# Difference between revisions of "1993 AJHSME Problems/Problem 18"

## 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.

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

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

 1993 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 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