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?

- 410⅓ + 7√3 cm
^{2} - 410⅔ + 49√3 cm
^{2} - 1232 + 49√3 cm
^{2} - 1232 + 7√3 cm
^{2} - 410⅔ + 49√3 cm
^{2}

#### Explanatory Answer

Video Explanation will be added soon### 4 steps to the answer

Use these hints to get the answer

- Compute the area of the sector formed at the points where the chord meets the circle.
- 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.
- Compute the area of the minor segment
- 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 = 60^{0}.

And ∠ OCD = 30^{0}.

So, Δ ODC is a 30-60-90 right triangle.

Sin(OCD) = Sin 30 = \\frac{\text{opposite side}}{\text{Hypotenuse}}) = \\frac{OD}{OC})

\\frac{1}{2}) = \\frac{7}{OC})

Or OC = 14.

The radius of the circle is 14 cm

### Step 2: Compute the area of the sector AOC

Chord AC subtends an angle of 120^{0} at the center of the circle.

So, area of sector AOC = \\frac{120}{360}) x Area of circle

Area of sector AOC = \\frac{120}{360}) * \\frac{22}{7}) * 14^{2} = \\frac{616}{3})

### Step 3: Compute area of isosceles triangle AOC

Area of a triangle = \\frac{1}{2}) * (product of two sides) * Sin (included angle)

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 cm^{2}

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 cm^{2}

So, area of major segment = 616 - (\\frac{616}{3}) - 49 √3)

= \\frac{1232}{3}) + 49 √3

= 410 ⅔ + 49 √3 cm^{2}