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Difference between revisions of "2002 AMC 12A Problems/Problem 7"

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The area of circle <math>A</math> is <math>{\left(\frac{2}{3}\right)}^2=\boxed{\textbf{(A) } 4/9}</math> that of circle <math>B</math>.
 
The area of circle <math>A</math> is <math>{\left(\frac{2}{3}\right)}^2=\boxed{\textbf{(A) } 4/9}</math> that of circle <math>B</math>.
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===Solution 3===
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We can use a very cheap way to find the answer based on the choices. Firstly, if a higher angle measure equals a lower angle measure, that means that the ratio of areas must be lower, cutting D and E. Then we realize that the ratio has to be with square numbers, as there are no square roots. Therefore, it is <math>(A)</math>
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-dragoon
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===Solution 4===
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*Written by a 6th grader, please don't mind*
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Set Circle A's arc to 45 degrees. In a full circle, there is 360 degrees, so 360/45 = 8. If we set the arc to be pi degrees, then in total there will be 8 pi in the circumference of this circle, yielding 4 as the radius and 16pi as the area.
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Since the arc of 30 degrees in Circle B is also pi, and there are 12 30 degree arcs in a circle, there will be 12pi in the circumference of Circle B, yielding 6 as the radius and 36pi as the area.
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Now, we put 16pi/36pi as our ratio, yielding 16/36=4/9. Therefore, it is <math>(A)</math>4/9.
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~MathKatana
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==Video Solution by Daily Dose of Math==
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https://youtu.be/BEn0GJOFbU8
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~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==

Latest revision as of 23:51, 25 July 2024

The following problem is from both the 2002 AMC 12A #7 and 2002 AMC 10A #7, so both problems redirect to this page.


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?

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

Solutions

Solution 1

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 $\frac{\theta}{360} \cdot 2\pi r$.

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 $\frac{30}{360} \cdot 2\pi{r_2}=\frac{r_2\pi}{6}$. 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 $\boxed{\textbf{(A) } 4/9}$.

Solution 2

The arc of circle $A$ is $\frac{45}{30}=\frac{3}{2}$ that of circle $B$.

The circumference of circle $A$ is $\frac{2}{3}$ that of circle $B$ ($B$ is the larger circle).

The radius of circle $A$ is $\frac{2}{3}$ that of circle $B$.

The area of circle $A$ is ${\left(\frac{2}{3}\right)}^2=\boxed{\textbf{(A) } 4/9}$ that of circle $B$.

Solution 3

We can use a very cheap way to find the answer based on the choices. Firstly, if a higher angle measure equals a lower angle measure, that means that the ratio of areas must be lower, cutting D and E. Then we realize that the ratio has to be with square numbers, as there are no square roots. Therefore, it is $(A)$

-dragoon

Solution 4

  • Written by a 6th grader, please don't mind*

Set Circle A's arc to 45 degrees. In a full circle, there is 360 degrees, so 360/45 = 8. If we set the arc to be pi degrees, then in total there will be 8 pi in the circumference of this circle, yielding 4 as the radius and 16pi as the area.

Since the arc of 30 degrees in Circle B is also pi, and there are 12 30 degree arcs in a circle, there will be 12pi in the circumference of Circle B, yielding 6 as the radius and 36pi as the area.

Now, we put 16pi/36pi as our ratio, yielding 16/36=4/9. Therefore, it is $(A)$4/9.

~MathKatana

Video Solution by Daily Dose of Math

https://youtu.be/BEn0GJOFbU8

~Thesmartgreekmathdude

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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
All AMC 12 Problems and Solutions
2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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
All AMC 10 Problems and Solutions

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