Geometry II — Circles and Quadrilaterals

What This Module Covers

This module covers the two most formula-heavy geometry topics on the GRE: circles and quadrilaterals. You'll learn how to work with arc length, sector area, and angle relationships inside circles, and how to confidently identify and compute areas for rectangles, squares, rhombuses, and trapezoids. By the end, you'll know exactly which formula to reach for and when.

Why It Matters on the GRE

Circle and quadrilateral questions appear on almost every GRE Quant section. The GRE loves to test whether you know the difference between a central angle and an inscribed angle, and whether you'll accidentally use diameter when the formula needs radius. These are high-frequency, high-leverage topics — getting them right reliably is a real score booster.

Core Concepts

Circles — The Fundamentals

Must Know: Circumference = 2πr (or πd). Area = πr². Always identify whether you have the radius or the diameter before plugging anything in.

A radius runs from the center of a circle to its edge. A diameter runs all the way across, passing through the center — it equals 2 times the radius and is always the longest chord in the circle. The GRE frequently gives you a diameter and expects you to halve it before using circle formulas. Don't skip that step.

Arc length is the portion of the circumference cut off by a central angle. Think of it as a fraction of the full circumference:

Arc length = (central angle / 360°) × 2πr

Sector area is the slice of the circle enclosed by two radii and an arc — like a pizza slice. It's the same fractional idea applied to area:

Sector area = (central angle / 360°) × πr²

Example — Arc and Sector:

1. A circle has radius 9 and a central angle of 80°.
2. Arc length = (80/360) × 2π(9) = (2/9) × 18π = 4π
3. Sector area = (80/360) × π(9²) = (2/9) × 81π = 18π
4. Verify: 80° is less than a quarter of 360° (90°), so the arc and sector should each be less than one-quarter of the full circumference and area. Full circumference = 18π, and 4π < 4.5π. ✓

Circle Angle Relationships

Must Know: A central angle equals the arc it intercepts. An inscribed angle equals half the arc it intercepts — so an inscribed angle is always half of the central angle that cuts the same arc.

A central angle has its vertex at the center of the circle. An inscribed angle has its vertex on the circle itself, with both sides as chords. Same arc, but the inscribed angle is exactly half the size of the central angle. This is the most commonly tested circle fact on the GRE.

Special case — the semicircle rule: If a triangle is inscribed in a circle and one of its sides is the diameter, the angle opposite the diameter is always 90°. The diameter forms a 180° arc, so the inscribed angle that intercepts it = 180°/2 = 90°. Whenever you see a triangle with one side as the diameter, you have a right triangle.

A tangent line touches a circle at exactly one point. At that point of contact, the tangent is always perpendicular to the radius — the angle between them is always 90°. This is a favorite setup for GRE problems that involve right triangles mixed with circles.

Example — Inscribed Angle:

1. A central angle of 100° intercepts arc AB.
2. An inscribed angle also intercepts arc AB.
3. Inscribed angle = 100° / 2 = 50°
4. Verify: the inscribed angle must be smaller than the central angle for the same arc. 50° < 100°. ✓

Parallelograms and Their Special Forms

Must Know: Every parallelogram has opposite sides that are parallel and equal, opposite angles that are equal, and diagonals that bisect each other. Area = base × height (not base × slant side).

The height is always the perpendicular distance between the two parallel bases — not the length of a slanted side. This distinction matters most in rhombus and general parallelogram problems.

Rectangle: A parallelogram with four right angles. Area = length × width. The diagonal cuts the rectangle into two right triangles, so diagonal = √(l² + w²) by the Pythagorean theorem.

Square: A rectangle where all four sides are equal. Area = s². The diagonal = s√2 (this comes from a 45-45-90 triangle). Memorize this — it shows up constantly.

Rhombus: A parallelogram with all four sides equal, but angles are not necessarily 90°. Its diagonals are perpendicular bisectors of each other. Area = (d₁ × d₂) / 2, where d₁ and d₂ are the lengths of the two diagonals. Don't confuse a rhombus with a rectangle — a square is the only shape that's both.

Trapezoid: Exactly one pair of parallel sides, called the bases. Area = (1/2)(b₁ + b₂) × height. You're averaging the two bases, then multiplying by the height. If you only use one base, you'll get the wrong answer.

Example — Trapezoid Area:

1. A trapezoid has parallel sides of length 5 and 9, and a height of 4.
2. Area = (1/2)(5 + 9) × 4
3. = (1/2)(14) × 4 = 7 × 4 = 28
4. Verify: the area should be between a rectangle with base 5 (area = 20) and one with base 9 (area = 36). 28 is squarely in that range. ✓

Interior Angles of Polygons

Must Know: Sum of interior angles of any n-sided polygon = (n − 2) × 180°. For a regular polygon, divide by n to get each angle.

A quadrilateral (4 sides) has (4 − 2) × 180° = 360° total. A pentagon has 540°. A hexagon has 720°. For a regular hexagon, each angle = 720° / 6 = 120°. These totals come up in GRE problems where you're given some angles and asked to find a missing one.

Common Traps

• Diameter vs. radius: Using the diameter directly in the area or circumference formula is the single most common circle error. If the problem gives you the diameter, divide by 2 before using A = πr² or C = 2πr.
• Arc length vs. sector area: Arc length is a one-dimensional measurement (units like cm). Sector area is two-dimensional (units like cm²). They use the same fractional setup but different base formulas — don't mix them.
• Inscribed angle confusion: An inscribed angle is half the central angle for the same arc — not equal to it. If you see an inscribed angle of 40°, the intercepted arc is 80°, not 40°.
• Rhombus vs. rectangle: Both are parallelograms, but a rhombus uses the diagonal formula for area, not base × height. Don't assume a rhombus has right angles unless stated.
• Trapezoid area with one base: The trapezoid formula averages both parallel sides. Forgetting to include both — or forgetting to divide by 2 — gives a wrong answer that often appears as a trap choice.

GRE Strategy

• Before solving any circle problem, write down whether you have r or d, and convert to r if needed. This one habit eliminates the most common circle error.
• When you see a triangle inscribed in a circle with one side as the diameter, mark the opposite angle as 90° immediately — this unlocks the Pythagorean theorem.
• For polygon interior angle problems, write (n − 2) × 180° in your scratch work right away. Don't try to recall the total from memory under pressure.
• In quadrilateral problems, identify the shape first: does it have four right angles (rectangle/square)? All sides equal (rhombus/square)? One pair of parallel sides (trapezoid)? The shape determines the formula.
• For arc length and sector area, set up the fraction (central angle / 360°) first, simplify it, then multiply. Simplifying early keeps the arithmetic clean.

Worked Example

Question: A circle has a radius of 6. A central angle of 120° cuts off an arc. What is the length of that arc?

(A) 2π (B) 4π (C) 6π (D) 8π (E) 12π

Solution:

Step 1 — Identify what's being asked. The question asks for arc length (not sector area). Arc length is the distance along the curved edge of the circle between the two points where the central angle meets the circle.

Step 2 — Write the arc length formula. Arc length = (central angle / 360°) × 2πr

Step 3 — Simplify the fraction first. 120° / 360° = 1/3

Step 4 — Plug in the radius. Arc length = (1/3) × 2π(6) = (1/3) × 12π = 4π

Answer: (B) 4π

Step 5 — Verify. A 120° central angle is exactly one-third of the full circle (360°). The full circumference = 2π(6) = 12π. One-third of 12π = 4π. ✓

Why the other choices are wrong:

• (A) 2π — This is what you'd get if you divided 12π by 6 instead of 3. A classic mis-simplification of the fraction.
• (C) 6π — This is half the circumference, which would correspond to a 180° central angle, not 120°.
• (D) 8π — There's no clean path to this answer; it likely results from computing (120/360) × πr² (mixing up arc length with sector area). Sector area = (1/3) × π(36) = 12π, not 8π — but the temptation to grab πr² instead of 2πr is real.
• (E) 12π — This is the full circumference of the circle. Choosing this means you ignored the central angle entirely and just computed C = 2πr. Always apply the fractional multiplier.