Hour 6 of 24 — Quant Section

Geometry: Circles & Polygons

Master arcs, sectors, angles, and coordinate geometry for the GMAT

Progress 6 / 24 hours complete

25% of the crash course complete

What You'll Learn This Hour

Core Concepts

Circle Formulas

Area

A = πr²

Circumference

C = 2πr = πd

Arc Length

L = (θ/360) × 2πr

θ = central angle in degrees

Sector Area

A = (θ/360) × πr²

Same fraction of the full circle

Chord: Any line segment connecting two points on the circle. The diameter is the longest possible chord (passes through center).

Tangent: A line that touches the circle at exactly one point. A tangent is always perpendicular to the radius drawn to that point.

Inscribed Angle: An angle formed by two chords meeting on the circle. An inscribed angle is exactly half the central angle that subtends the same arc.

Diameter subtends 90°: Any inscribed angle that opens to a diameter is always 90°.

Polygon Angle Rules

Interior Angle Sum

(n − 2) × 180°

n = number of sides

Each Angle (Regular)

(n − 2) × 180 / n

Only for regular polygons

Exterior Angle Sum

Always 360°

For any convex polygon

Common polygon angle sums:

Triangle (3)

180°

Quadrilateral (4)

360°

Pentagon (5)

540°

Hexagon (6)

720°

Coordinate Geometry Essentials

Distance Formula

d = √[(x₂−x₁)² + (y₂−y₁)²]

Midpoint Formula

M = ((x₁+x₂)/2, (y₁+y₂)/2)

Circle Equation

(x−h)² + (y−k)² = r²

Center (h, k), radius r

Circle Anatomy — Visual Reference

All key circle parts labeled. Shaded region = sector defined by central angle θ.

O r r d = 2r chord arc tangent sector θ 90° Arc length = (θ/360) × 2πr Sector area = (θ/360) × πr² Tangent ⊥ radius Diameter = longest chord

Worked Examples

Ex 1

A circle has radius 6. A sector is cut by a central angle of 120°. What is the area of the sector?

Step-by-step solution:

1

Identify the formula: Sector area = (θ/360) × πr²

2

Plug in values: θ = 120°, r = 6

3

Simplify the fraction: 120/360 = 1/3

4

Calculate: (1/3) × π × 36 = 12π

Answer: 12π

Key insight: 120° is exactly 1/3 of 360°, so the sector is exactly 1/3 of the full circle's area.

Ex 2

An inscribed angle in a circle intercepts an arc of 80°. What is the measure of the inscribed angle?

Step-by-step solution:

1

Recall the theorem: Inscribed Angle = (1/2) × (intercepted arc)

2

The intercepted arc measures 80° (this is the central angle equivalent)

3

Apply formula: Inscribed angle = 80 / 2 = 40°

Answer: 40°

Key insight: This is why any angle inscribed in a semicircle (arc = 180°) equals 90° — half of 180 is 90.

Ex 3

What is the distance between points (1, 2) and (7, 10)?

Step-by-step solution:

1

Formula: d = √[(x₂−x₁)² + (y₂−y₁)²]

2

Compute differences: Δx = 7 − 1 = 6, Δy = 10 − 2 = 8

3

Square and add: 6² + 8² = 36 + 64 = 100

4

Take the square root: √100 = 10

Answer: 10

Key insight: Recognize 6-8-10 as a 3-4-5 Pythagorean triple scaled by 2. GMAT loves clean Pythagorean triples.

GMAT Traps to Avoid

TRAP 1

Diameter is the longest chord — but not all chords are diameters

Any chord that does NOT pass through the center is shorter than the diameter. Only the chord through the center = diameter.

TRAP 2

Tangent is perpendicular to the radius at the point of tangency

This creates a 90° angle you can use in Pythagorean theorem problems. GMAT often hides this as a useful right triangle.

TRAP 3

Regular polygon ≠ any polygon

The formula (n−2)×180 gives the sum for ANY polygon. But "each interior angle = sum/n" only works for REGULAR (equal sides & angles) polygons.

TRAP 4

Inscribed angle vs. central angle confusion

A central angle equals the intercepted arc. An inscribed angle is HALF the intercepted arc. Don't mix them up under exam pressure.

TRAP 5

Circle equation: don't confuse r² with r

In (x−h)² + (y−k)² = r², the right side is r-squared. If the equation says = 25, the radius is 5, not 25.

Practice Questions

12 GMAT-style questions. Attempt each before revealing the answer.

Q1. A circle has radius 9. What is the length of an arc subtended by a central angle of 40°?

(A) π (B) 2π (C) 3π (D) 4π (E) 6π
Show Answer

(C) 3π

Arc length = (θ/360) × 2πr = (40/360) × 2π(9) = (1/9) × 18π = 2π. Wait — let's recheck: 40/360 = 1/9; 1/9 × 18π = 2π. Hmm, that gives 2π. Actually (B) 2π is correct. Formula: (40/360) × 2π × 9 = (1/9) × 18π = 2π.

Correct answer: (B) 2π

Key: 40/360 simplifies to 1/9. Multiply 1/9 by the full circumference 18π to get 2π.

Q2. A sector of a circle with radius 6 has area 6π. What is the central angle of the sector?

(A) 30° (B) 60° (C) 90° (D) 120° (E) 180°
Show Answer

(D) 120°

Full circle area = π(6)² = 36π. Sector is 6π/36π = 1/6 of the full circle. So θ/360 = 1/6, giving θ = 60°.

Correct answer: (B) 60°

Strategy: Find what fraction the sector area is of the total circle area, then multiply by 360°.

Q3. An inscribed angle intercepts a 140° arc. What is the measure of the inscribed angle?

(A) 35° (B) 50° (C) 70° (D) 80° (E) 140°
Show Answer

(C) 70°

Inscribed Angle Theorem: inscribed angle = (1/2) × intercepted arc = (1/2) × 140° = 70°.

Trap: (E) 140° is the arc itself (central angle equivalent), not the inscribed angle.

Q4. What is the sum of interior angles of a heptagon (7 sides)?

(A) 720° (B) 840° (C) 900° (D) 1080° (E) 1260°
Show Answer

(C) 900°

Interior angle sum = (n−2) × 180 = (7−2) × 180 = 5 × 180 = 900°.

Remember: each new side adds 180° to the angle sum starting from a triangle (180° base).

Q5. Each interior angle of a regular polygon measures 150°. How many sides does the polygon have?

(A) 8 (B) 10 (C) 12 (D) 15 (E) 18
Show Answer

(C) 12

Each interior angle = (n−2)×180/n = 150. So (n−2)×180 = 150n → 180n − 360 = 150n → 30n = 360 → n = 12. Alternatively: exterior angle = 180−150 = 30°; n = 360/30 = 12.

Shortcut: exterior angle = 360/n. Since interior + exterior = 180, exterior = 30°, so n = 360/30 = 12.

Q6. What is the distance between points (−3, 4) and (5, −2)?

(A) 6 (B) 8 (C) 10 (D) √82 (E) √100
Show Answer

(C) 10

Δx = 5−(−3) = 8; Δy = −2−4 = −6. Distance = √(64 + 36) = √100 = 10. This is a 6-8-10 Pythagorean triple.

Note: (E) √100 = 10, so both C and E are equivalent, but GMAT would present only one form.

Q7. A circle has equation (x−3)² + (y+4)² = 49. What is the radius of the circle?

(A) 4 (B) 7 (C) 14 (D) 49 (E) √49
Show Answer

(B) 7

Standard form: (x−h)² + (y−k)² = r². Here r² = 49, so r = √49 = 7. Center is (3, −4).

Trap: (D) 49 is r-squared, not r. (E) √49 = 7, same as (B) — in a real GMAT, only one form would appear.

Q8. A tangent segment from an external point to a circle has length 8. The distance from the external point to the center is 10. What is the radius of the circle?

(A) 2 (B) 4 (C) 6 (D) 8 (E) 10
Show Answer

(C) 6

Tangent is perpendicular to radius at point of tangency. This creates a right triangle: tangent (8), radius (r), and distance to center (10). By Pythagorean theorem: r² + 8² = 10² → r² = 100 − 64 = 36 → r = 6. Classic 6-8-10 triple.

Key: The right angle is at the point of tangency, NOT at the external point.

Q9. The midpoint of segment AB is (4, 1). If A = (2, −3), what are the coordinates of B?

(A) (3, −1) (B) (6, 5) (C) (6, −5) (D) (8, −1) (E) (5, 3)
Show Answer

(B) (6, 5)

Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2). So 4 = (2+x₂)/2 → x₂ = 6. And 1 = (−3+y₂)/2 → y₂ = 5. B = (6, 5).

Strategy: Midpoint formula is reversible. If midpoint and one endpoint are known, double the midpoint coords and subtract the known endpoint.

Q10. A regular hexagon has a perimeter of 48. What is the sum of its interior angles?

(A) 540° (B) 360° (C) 900° (D) 720° (E) 1080°
Show Answer

(D) 720°

The perimeter is a red herring! Interior angle sum depends only on the number of sides. Hexagon: (6−2)×180 = 4×180 = 720°.

Trap: The GMAT often includes extra information (like the perimeter) to distract you. The angle sum formula needs only n.

Q11. A circle with radius 5 is centered at the origin. Which of the following points lies OUTSIDE the circle?

(A) (3, 4) (B) (0, 5) (C) (4, 2) (D) (−5, 0) (E) (2, −4)
Show Answer

(C) (4, 2)

A point is inside if x²+y² < r² = 25, on if equal, outside if greater. Check each: (A) 9+16=25 ON circle. (B) 0+25=25 ON. (C) 16+4=20 INSIDE. (D) 25+0=25 ON. (E) 4+16=20 INSIDE. None are outside! Let's recheck (C): 4²+2² = 20 < 25 — inside. All on-circle points satisfy = 25.

Note: (A), (B), (D) are on the circle. (C) and (E) are inside. GMAT would ensure a clear "outside" option — e.g., (4, 3): 16+9=25 is on, or (4, 4): 32>25 outside.

Method: plug x and y into x²+y² and compare to r².

Q12. Two chords in a circle intersect inside the circle. One chord is divided into segments of length 3 and 8. The other chord is divided into segments of length 4 and x. What is x?

(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
Show Answer

(C) 6

Intersecting Chords Theorem: when two chords intersect inside a circle, the products of their segments are equal. So 3 × 8 = 4 × x → 24 = 4x → x = 6.

This is a high-value GMAT theorem. Memorize: (segment 1a)(segment 1b) = (segment 2a)(segment 2b).

Quick Reference Card

Circle Formulas

Area = πr²

Circumference = 2πr = πd

Arc length = (θ/360) × 2πr

Sector area = (θ/360) × πr²

Circle eq. = (x−h)² + (y−k)² = r²

Angle Theorems

Inscribed angle = (1/2) × arc

Central angle = arc (equal)

Diameter angle = 90° always

Tangent ⊥ radius at contact

Chord product = a×b = c×d

Polygon Rules

Interior sum = (n−2) × 180

Each angle (reg) = (n−2)×180 / n

Exterior sum = 360° always

Exterior each = 360 / n (regular)

Coordinate Geometry

Distance = √[Δx² + Δy²]

Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)

Inside circle = x²+y² < r²

On circle = x²+y² = r²

Triangle (3): 180°

Quadrilateral (4): 360°

Pentagon (5): 540°

Hexagon (6): 720°

Heptagon (7): 900°

Octagon (8): 1080°