2018 AMC 8 Problem 15

Below is the video solution and professionally curated solution for Problem 15 of the 2018 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 8 solutions, or check the answer key.

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Concepts:circle areapower scaling of length, area, and volume

Difficulty rating: 1070

15.

In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of 11 square unit, then what is the area of the shaded region, in square units?

14 \dfrac{1}{4}

13 \dfrac{1}{3}

12 \dfrac{1}{2}

1 1

π2 \dfrac{\pi}{2}

Video solution:
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Written solution:

Let AA be the area of the large circle.

Since the diameter of each of the two smaller circles is itself the radius of the larger circle, the radius of each smaller circle is half that of the larger circle.

Symbolically, if we allow rlr_l to be the radius of the large circle and rsr_s to be the radius of each of the smaller circles: rs=12rlr_s = \dfrac12 r_l As the area of the larger circle is equal to A=πrl2,A=\pi r_l^2, the area of the smaller circles are equal to πrs2=π(12rl)2=14(πrl2)=14A.\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A.\end{align*} As the area of two of these smaller circles combined is equal to 1 square unit, then it follows that 214A=12\cdot \dfrac14 A=1 square unit, implying that A=2A=2 square units.

As the area of the shaded region is equal to the area of the larger circle (A)(A) minus the combined area of the two smaller circles (1),(1), the area of the shaded region is A1=21=1 A - 1=2-1=1 square unit.

Thus, the correct answer is D

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