AMC-8 Math Olympiad USA Math Olympiad

Radius of semicircle | AMC-8, 2013 | Problem 23

Try this beautiful problem from Geometry: Radius of the semicircle

Radius of the semicircle- AMC-8, 2013- Problem 23

Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$, and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$. What is the radius of the semicircle on $\overline{BC}$?

radius of semicircle
  • \(9\)
  • \(7.5\)
  • \(6\)

Key Concepts




Check the Answer

Answer: \(7.5\)

AMC-8 (2013) Problem 23

Pre College Mathematics

Try with Hints

We have to find out the radius of the semi-circle on ${BC}$? Now ABC is a Right angle if you find out AB and AC then BC will be easily calculated…..

Can you now finish the problem ……….

To find the value of AB and AC, notice that area of the semi-circle on AB is given and length of the arc of AC is given….

can you finish the problem……..

radius of semicircle

Let the length of AB=\(2x\),So the radius of semi-circle on AB=\(x\).Therefore the area =\(\frac{1}{2}\times \pi (\frac{x}{2})^2=8\pi\)\(\Rightarrow x=4\)

Theregfore length of AB=\(2x\)=8

Given that the arc of the semi-circle on $\overline{AC}$ has length $8.5\pi$,let us take the radius of the semicircle on AB =\(r\).now length of the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$ i.e half perimeter=\(8.5\)

so \(\frac{1}{2}\times {2\pi r}=8.5\pi\)\(\Rightarrow r=8.5\)\(\Rightarrow 2r=17\).so AC=17

The triangle ABC is a Right Triangle,using pythagorean theorm……

\((AB)^2 + (BC)^2=(AC)^2\)\(\Rightarrow BC=\sqrt{(AC)^2 -(AB)^2}\)\(\Rightarrow BC =\sqrt{(17)^2 -(8)^2}\)\(\Rightarrow BC= \sqrt{289 -64 }\)\(\Rightarrow BC =\sqrt {225}=15\)

Therefore the radius of semicircle on BC =\(\frac{15}{2}=7.5\)

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