1988 AMC 8 Problem 17

Below is the professionally curated solution for Problem 17 of the 1988 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 8 solutions, or check the answer key.

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Concepts:area inclusion-exclusion

Difficulty rating: 960

17.

The shaded area is formed by two intersecting perpendicular rectangles, with dimensions as shown. Its area, in square units, is

$23$

$38$

$44$

$46$

unable to be determined from the information given

Solution:

The horizontal rectangle has area $10 \times 2 = 20,$ and the vertical rectangle has area $3 \times 8 = 24.$ Their overlap, a $3 \times 2$ rectangle, has area $6.$

Adding the two rectangles counts the overlap twice, so the shaded area is $20 + 24 - 6 = 38.$

Thus, the correct answer is B .

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