AMC 8 Daily Practice - Pythagorean Theorem

Complete problem set with solutions and individual problem pages

Problem 7 Easy

In the rectangle, the arcs of the two sectors are tangent to each other. What is the area of the shaded part?

  • A.

    6\pi

  • B.

    7\pi

  • C.

    8\pi

  • D.

    9\pi

  • E.

    10\pi

Answer:E

As shown in the figure, connect the centers of the semicircle and the sector.

Let the radius of the semicircle be r.

Since AB = DC, the radius of the sector is 2r.

In \triangle OBC, by the Pythagorean theorem:

r^2 + 4^2 = (r + 2r)^2

r^2 + 16 = 9r^2

8r^2 = 16 \implies r^2 = 2

The area of the shaded part is: \pi r^2 + \pi (2r)^2 = \pi r^2 + 4\pi r^2 = 5\pi r^2 = 5\pi \times 2 = 10\pi

Related Problems
Problem 4 AMC 8 Daily Practice - Pythagorean Theorem
Problem 5 AMC 8 Daily Practice - Pythagorean Theorem
Problem 6 AMC 8 Daily Practice - Pythagorean Theorem
Problem 8 AMC 8 Daily Practice - Pythagorean Theorem
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