2007 AMC 8 Problem 23

Attempt Problem 23 of the 2007 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

What is the area of the shaded pinwheel shown in the 5×55 \times 5 grid?
44
66
88
1010
1212
Answer: B
Concepts:area decompositiontriangle area
Small hint:
It is easier to subtract the unshaded area from the whole grid.
Big hint:
The unshaded part consists of four corner squares and four congruent triangles.
Solution:
We can find the area of the shaded part by subtracting the area of the unshaded part from the whole area.
There are 44 squares with side length 1,1, contributing 11 each to the unshaded area, for a total of 4.4.
There are also 44 triangles with base 33 and height 52.\dfrac{5}{2}. This contributes a total area of 4⋅12⋅3⋅52=15. 4 \cdot \dfrac{1}{2} \cdot 3 \cdot \dfrac{5}{2} = 15.
The total unshaded area is therefore 4+15=19.4 + 15 = 19.
The total area is 52=25,5^2 = 25, and subtracting the unshaded area yields 25−19=625 - 19 = 6 as the area of the shaded region.
Thus, B is the correct answer.
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