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

## Check the Answer

But try the problem first…

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

AMC-8, 2016 problem 25

Challenges and Thrills of Pre College Mathematics

## Try with Hints

First hint

Draw a perpendicular from the point C on base AB

Can you now finish the problem ……….

Second Hint

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

Find AC and area

can you finish the problem……..

Final Step

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 $x=\frac {120}{7}$

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