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# Radius of a Semi Circle -AMC 8, 2017 - Problem 22

Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.

## AMC-8(2017) - Geometry (Problem 22)

In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?

• $\frac{7}{6}$
• $\frac{10}{3}$
• $\frac{9}{8}$

### Key Concepts

Geometry

congruency

similarity

Answer:$\frac{10}{3}$

AMC-8(2017)

Pre College Mathematics

## Try with Hints

Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.

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

Now the $\triangle ODB$and $\triangle OCB$ are congruent

can you finish the problem........

Let x be the radius of the semi circle

Now the $\triangle ODB$ and $\triangle OCB$ we have

OD=OC

OB=OB

$\angle ODB$=$\angle OCB$= 90 degree

so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)

BD=BC=5

And also $\triangle ODA$ and $\triangle BCA$ are similar....

$\frac{8}{12}$=$\frac{x}{5}$

i.e x =$\frac{10}{3}$

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Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.

## AMC-8(2017) - Geometry (Problem 22)

In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?

• $\frac{7}{6}$
• $\frac{10}{3}$
• $\frac{9}{8}$

### Key Concepts

Geometry

congruency

similarity

Answer:$\frac{10}{3}$

AMC-8(2017)

Pre College Mathematics

## Try with Hints

Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.

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

Now the $\triangle ODB$and $\triangle OCB$ are congruent

can you finish the problem........

Let x be the radius of the semi circle

Now the $\triangle ODB$ and $\triangle OCB$ we have

OD=OC

OB=OB

$\angle ODB$=$\angle OCB$= 90 degree

so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)

BD=BC=5

And also $\triangle ODA$ and $\triangle BCA$ are similar....

$\frac{8}{12}$=$\frac{x}{5}$

i.e x =$\frac{10}{3}$

## Subscribe to Cheenta at Youtube

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