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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of Isosceles Triangle | AMC 8, 2005 | Problem 23

Try this beautiful problem from AMC-8, 2005, Problem-23 based on the area of an isosceles triangle. You may use sequential hints to solve the problem

Try this beautiful problem from AMC-8, 2005, Problem-23 based on the area of an isosceles triangle.

Area of the Triangle- AMC- 8, 2005 – Problem 23


Isosceles right triangle ABC encloses a semicircle of area \(2\pi\) . The circle has its center O on hypotenuse AB and is tangent to sides AC and BC . What is the area of triangle ABC ?

Area of an isoceles triangle
  • $6$
  • $8$
  • $10$

Key Concepts


Geometry

Triangle

Semi-circle

Check the Answer


Answer:$8$

AMC-8 (2005) Problem 23

Pre College Mathematics

Try with Hints


area of triangle - solution

Join Oand D

Can you now finish the problem ……….

This is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments

can you finish the problem……..

area of triangle - solution

Given that AB and AC are the two sides of the Isosceles right triangle ABC and encloses a semicircle of area \(2\pi\), center of the semicircle is O.

Let OD=r be the radius of the semi-circle

then area of semi-circle be \(\frac{\pi r^2}{2}\)

Now \(\frac{\pi r^2}{2}\) = \(2\pi\)

\(\Rightarrow r^2=4\)

\(\Rightarrow r=2\)

this is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments

They also create a square in the top left corner. From this, we can conclude the legs of the triangle are twice the length of the radii, 4

The area of the triangle is \(\frac{1}{2} \times 4 \times 4\)=8

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