### All GRE Math Resources

## Example Questions

### Example Question #1 : Geometry

O is the center of the circle above.

The length of is .

Quantity A: The area of the circle.

Quantity B:

Which of the following is true?

**Possible Answers:**

The relationship cannot be determined.

The two quantities are equal.

Quantity B is greater.

Quantity A is greater.

**Correct answer:**

Quantity A is greater.

O is the center of the circle above.

The length of is .

Quantity A: The area of the circle.

Quantity B:

Do not be tricked by this question. It is true that can be split into halves, each of which are in length. These halves are not, however, radii to the circle. Since this does not go through the center of the circle, its length is shorter than the diameter. This means that the radius of the circle must be greater than . Now, if it were , the area would be . Since it is larger than , the area must be larger than . Quantity A is larger than quantity B.

### Example Question #2 : How To Find The Length Of A Chord

O is the center of the circle above.

The circumference of the circle above is .

Quantity A: The length of .

Quantity B:

Which of the following is true?

**Possible Answers:**

Quantity B is larger.

The two quantities are equal.

Quantity A is larger.

The relationship cannot be determined.

**Correct answer:**

Quantity B is larger.

Now, we know that the circumference of a circle is:

or

This means that the diameter of our circle is must be . Given this, we know that the must be shorter than , for the diameter is the longer than any chord that does not pass through the center of the circle. Quantity B is larger than quantity A.

### Example Question #2 : Geometry

What is the circumference of a circle with an area of 36π?

**Possible Answers:**

12π

None of the other answers

6π

32

15π

**Correct answer:**

12π

We know that the area of a circle can be expressed: a = πr^{2}

If we know that the area is 36π, we can substitute this into said equation and get: 36π = πr^{2}

Solving for r, we get: 36 = r^{2}; (after taking the square root of both sides:) 6 = r

Now, we know that the circuference of a circle is expressed: c = πd. Since we know that d = 2r (two radii, placed one after the other, make a diameter), we can rewrite the circumference equation to be: c = 2πr

Since we have r, we can rewrite this to be: c = 2π*6 = 12π

### Example Question #3 : Geometry

Which is greater: the circumference of a circle with an area of , or the perimeter of a square with side length inches?

**Possible Answers:**

The relationship cannot be determined from the information given.

The circumference of the circle is greater.

The perimeter of the square is greater.

The two quantities are equal.

**Correct answer:**

The circumference of the circle is greater.

Starting with the circle, we need to find the radius in order to get the circumference. Find by plugging our given area into the equation for the area of a circle:

Then calculate circumference:

(approximating as 3.14)

To find the perimeter of the square, we can use , where is the perimeter and is the side length:

, so the circle's circumference is greater.

### Example Question #4 : Geometry

Circle A has an area of . What is the perimeter of an enclosed semi-circle with half the radius of circle A?

**Possible Answers:**

**Correct answer:**

Based on our information, we know that the 121π = πr^{2}; 121 = r^{2}; r = 11.

Our other circle with half the radius of A has a diameter equal to the radius of A. Therefore, the circumference of this circle is 11π. Half of this is 5.5π. However, since this is a semi circle, it is enclosed and looks like this:

Therefore, we have to include the diameter in the perimeter. Therefore, the total perimeter of the semi-circle is 5.5π + 11.

### Example Question #5 : Geometry

Quantity A: The circumference of a circle with radius

Quantity B: The area of a circle with a diameter one fourth the radius of the circle in Quantity A

Which of the following is true?

**Possible Answers:**

Quantity A is larger.

The relationship between the two values cannot be determined.

The two quantities are equal.

Quantity B is larger.

**Correct answer:**

The relationship between the two values cannot be determined.

Let's compute each value separately. We know that the radii are positive numbers that are greater than or equal to . This means that we do not need to worry about the fact that the area could represent a square of a decimal value like .

**Quantity A**

Since , we know:

**Quantity B**

If the diameter is one-fourth the radius of A, we know:

Thus, the radius must be half of that, or .

Now, we need to compute the area of this circle. We know:

Therefore,

Now, notice that if , Quantity A is larger.

However, if we choose a value like , we have:

Quantity A:

Quantity B:

Therefore, the relation cannot be determined!

### Example Question #3 : Geometry

Circle has a center in the center of Square .

The area of Square is .

What is the circumference of Circle ?

**Possible Answers:**

**Correct answer:**

Since we know that the area of Square is , we know , where is the length of one of its sides. From this, we can solve for by taking the square root of both sides. You will have to do this by estimating upward. Therefore, you know that is . By careful guessing, you can quickly see that is . From this, you know that the diameter of your circle must be half of , or (because it is circumscribed). Therefore, you can draw:

The circumference of this circle is defined as:

or, for your values:

(You could also compute this from the diameter, but many students just memorize the formula above.)

### Example Question #5 : Geometry

What is the area of a circle, one-quarter of the circumference of which is 5.5 inches?

**Possible Answers:**

121/π

π/3

225π

121π

**Correct answer:**

121/π

Here, you need to “solve backward” from the data you have been given. We know that 0.25C = 5.5; therefore, C = 22. In order to solve for the area, we will need the radius of the circle. This can be obtained by recalling that C = 2πr. Replacing 22 for C, we get 22 = 2πr.

Solve for r: r = 22 / 2π = 11 / π.

Now, we solve for the area: A = πr^{2}. Replacing 11 / π for r: A = π (11 / π)^{2} = (121π) / (π^{2}) = 121 / π.

### Example Question #6 : Geometry

In the diagram above, square ABCD is inscribed in the circle. If the area of the square is 9, what is the area of the circle?

**Possible Answers:**

3√(2)π

18π

4.5π

3π

9π

**Correct answer:**

4.5π

If the area of the square is 9, then s^{2} = 9 and s = 3. If the sides thus equal 3, we can calculate the diagonals (either CB or AD) by using the 45-45-90 triangle ratio. For a side of 3, the diagonal will be 3√(2). Note that since the square is inscribed in the circle, this diagonal is also the diameter of the circle. If it is such, the radius is one half of that or 1.5√(2).

Based on that value, we can computer the circle’s area:

A = πr^{2} = π(1.5√(2))^{2} = (2.25 * 2)π = 4.5π

### Example Question #7 : Geometry

Quantitative Comparison

Quantity A: Area of a circle with radius r

Quantity B: Perimeter of a circle with radius r

**Possible Answers:**

Quantity A is greater.

Quantity B is greater.

The relationship cannot be determined from the information given.

The two quantities are equal.

**Correct answer:**

The relationship cannot be determined from the information given.

Try different values for the radius to see if a pattern emerges. The formulas needed are Area = *π**r*^{2} and Perimeter = 2*πr*.

If *r* = 1, then the Area = *π* and the Perimeter = 2*π*, so the perimeter is larger.

If r = 4, then the area = 16*π* and the perimeter = 8*π*, so the area is larger.

Therefore the relationship cannot be determined from the information given.