$O$ is the center of the circle, and the perimeter of $△AOB$ is $6$.
Quantity A | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | Quantity B |
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | ||
The circumference of the circle | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | $12$ |
- Quantity A is greater.
- Quantity B is greater.
- The two quantities are equal.
- The relationship cannot be determined from the information given
So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 1 of the first Quantitative section on Practice Test 1. Those Circles questions can be kind of tricky, but never fear, PrepScholar has got your back!
Survey the Question
Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on your paper.
We see that we have a triangle inside of a circle. So this question likely tests what we’ve learned about Triangles and Circles from geometry.
What Do We Know?
Let’s carefully read through the question and make a list of the things that we know.
- We have a triangle drawn inside a circle
- Two of the triangle’s sides $(OA \and OB)$ are radii of the circle
- One of the triangle’s angles $(∠AOB)$ is $60$ degrees
- We know the perimeter of the triangle is $18$
- We want to compare the circumference of the circle to the value $12$
Develop a Plan
We are given the perimeter of $△AOB$ and we want to find the circumference of circle $O$, but it’s hard to see immediately how they’re connected. Let’s start with a top-down approach, where we will begin with what we’re looking for and work down to the details of what we’re given in this question. We want to compare the circumference of circle $O$ to the value $12$. Let’s think of what math equation we have for the circumference of a circle.
$$\Circumference =2π·\Radius$$
Okay, so if we can find the radius of the circle, then we could definitely calculate its circumference. In Circle $O$, we can see two radii drawn: $OA$ and $OB$. We need their lengths! These radii are also part of $△AOB$, which is nice since we know that the perimeter of a triangle is the sum of its three sides.
$$OA+OB+AB=6$$
We know that in triangles, if two sides are equal, then the angles opposite of those sides must be equal. Since the two radii must have the same length $(OA=OB)$, then $∠OAB=∠OBA$. We also know that the sum of the angles in a triangle is $180°$ .
$$∠OAB + ∠OBA + ∠AOB=180°$$
The figure shows us that $∠AOB=60°$. Let’s put this into our “sum of the angles” equation, also substituting in $∠OBA$ for $∠OAB$. Then we can solve for $∠OAB$!
$∠OAB + ∠OBA + ∠AOB$ | $=$ | $180°$ |
$ $ | $ $ | $ $ |
$∠OAB + ∠OAB + 60°$ | $=$ | $180°$ |
$ $ | $ $ | $ $ |
$2·∠OAB$ | $=$ | $180°-60°$ |
$ $ | $ $ | $ $ |
$2·∠OAB$ | $=$ | $120°$ |
$ $ | $ $ | $ $ |
$∠OAB$ | $=$ | $60°$ |
Well that’s interesting! $∠OAB$ and $∠AOB$ are both $60°$. Also, since $∠OAB=∠OBA$, then all three of the angles in $△AOB$ are $60°$. Ladies and gentlemen, we have ourselves an equilateral triangle!
What do we know about equilateral triangles….ah…all of their sides are equal! So if the sum of the three sides of the triangle is $6$, then each side must be: $6/3=2$. And since the radius of the circle is a side in this triangle, then the radius of the circle must be $2$!
And that is how we make a plan! Alright, so all we have to do now is plug our radius into the equation for the circumference of the circle, then compare our two quantities.
Solve the Question
Let’s solve for the circumference of circle $O$ now.
Circumference of Circle $O$ | $=$ | $2π·\Radius$ |
$ $ | $ $ | $ $ |
Circumference of Circle $O$ | $=$ | $2π·2$ |
$ $ | $ $ | $ $ |
Circumference of Circle $O$ | $=$ | $4π$ |
Now to compare the two quantities. Let’s write them on our paper now.
Quantity A | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | Quantity B |
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | ||
$4π$ | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | $12$ |
A clever PrepScholar trick in Quantitative Comparison questions is to simplify quantities by dividing both of them by the same positive number. Both quantities have $4$ as a factor, so let’s divide both of them by $4$ then continue our comparison.
Quantity A | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | Quantity B |
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | ||
${4π}/4$ | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | $12/4$ |
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | ||
$π$ | $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ | $3$ |
Excellent! Since $π$ is slightly greater than $3$, the correct answer is {A}, Quantity A is greater.
What Did We Learn
For triangles drawn inside circles, pay close attention to if any of the sides of the triangles are radii of the circle. They key to this question was realizing that two of the sides of our triangle were radii of our circle.
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