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

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 ?

- $6$
- $8$
- $10$

Geometry

Triangle

Semi-circle

But try the problem first...

Answer:$8$

Source

Suggested Reading

AMC-8 (2005) Problem 23

Pre College Mathematics

First hint

Join Oand D

Can you now finish the problem ..........

Second Hint

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........

Final Step

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

- https://www.cheenta.com/ratio-of-the-area-between-square-and-pentagon-amc-8-2013-problem-24/
- https://www.youtube.com/watch?v=y8jLtbhc3iU

Contents

[hide]

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

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 ?

- $6$
- $8$
- $10$

Geometry

Triangle

Semi-circle

But try the problem first...

Answer:$8$

Source

Suggested Reading

AMC-8 (2005) Problem 23

Pre College Mathematics

First hint

Join Oand D

Can you now finish the problem ..........

Second Hint

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........

Final Step

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

- https://www.cheenta.com/ratio-of-the-area-between-square-and-pentagon-amc-8-2013-problem-24/
- https://www.youtube.com/watch?v=y8jLtbhc3iU

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