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## Circles and semi-circles| AMC 8, 2010|Problem 23

Try this beautiful problem from Geometry based on Ratio of the area of circle and semi-circles.

## Area of circles and semi-circles – AMC-8, 2010 – Problem 23

Semicircles POQ and ROS pass through the center O. What is the ratio of the combined areas of the two semicircles to the area of circle O?

• $\frac{1}{2}$
• $\frac{2}{\pi}$
• $\frac{3}{2}$

### Key Concepts

Geometry

Circle

co-ordinate geometry

Answer:$\frac{1}{2}$

AMC-8 (2010) Problem 23

Pre College Mathematics

## Try with Hints

Find the radius of the circle

Can you now finish the problem ……….

Join O and Q

can you finish the problem……..

The co-ordinate of Q is (1,1), So OB=1 and BQ=1

By the Pythagorean Theorem, the radius of the larger circle i.e OQ=$\sqrt{1^2+1^2}$=$\sqrt 2$.

Therefore the area of the larger circle be $\pi (\sqrt 2)^2=2\pi$

Now for the semicircles, radius OB=OC=1(as co-ordinate of P=(1,1) and S=(1,-1))

So, the area of the two semicircles is  $2\times\frac{\pi(1)^2}{2}=\pi$

Finally, the ratio of the combined areas of the two semicircles to the area of circle O is

$\frac{\pi}{2\pi}$=$\frac{1}{2}$

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## Radius of a Semicircle | AMC 8, 2016 | Problem 25

Try this beautiful problem from Geometry based on Radius of a semicircle inscribed in an isosceles triangle.

## Radius of a Semi circle – AMC-8, 2016 – Problem 25

A semicircle is inscribed in an isoscles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle as shown .what is the radius of the semicircle?

• $\frac{110}{19}$
• $\frac{120}{17}$
• $\frac{9}{5}$

### Key Concepts

Geometry

Area

pythagoras

Answer:$\frac{120}{17}$

AMC-8, 2016 problem 25

Challenges and Thrills of Pre College Mathematics

## Try with Hints

Draw a perpendicular from the point C on base AB

Can you now finish the problem ……….

D be the midpoint of the AB(since $\triangle ABC$ is an isoscles Triangle)

Find AC and area

can you finish the problem……..

Area of the $\triangle ABC= \frac{1}{2} \times AB \times CD$

= $\frac{1}{2} \times 16 \times 15$

=120 sq.unit

Using the pythagoras th. $AC^2= AD^2+CD^2$

i.e $AC^2=(8)^2+(15)^2$

i.e $AC=17$

Let$ED = x$ be the radius of the semicircle

Therefore Area of $\triangle CAD = \frac{1}{2} \times AC \times ED$=$\frac {1}{2} area of \triangle ABC$

i.e $\frac{1}{2} \times AC \times ED$=60

i.e $\frac{1}{2} \times 17 \times x$ =60

i.e $x=\frac {120}{7}$