Difference between revisions of "2002 AMC 10A Problems/Problem 7"

(New page: ==Problem== A <math>45^\circ</math> arc of circle A is equal in length to a <math>30^\circ</math> arc of circle B. What is the ratio of circle A's area and circle B's area? <math>\text{(...)
 
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Using that here, the arc of circle A has length <math>\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}</math>. The arc of circle B has length <math>\frac{30}{360}\cdot{2\pi{r_2}=\frac{r_2\pi}{6}</math>. We know that they are equal, so <math>\frac{r_1\pi}{4}=\frac{r_2\pi}{6}</math>, so we multiply through and simplify to get <math>\frac{r_1}{r_2}=\frac{2}{3}</math>. As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is <math>\boxed{\text{(A)}\ 4/9}</math>.
 
Using that here, the arc of circle A has length <math>\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}</math>. The arc of circle B has length <math>\frac{30}{360}\cdot{2\pi{r_2}=\frac{r_2\pi}{6}</math>. We know that they are equal, so <math>\frac{r_1\pi}{4}=\frac{r_2\pi}{6}</math>, so we multiply through and simplify to get <math>\frac{r_1}{r_2}=\frac{2}{3}</math>. As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is <math>\boxed{\text{(A)}\ 4/9}</math>.
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==See Also==
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{{AMC10 box|year=2002|ab=A|num-b=6|num-a=8}}
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[[Category:Introductory Geometry Problems]]

Revision as of 18:33, 26 December 2008

Problem

A $45^\circ$ arc of circle A is equal in length to a $30^\circ$ arc of circle B. What is the ratio of circle A's area and circle B's area?

$\text{(A)}\ 4/9 \qquad \text{(B)}\ 2/3 \qquad \text{(C)}\ 5/6 \qquad \text{(D)}\ 3/2 \qquad \text{(E)}\ 9/4$

Solution

Let $r_1$ and $r_2$ be the radii of circles A and B, respectively.

It is well known that in a circle with radius r, a subtended arc opposite an angle of $\theta$ degrees has length $\frac{\theta}{360}\cdot{2\pi{r}$ (Error compiling LaTeX. Unknown error_msg).

Using that here, the arc of circle A has length $\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}$. The arc of circle B has length $\frac{30}{360}\cdot{2\pi{r_2}=\frac{r_2\pi}{6}$ (Error compiling LaTeX. Unknown error_msg). We know that they are equal, so $\frac{r_1\pi}{4}=\frac{r_2\pi}{6}$, so we multiply through and simplify to get $\frac{r_1}{r_2}=\frac{2}{3}$. As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is $\boxed{\text{(A)}\ 4/9}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AMC 10 Problems and Solutions