# 2004 AMC 8 Problems/Problem 24

## Problem

In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\overline{HE}$ and $\overline{FG}$?

$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(10,0), B=(10,8), A=(0,8); pair E=(4,8), F=(10,3), G=(6,0), H=(0,5); draw(A--B--C--D--cycle); draw(E--F--G--H--cycle); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("E",E,N); label("F",(10.8,3)); label("G",G,S); label("H",H,W); label("4",A--E,N); label("6",B--E,N); label("5",(10.8,5.5)); label("3",(10.8,1.5)); label("4",G--C,S); label("6",G--D,S); label("5",D--H,W); label("3",A--H,W); draw((3,7.25)--(7.56,1.17)); label("d",(3,7.25)--(7.56,1.17), NE); [/asy]$

$\textbf{(A)}\ 6.8\qquad \textbf{(B)}\ 7.1\qquad \textbf{(C)}\ 7.6\qquad \textbf{(D)}\ 7.8\qquad \textbf{(E)}\ 8.1$

## Excellent Solution

The area of the parallelogram can be found in two ways. The first is by taking rectangle $ABCD$ and subtracting the areas of the triangles cut out to create parallelogram $EFGH$. This is $$(4+6)(5+3) - 2 \cdot \frac12 \cdot 6 \cdot 5 - 2 \cdot \frac12 \cdot 3 \cdot 4 = 80 - 30 - 12 = 38$$ The second way is by multiplying the base of the parallelogram such as $\overline{FG}$ with its altitude $d$, which is perpendicular to both bases. $\triangle FGC$ is a $3-4-5$ triangle so $\overline{FG} = 5$. Set these two representations of the area together. $$5d = 38 \rightarrow d = \boxed{\textbf{(C)}\ 7.6}$$

~Ak

~ pi_is_3.14

## See Also

 2004 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 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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