1985 AJHSME Problems/Problem 4

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Problem

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

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

Solution

Solution 1

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

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

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 $\begin{align*}\langle ABCDEF \rangle &= \langle ABCG \rangle - \langle GFED \rangle \\&= 54-8 \\&= 46 \\\end{alig...$

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

1985 AJHSME (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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