2016 AMC 12A Problem 8

Below is the professionally curated solution for Problem 8 of the 2016 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12A solutions, or check the answer key.

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Concepts:triangle areasymmetry

Difficulty rating: 1350

8.

What is the area of the shaded region of the given 8×58\times 5 rectangle?

4344\tfrac{3}{4}

55

5145\tfrac{1}{4}

6126\tfrac{1}{2}

88

Solution:

The diagonal of the rectangle from the upper-left corner to the lower-right corner divides the shaded region into four triangles, all meeting at the center of the rectangle.

Two of these triangles have a horizontal base of length 11 and altitude 125=52,\frac12\cdot 5=\frac52, and the other two have a vertical base of length 11 and altitude 128=4.\frac12\cdot 8=4. The total area is 212152+21214=52+4=132. 2\cdot\dfrac12\cdot 1\cdot\dfrac52+2\cdot\dfrac12\cdot 1\cdot 4=\dfrac52+4=\dfrac{13}{2}.

Thus, the correct answer is D.

Problem 8 in Other Years