### All GRE Math Resources

## Example Questions

### Example Question #1 : How To Find The Area Of A Sector

If the outer arc of 1/12th of a circular pie is 7π, what is the area of 1/4th of the pie?

**Possible Answers:**

10.5π

441π

None of the other answers

1764π

21π

**Correct answer:**

441π

Our initial data tells us that (1/12)c = 7π or (1/12)πd = 7π. This simplifies to (1/12)d = 7 or d = 84. Furthermore, we know that r is 42. Given this, we can ascertain the area of a quarter of the whole pie by taking one fourth of the whole area or 0.25 * π * 42^{2} = 0.25 * 1764 * π = 441π

### Example Question #41 : Circles

What is the area of a sector of a circle of radius 4 that spans an arc of 86 degrees?

**Possible Answers:**

1/4 * ∏

12/53 * ∏

1/5 * ∏

172/45 * ∏

17/42 * ∏

**Correct answer:**

172/45 * ∏

sector area = ∏ * r^{2} * degrees/360 = ∏ * 4^{2} * 86/360 = 172/45 * ∏

### Example Question #1 : How To Find The Area Of A Sector

The circle above has a center of A, and points B and C lie on the circumference.

What is the area of the shaded region?

**Possible Answers:**

None of the other answers

**Correct answer:**

We need to begin by finding the area of the following sector:

If ∠BAC = 120°, then the area of sector BAC is equal to 120 / 360 = 1/3 of the entire circle. Since AC is 12 and is a radius, we know the total area is pi *12^{2} = 144*pi. The sector is then 144*pi / 3 = 48*pi.

Now, we need to find the area of the triangle ABC. Since AB and AC are equal (both being radii of our circle), we have an icoseles triangle. If we drop a height from ∠BAC, we have the following triangle ABC:

We can use the 30-60-90 rule to find the height and base. One half of the base (x) is found by the following ratio:

√(3) / 2 = x / 12

Solving for x, we get: x = 6 * √(3). Therefore, the base is 12 * √(3).

To find the height (y), we use the following ratio:

1 / 2 = y / 12

Solving for y, we get y = 6.

Therefore, the area of the triangle = 0.5 * 6 * 12 * √(3) = 3 * 12 * √(3) = 36 * √(3)

The area of the shaded region is then 48*pi - 36 * √(3).

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