# 1995 AHSME Problems/Problem 26

## Problem

In the figure, $\overline{AB}$ and $\overline{CD}$ are diameters of the circle with center $O$, $\overline{AB} \perp \overline{CD}$, and chord $\overline{DF}$ intersects $\overline{AB}$ at $E$. If $DE = 6$ and $EF = 2$, then the area of the circle is $[asy]size(120); defaultpen(linewidth(0.7)); pair O=origin, A=(-5,0), B=(5,0), C=(0,5), D=(0,-5), F=5*dir(40), E=intersectionpoint(A--B, F--D); draw(Circle(O, 5)); draw(A--B^^C--D--F); dot(O^^A^^B^^C^^D^^E^^F); markscalefactor=0.05; draw(rightanglemark(B, O, D)); label("A", A, dir(O--A)); label("B", B, dir(O--B)); label("C", C, dir(O--C)); label("D", D, dir(O--D)); label("F", F, dir(O--F)); label("O", O, NW); label("E", E, SE);[/asy]$ $\mathrm{(A) \ 23 \pi } \qquad \mathrm{(B) \ \frac {47}{2} \pi } \qquad \mathrm{(C) \ 24 \pi } \qquad \mathrm{(D) \ \frac {49}{2} \pi } \qquad \mathrm{(E) \ 25 \pi }$

## Solution

Solution 1

Let the radius of the circle be $r$ and let $x=\overline{OE}$.

By the Pythagorean Theorem, $OD^2+OE^2=DE^2 \Rightarrow r^2+x^2=6^2=36$.

By Power of a point, $AE \cdot EB = DE \cdot EF \Rightarrow (r+x)(r-x)=r^2-x^2=6\cdot2=12$.

Adding these equations yields $2r^2=48 \Rightarrow r^2 = 24$.

Thus, the area of the circle is $\pi r^2 = 24\pi \Rightarrow C$.

Solution 2

Let the radius of the circle be $r$.

We can see that $\triangle CFD$ has a right angle at $F$ and that $\triangle EOD$ has a right angle at $O$.

Both triangles also share $\angle ODE$, so $\triangle CFD$ and $\triangle EOD$ are similar.

This means that $\frac {DE}{OD}=\frac {CD}{FD}$.

So, $\frac {6}{r}=\frac {2r}{8}$. Simplifying, $r^2 = 24$.

This means the area is $24\pi \Rightarrow C$.

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