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

Geometry
Area
pythagoras
Answer:$\frac{120}{17}$
AMC-8, 2016 problem 25
Challenges and Thrills of Pre College Mathematics
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 $x=\frac {120}{7}$

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