1985 AJHSME 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

Notice that the polygon is made up of two smaller rectangles with dimensions $5 \times 6$ and $4 \times 4.$ That makes the area $30 + 16 = 46,$ so the answer is $\text{(C) 46.}$

Easier Method

Think of the polygon as a rectangle with a missing piece. We need to find the area of the rectangle, and the missing area. The polygon's length and width is 9 and 6, making the total area $9 \times 6$ = 54. Now, we need to find the missing area. We see that the missing side's lengths are 2 and 4, making the area $4 \times 2$ = 8. We got both the total area of the rectangle, and the missing area. $54 - 8$ = 46, so the answer is $\text{(C) 46.}$

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1985 AJHSME Problem 4

Problem

Solution

Easier Method