1988 AJHSME Problems/Problem 17


The shaded region formed by the two intersecting perpendicular rectangles, in square units, is

[asy] fill((0,0)--(6,0)--(6,-3.5)--(9,-3.5)--(9,0)--(10,0)--(10,2)--(9,2)--(9,4.5)--(6,4.5)--(6,2)--(0,2)--cycle,black); label("2",(0,.9),W); label("3",(7.3,4.5),N); draw((0,-3.3)--(0,-5.3),linewidth(1)); draw((0,-4.3)--(3.7,-4.3),linewidth(1)); label("10",(4.7,-3.7),S); draw((5.7,-4.3)--(10,-4.3),linewidth(1)); draw((10,-3.3)--(10,-5.3),linewidth(1)); draw((11,4.5)--(13,4.5),linewidth(1)); draw((12,4.5)--(12,2),linewidth(1)); label("8",(11.3,1),E); draw((12,0)--(12,-3.5),linewidth(1)); draw((11,-3.5)--(13,-3.5),linewidth(1)); [/asy]

$\text{(A)}\ 23 \qquad \text{(B)}\ 38 \qquad \text{(C)}\ 44 \qquad \text{(D)}\ 46 \qquad \text{(E)}\ \text{unable to be determined from the information given}$


Looking at the diagram, the shaded region is the union of two rectangles, with a small rectangle as overlap. If we just add the areas of the two rectangles, then we'll overcount the small rectangle, so we must subtract that area to get the desired area.

The areas of the two larger rectangles are $2\cdot 10=20$ and $3\cdot 8=24$, and the area of the small rectangle is $2\cdot 3=6$. The desired area is thus $20+24-6=38 \rightarrow \boxed{\text{B}}$.

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png