1985 AJHSME Problems/Problem 4

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Problem

The area of polygon $ABCDEF$, in square units, is

$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Solution 1

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((2,4)--(6,4),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(6,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of.

If we continue segment $\overline{FE}$ until it reaches the right side at $G$, we create two rectangles - one on the top and one on the bottom.

We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\times5 = 30$. For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.

Note that $GC+GB=9$, and $GB=AF=5$, so we must have \[GC+5=9\Rightarrow GC=4\]

The area of the bottom rectangle is then \[(DC)(GC)=4\times 4=16\]

Finally, we just add the areas of the rectangles together to get $16 + 30 = 46$.

$\boxed{\text{C}}$

Solution 2

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(0,0),dashed); draw((0,0)--(2,0),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(0,0),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Let $\langle ABCDEF \rangle$ be the area of polygon $ABCDEF$. Also, let $G$ be the intersection of $DC$ and $AF$ when both are extended.

Clearly, \[\langle ABCDEF \rangle = \langle ABCG \rangle - \langle GFED \rangle\]

Since $AB=6$ and $BC=9$, $\langle ABCG \rangle =6\times 9=54$.

To compute the area of $GFED$, note that \[AB=GD+DC\] \[BC=GF+FA\]

We know that $AB=6$, $DC=4$, $BC=9$, and $FA=5$, so \[6=GD+4\Rightarrow GD=2\] \[9=GF+5\Rightarrow GF=4\]

Thus $\langle GFED \rangle = 4\times 2=8$

Finally, we have \begin{align*} \langle ABCDEF \rangle &= \langle ABCG \rangle - \langle GFED \rangle \\ &= 54-8 \\ &= 46 \\ \end{align*}

This is answer choice $\boxed{\text{C}}$

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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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