AMC 8 Daily Practice - Circles

Complete problem set with solutions and individual problem pages

Problem 5 Easy

Given that the side length of square ABCD is 7 and the side length of square EFGH is 5, and square EFGH has an inscribed circle O, find the area of the circle.

  • A.

    \pi

  • B.

    4

  • C.

    \frac{3}{2}\pi

  • D.

    6

  • E.

    2\pi

Answer:A

It is known that the four right triangles are congruent.

The area of one such right triangle is: \frac{7^2 - 5^2}{4} = \frac{49 - 25}{4} = \frac{24}{4} = 6

Let the radius of circle O be r.

Since circle O is the inscribed circle (incircle) of the right triangle, connecting the three vertices of the triangle to the center O divides the triangle into three smaller triangles, each with a height equal to r.

\frac{1}{2} \times r \times HD + \frac{1}{2} \times r \times GD + \frac{1}{2} \times r \times FG = 6

Factoring out \frac{1}{2}r, we get: \frac{1}{2}r \times (HD + GD + FG) = 6

Notice that HD + GD + FG is equal to the sum of the side length of the large square and the side length of the small square, 7 + 5 = 12.

Substituting this in:

\frac{1}{2}r \times 12 = 6

6r = 6

r = 1

Thus, the area of circle O is: \pi r^2 = \pi \times 1^2 = \pi

Answer: The area of the circle is \pi square units.

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Problem 3 AMC 8 Daily Practice - Circles
Problem 4 AMC 8 Daily Practice - Circles
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