1995 AHSME Problems/Problem 26

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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

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$\mathrm{(A) \ 23 \pi } \qquad \mathrm{(B) \ \frac {47}{2} \pi } \qquad \mathrm{(C) \ 24 \pi } \qquad \mathrm{(D) \ \frac {49}{2} \pi } \qquad \mathrm{(E) \ 25 \pi }$

Solution

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 $\pi r^2 = 24\pi \Rightarrow C$ .

1995 AHSME (Problems)
Preceded by Problem 25	Followed by Problem 27
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1995 AHSME Problems/Problem 26

From AoPSWiki

Problem

Solution

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