Polygons & Circles

Master the essential properties of polygons and circles: interior angles, area formulas, arc length, sector area, and inscribed figures for GRE Quantitative success.

GRE Quantitative

What Is a Polygon?

A polygon is a closed, flat shape formed by straight line segments (called sides) connected end to end. The simplest polygon is a triangle (\(n = 3\)), and the GRE frequently tests polygons from \(n = 3\) up to \(n = 8\) or beyond.

Regular Polygon

All sides equal length, all interior angles equal measure.

Irregular Polygon

Sides and angles may differ: only the sum of angles is fixed.

Circle Fundamentals

Why This Matters on the GRE

Polygon and circle problems appear in both Quantitative Comparison and Problem Solving sections. The GRE tests your ability to combine properties: a regular hexagon inscribed in a circle, or finding the area of a shaded region between a polygon and a circle.

Angle Sums

Interior angle formulas for any \(n\)-gon

Arc & Sector

Fractional parts of a circle's perimeter and area

Inscribed Figures

Polygons inside circles and circles inside polygons

Interior Angle Sum Theorem

Arc Length & Sector Area

Central vs Inscribed Angle

How to Apply These Formulas

1

Identify the shape. Is it a polygon, a circle, or a combined figure (inscribed/circumscribed)?

2

Extract known values. Look for side count \(n\), radius \(r\), angle \(\theta\), or side length \(s\).

3

Select the right formula. Angle sum: \((n-2) \times 180^\circ\). Arc length: \(\frac{\theta}{360^\circ} \times 2\pi r\). Sector area: \(\frac{\theta}{360^\circ} \times \pi r^2\).

4

Solve and simplify. Leave answers in terms of \(\pi\) unless the problem asks for a decimal.

5

Check reasonableness. A sector area must be less than \(\pi r^2\). An interior angle must be less than \(180^\circ\).

Example 1: Hexagon Interior Angle

Example 2: Finding Arc Length

A circle has radius 10. A central angle measures \(72^\circ\). Find the arc length.

Example 3: Sector Area

A pizza slice is a sector with radius 6 and central angle \(90^\circ\). Find its area.

Example 4: Square Inscribed in a Circle

A square is inscribed in a circle of radius 5. Find the side length and area of the square.

Exterior Angle Sum

Common Mistakes

✕ Wrong angle formula

Using \(n \times 180^\circ\) instead of \((n-2) \times 180^\circ\) for interior angle sum.

✕ Confusing \(r\) and \(d\)

Plugging diameter into \(\pi r^2\). The area formula requires radius, not diameter.

✕ Inscribed angle = central

Forgetting the inscribed angle is half the central angle for the same arc.

Formula Reference Table

Property	Formula	Example
Interior angle sum	\((n-2) \times 180^\circ\)	Hexagon: \(720^\circ\)
Each interior angle (regular)	\(\frac{(n-2) \times 180^\circ}{n}\)	Hexagon: \(120^\circ\)
Each exterior angle (regular)	\(\frac{360^\circ}{n}\)	Hexagon: \(60^\circ\)
Circumference	\(2\pi r\)	\(r=7\): \(14\pi\)
Circle area	\(\pi r^2\)	\(r=7\): \(49\pi\)
Arc length	\(\frac{\theta}{360^\circ} \times 2\pi r\)	\(\theta=90^\circ, r=4\): \(2\pi\)
Sector area	\(\frac{\theta}{360^\circ} \times \pi r^2\)	\(\theta=90^\circ, r=4\): \(4\pi\)

Question 1 of 5

What is the sum of interior angles of a regular octagon?

An octagon has \(n = 8\). Interior angle sum \(= (8-2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ\).

Question 2 of 5

A circle has radius 8. A central angle of \(45^\circ\) cuts off an arc. What is the arc length?

Arc length \(= \frac{45^\circ}{360^\circ} \times 2\pi(8) = \frac{1}{8} \times 16\pi = 2\pi\).

Question 3 of 5

Each interior angle of a regular polygon measures \(150^\circ\). How many sides does it have?

Each exterior angle \(= 180^\circ - 150^\circ = 30^\circ\). Number of sides \(= \frac{360^\circ}{30^\circ} = 12\). It is a regular dodecagon.

Question 4 of 5

An inscribed angle in a circle intercepts an arc of \(160^\circ\). What is the measure of the inscribed angle?

An inscribed angle is always half its intercepted arc: \(\frac{160^\circ}{2} = 80^\circ\).

Question 5 of 5

A square is inscribed in a circle of radius 6. What is the area of the square?

Diagonal of the square \(= \) diameter \(= 12\). Since diagonal \(= s\sqrt{2}\), we get \(s = \frac{12}{\sqrt{2}} = 6\sqrt{2}\). Area \(= s^2 = (6\sqrt{2})^2 = 72\).