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AMC 8 Math Olympiad USA Math Olympiad

Problem on Semicircle | AMC 8, 2013 | Problem 20

Try this beautiful problem from AMC-8, 2013, (Problem-20) based on area of semi circle.You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on Semicircle.

Area of the Semicircle – AMC 8, 2013 – Problem 20


A $1\times 2$ rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?

  • \( \frac{\pi}{2}\)
  • \(\pi\)
  • \( \frac{\pi}{3}\)

Key Concepts


Geometry

Square

semi circle

Check the Answer


Answer:\(\pi\)

AMC-8 (2013) Problem 20

Pre College Mathematics

Try with Hints


Problem on Semicircle

At first we have to find out the radius of the semicircle for the area of the semicircle.Now in the diagram,AC is the radius of the semicircle and also AC is the hypotenuse of the right Triangle ABC.

Can you now finish the problem ……….

Now AB=1 AND AC=1 (As ABDE $1\times 2$ rectangle .So using pythagorean theorm we can eassily get the value of AC .and area of semicircle =\(\frac{\pi r^2}{2}\)

can you finish the problem……..

Problem on Semicircle

Given that ABDE is a square whose AB=1 and BD=2

Therefore BC=1

Clearly AC be the radius of the given semi circle

From the \(\triangle ABC\),\((AB)^2+(BC)^2=(AC)^2\Rightarrow AC=\sqrt{(1^2+1^2)}=\sqrt2\)

Therefore the area of the semicircle=\(\frac{1}{2}\times \pi(\sqrt 2)^2\)=\(\pi\)

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