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 \(CE\)?

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 ..........

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........

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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