1985 AJHSME Problems/Problem 4

Solution

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 $FE$ until it reaches the right side, 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 $6\times5 = 30$ . For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other, so we'll call it $x$ . And we'll also color in the part of the right side that it does cover.

We realize that the space that $x$ doesn't cover is 5, so $x$ must be $9 - 5 = 4$ . So the area of the bottom rectangle is $4\times4$ , or $16$ .

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

$C$

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1985 AJHSME Problems/Problem 4

Solution