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AMC-8 Geometry Math Olympiad

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

Try this beautiful problem from Geometry based radius of a semicircle inscribed in a isosceles triangle. You may use sequential hints to solve the problem

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?

Semicircle in an isosceles triangle

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

Key Concepts


Geometry

Area

pythagoras

Check the Answer


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

semicircle inscribed in an isosceles triangle

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……..

radius of a semicircle in an isosceles triangle

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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