the two right triangles $\triangle BCD$ and $\triangle EFD$ subtracted from the area of the rectangle $ABCD$. Because $B$ and $F$ are midpoints, we know the dimensions of the two right triangles.

[asy] pair A,B,C,D,EE,F; A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10); draw(A--C--D--EE--cycle); draw(B--D--F); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,W); label("$16$",A--B,N); label("$16$",B--C,N); label("$32$",E--D,S); label("$10$",E--F,W); label("$10$",A--F,W); label("$20$",C--D,E); [/asy]

\[(20)(32)-\frac{(16)(20)}{2}-\frac{(10)(32)}{2} = 640-160-160 = \boxed{\text{(A)}\ 320}\]

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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