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May 2, 2020

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

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 \(CE\)?

Length of a Tangent - Problem
  • \(\frac{4}{3}\)
  • \(\frac{3}{2}\)
  • \(\sqrt 3\)
  • \(\frac{5}{2}\)
  • \(1+\sqrt 3\)

Key Concepts




Check the Answer

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

AMC-10A (2004) Problem 22

Pre College Mathematics

Try with Hints

Length of a Tangent - Problem figure

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 ..........

Shaded figure 2
Shaded figure 1

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........

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

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