2002 AMC 10A Problems/Problem 7

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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