Try this beautiful problem from Geometry: Radius of the semicircle
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 $\overline{BC}$?
Geometry
Triangle
Semi-circle
But try the problem first...
Answer: \(7.5\)
AMC-8 (2013) Problem 23
Pre College Mathematics
First hint
We have to find out the radius of the semi-circle on ${BC}$? Now ABC is a Right angle Triangle.so if you find out AB and AC then BC will be easily calculated.....
Can you now finish the problem ..........
Second Hint
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........
Final Step
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 =\(\frac{15}{2}=7.5\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
Try this beautiful problem from Geometry: Radius of the semicircle
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 $\overline{BC}$?
Geometry
Triangle
Semi-circle
But try the problem first...
Answer: \(7.5\)
AMC-8 (2013) Problem 23
Pre College Mathematics
First hint
We have to find out the radius of the semi-circle on ${BC}$? Now ABC is a Right angle Triangle.so if you find out AB and AC then BC will be easily calculated.....
Can you now finish the problem ..........
Second Hint
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........
Final Step
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 =\(\frac{15}{2}=7.5\)
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
