# GRE® Quant Practice | Geometry

###### GRE Sample Questions | Area of sector | Minor segment & Major segment of circle

This GRE quant practice question is a problem solving question in geometry - circle. 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?

1. 410 $$frac{1}{3}$ + 7√3 cm2 2. 410 $\frac{1}{3}$ + 49√3 cm2 3. 1232 + 49√3 cm2 4. 1232 + 7√3 cm2 5. 410 $\frac{2}{3}$ + 49√3 cm2 ## Get to 170 in GRE Quant #### Online GRE Course @ INR 3000 ### Video Explanation ### Explanatory Answer | GRE Geometry Practice Question 3 #### Step 1 of solving this question: Compute the radius of the circle 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)

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

#### Step 2 of solving this question: Compute the area of the sector AOC

Area of sector = $$frac{θ}{360}$ × π r2, where θ is the angle subtended by the chord at the centre of the circle Chord AC subtends an angle of 120° at the center of the circle. So, area of sector AOC = $\frac{120}{360}$ × $\frac{22}{7}$ × 14 × 14 Area of sector AOC = $\frac{616}{3}$ cm2 #### Step 3 of solving this question: Compute the area of triangle AOC Area of a triangle = $\frac{1}{2}$ × base × height Base of the triangle, AC = 2 × CD In step 1, we computed CD = 7√3. So, AC = 14√3 Area of triangle AOC = $\frac{1}{2}$ × 14√3 × 7 = 49√3 #### Step 4 of solving this question: Compute the area of minor segment Area of minor segment = Area of sector ABC – Area of triangle AOC Area of minor segment = $\frac{616}{3}$ - 49√3 cm2 #### Step 5 of solving this question: Compute the area of major segment Area of major segment = Area of circle – Area of minor segment Area of a circle = π r2 = $\frac{22}{7}$ × 14 × 14 = 616 cm2 Area of minor segment = $\frac{616}{3}$ - 49√𝟑 cm2 Area of major segment = 616 -$$$frac{616}{3}$ - 49√3 cm2) = $\frac{1232}{3}$ + 49√𝟑 Area of major segment = 410$\frac{2}{3}$ + 49√3 ##### Choice E is the correct answer ###### GRE Online Course - QuantTry it free! Register in 2 easy steps and Start learning in 5 minutes! ###### Already have an Account? ##### Where is Wizako located? Wizako - GMAT, GRE, SAT Prep An Ascent Education Initiative 48/1 Ramagiri Nagar Velachery Taramani Link Road., Velachery, Chennai 600 042. India ##### How to reach Wizako? Phone:$91) 44 4500 8484
Mobile: (91) 93800 48484
WhatsApp: WhatsApp Now
Email: learn@wizako.com