AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 15 Easy

In the figure, a semicircle with diameter EF intersects each side of rectangle ABCD at exactly one point. The lengths of segments EF and FC are 20 centimeters and 2 centimeters, respectively. The area of the shaded region is a- b \cdot \pi square centimeters. Find the value of a+b. (Assuming the value of \pi is 3.14.)

  • A.

    326

  • B.

    338

  • C.

    348

  • D.

    444

  • E.

    500

Answer:B

Let EF be a line segment with midpoint O. Draw a perpendicular line OP from O to CD at point P. Connect OM and ON.

Then, OM is perpendicular to AB, and ON is perpendicular to BC. Given that EF has a length of 20\text{ cm}, we can deduce that OE = OF = OM = ON = 10\text{ cm}. Therefore, PF can be calculated as 10 - 2 = 8\text{ cm}.

The length of OP is 6\text{ cm}, and since OP bisects CD, we have DP = PF = 8\text{ cm}. Consequently, AD = MP = 16\text{ cm}.

The length of BC is given as 8 + 8 + 2 = 18\text{ cm}.

Now, we want to find the area of the shaded region, denoted as S_{\text{shaded}}. This can be obtained by subtracting the area of the semicircle from the area of quadrilateral ABCD. The area of quadrilateral ABCD can be calculated as AD \times BC = 16 \times 18. The area of the semicircle can be calculated as \frac{1}{2}\pi \times r^2, where the radius r is \frac{1}{2} \times EF = 10\text{ cm}. Thus, the area of the shaded region is:

S_{\text{shaded}} = 16 \times 18 - \frac{1}{2} \times \pi \times 10^2 = 288 - 50\pi.

Therefore, the answer is 288 + 50 =\boxed{\textbf{(B)} 338}.

Related Problems
Problem 13 AMC 10 Daily Practice Round 4
Problem 14 AMC 10 Daily Practice Round 4
Problem 16 AMC 10 Daily Practice Round 4
Problem 17 AMC 10 Daily Practice Round 4
Think Academy Powered by ChatGPT