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

Question 3 : 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?

- 410 \\frac{1}{3}) + 7√3 cm
^{2} - 410 \\frac{1}{3}) + 49√3 cm
^{2} - 1232 + 49√3 cm
^{2} - 1232 + 7√3 cm
^{2} - 410 \\frac{2}{3}) + 49√3 cm
^{2}

From INR 2999

In the figure AC is the chord and the angle that it subtends at the center of the circle is ∠AOC.

OA and OC are radii to the circle.

Draw OD perpendicular to AC.

ΔODC and ΔODA are congruent right triangles. (RHS test)

Hence, AD equals DC

And ∠AOD = ∠COD = 60°

Because ∠COD = 60° and ∠CDO = 90°, ∠OCD = 30°

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

The sides opposite 30° – 60° – 90° will be in the ratio 1 : √3 : 2

The side opposite 30° is OD = 7 cm

So, the side opposite 90° ⇒ OC = 2 × 7 = 14 cm

The radius OC = 14 cm.

Let us also compute CD = √3 × 7 = 7√3 cm

Area of sector = \\frac{θ}{360}) × π r^{2}, where θ is the angle subtended by the chord at the centre of the circle

Chord AC subtends an angle of 120° at the center of the circle.

So, area of sector AOC = \\frac{120}{360}) × \\frac{22}{7}) × 14 × 14

Area of sector AOC = \\frac{616}{3}) cm^{2}

Area of a triangle = \\frac{1}{2}) × base × height

Base of the triangle, AC = 2 × CD

In step 1, we computed CD = 7√3. So, AC = 14√3

Area of triangle AOC = \\frac{1}{2}) × 14√3 × 7 = 49√3

**Area of minor segment = Area of sector ABC – Area of triangle AOC**

Area of minor segment = \\frac{616}{3}) - 49√3 cm^{2}

**Area of major segment = Area of circle – Area of minor segment**

Area of a circle = π r^{2} = \\frac{22}{7}) × 14 × 14 = 616 cm^{2}

Area of minor segment = \\frac{616}{3}) - 49√𝟑 cm^{2}

Area of major segment = 616 - (\\frac{616}{3}) - 49√3 cm^{2}) = \\frac{1232}{3}) + 49√𝟑

Area of major segment = **410\\frac{2}{3}) + 49√3**

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