1985 AJHSME Problems/Problem 4

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 $\boxed{\text{C}}$

Video Solution by BoundlessBrain!

https://youtu.be/-_r5GzafCUY

Video Solution

https://youtu.be/9OHRebmMyz4

~savannahsolver

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AJHSME/AMC 8 Problems and Solutions

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