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 $24\pi \Rightarrow C$ .

1995 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1995 AHSME Problems/Problem 26

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