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Difference between revisions of "1995 AHSME Problems/Problem 26"
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<math> \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 }  </math>
 
<math> \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 }  </math>
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== Solution ==
 
== Solution ==
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'''Solution 1'''
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Let the radius of the circle be <math>r</math> and let <math>x=\overline{OE}</math>.  
 
Let the radius of the circle be <math>r</math> and let <math>x=\overline{OE}</math>.  
  
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Thus, the area of the circle is <math>\pi r^2 = 24\pi \Rightarrow C</math>.
 
Thus, the area of the circle is <math>\pi r^2 = 24\pi \Rightarrow C</math>.
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'''Solution 2'''
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Let the radius of the circle be <math>r</math>.
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We can see that <math>\triangle CFD</math> has a right angle at <math>F</math> and that <math>\triangle EOD</math> has a right angle at <math>O</math>. 
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Both triangles also share <math>\angle ODE</math>, so <math>\triangle CFD</math> and <math>\triangle EOD</math> are [[Similar(geometry)|similar]].
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This means that <math>\frac {DE}{OD}=\frac {CD}{FD}</math>.
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So,  <math>\frac {6}{r}=\frac {2r}{8}</math>. Simplifying, <math>r^2 = 24</math>.
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This means the area is <math>24\pi \Rightarrow C</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 19:01, 23 January 2017

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 $24\pi \Rightarrow C$.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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