# GRE® Geometry Circles

Concept: Area of sector, minor segment, major segment of circle

This GRE quant practice question is a problem solving question in geometry - circle. Concept tested: Area of a sector of a circle, area of minor segment, and area of major segment of a circle.

#### Question: Chord AC at a distance of 7 cm from the center of a circle subtends an angle of 120 degrees at the center. What is the area of major segment?

1. 410⅓ + 7√3 cm2
2. 410⅔ + 49√3 cm2
3. 1232 + 49√3 cm2
4. 1232 + 7√3 cm2
5. 410⅔ + 49√3 cm2

Video Explanation will be added soon

### 4 steps to the answer

Use these hints to get the answer

1. Compute the area of the sector formed at the points where the chord meets the circle.
2. Find the area of the triangle formed by joining the chord and the two radii that meet the chord at the circumference of the circle.
3. Compute the area of the minor segment
4. Compute the area of the major segment

### Step 1: Compute the radius of the circle

The diagram given below shows chord AC and the angle that it subtends at the center of the circle.

OA and OC are radii to the circle. ∴ AOC is an isosceles triangle. Sides OA and OC are equal.
In an isosceles triangle, the median to the side that is not equal (side AC is the side that is not equal) is the altitude, perpendicular bisector and the corresponding angle bisector. (Proof: Δ ODC and Δ ODA are two congruent right triangles. Hence, AD equals DC)
Therefore, OD will bisect ∠ AOC into two equal halves.
∠ DOC = 600.
And ∠ OCD = 300.
So, Δ ODC is a 30-60-90 right triangle.
Area of triangle AOC = $$frac{1}{2}$ x OA x OC x Sin 120 Area = $\frac{1}{2}$ x 14 x 14 x $\frac{\sqrt3}{2}$ = 49 √3 ### Step 4: Compute area of minor segment; then compute area of major segment Area of minor segment = Area of sector AOC - area of Δ AOC $\frac{616}{3}$ - 49 √3 cm2 Area of the major segment = Area of circle - area of minor segment Area of circle = π x 14 x 14 = 196 π = $\frac{22}{7}$ x 196 = 616 cm2 So, area of major segment = 616 -$$\frac{616}{3}$ - 49 √3)
= $\frac{1232}{3}$ + 49 √3
= 410 ⅔ + 49 √3 cm2