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# Length of a Tangent | AMC-10A, 2004 | Problem 22

Try this beautiful problem from AMC-10A, 2004 based on Triangle. You may use sequential hints to solve the problem.

Try this beautiful problem from American Mathematics Competitions, AMC 10A, 2004 based on Geometry: Length of a Tangent.

## Length of a Tangent – AMC-10A, 2004- Problem 22

Square $ABCD$ has a side length $2$. A Semicircle with diameter $AB$ is constructed inside the square, and the tangent  to the semicircle from $C$ intersects side $AD$at $E$. What is the length of $CE$?

• $\frac{4}{3}$
• $\frac{3}{2}$
• $\sqrt 3$
• $\frac{5}{2}$
• $1+\sqrt 3$

### Key Concepts

Square

Semi-circle

Geometry

But try the problem first…

Answer: $\frac{5}{2}$

Source

AMC-10A (2004) Problem 22

Pre College Mathematics

## Try with Hints

First hint

We have to find out length of $CE$.Now $CE$ is a tangent of inscribed the semi circle .Given that length of the side is $2$.Let $AE=x$.Therefore $DE=2-x$. Now $CE$ is the tangent of the semi-circle.Can you find out the length of $CE$?

Can you now finish the problem ……….

Second Hint

Since $EC$ is tangent,$\triangle COF$ $\cong$ $\triangle BOC$ and $\triangle EOF$ $\cong$ $\triangle AOE$ (By R-H-S law).Therefore $FC=2$ & $EC=x$.Can you find out the length of $EC$?

can you finish the problem……..

Final Step

Now the $\triangle EDC$ is a Right-angle triangle……..

Therefore $ED^2+ DC^2=EC^2$ $\Rightarrow (2-x)^2 + 2^2=(2+x)^2$ $\Rightarrow x=\frac{1}{2}$

Hence $EC=EF+FC=2+\frac{1}{2}=\frac{5}{2}$

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