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

Contents

Length of a Tangent - AMC-10A, 2004- Problem 22
- Key Concepts
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Try with Hints
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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

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But try the problem first...

Answer: \(\frac{5}{2}\)

Source

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

Shaded Triangle to find the length of the tangent

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 \(EC=EF+FC=2+\frac{1}{2}=\frac{5}{2}\)