Use properties of circles to reason about radii and chords. - SSAT Middle Level Quantitative

SECTION 1

Why People Started Studying Circles

Circles are everywhere — wheels, clocks, coins, and even the moon. Thousands of years ago, people noticed that circles had special patterns. Ancient mathematicians wanted to understand those patterns so they could build better buildings, measure land, and study the stars.

The story of circles stretches across many cultures. Let's look at some key moments in the history of circle geometry.

~2000 BC • Ancient Babylon

Babylonian mathematicians discovered that the distance around a circle is always a little more than three times the distance across. They carved their findings into clay tablets.

~300 BC • Euclid's Elements

The Greek mathematician Euclid wrote a famous textbook called Elements. He defined circles, radii, and chords, and proved rules about them that we still use today.

~250 BC • Archimedes Measures π

Archimedes figured out a more accurate value for pi (π) by drawing shapes inside and outside circles. His work helped people calculate distances in circles more precisely.

~600 AD • Indian Mathematicians

Scholars in India, including Aryabhata, developed new ways to calculate with circles. They used chord lengths in early versions of trigonometry.

All of these mathematicians asked similar questions: What are the parts of a circle? How are they related? Understanding radii (the lines from the center to the edge) and chords (lines that cut across the inside) is the key to answering those questions — and to solving circle problems on the SSAT.

SECTION 2

Core Definitions and Rules

Before we solve problems, let's nail down the vocabulary. Every circle problem on the SSAT uses these terms, so knowing them well gives you a head start.

Radius

A radius is a line segment from the center of the circle to any point on the circle. All radii (plural of radius) in the same circle have the same length.

Chord

A chord is a line segment whose two endpoints are both on the circle. It stretches across the inside of the circle.

Diameter

A diameter is a special chord that passes through the center. It is the longest possible chord and equals twice the radius.

Perpendicular from Center

When a line from the center meets a chord at a 90° angle (perpendicular), that line cuts the chord exactly in half. This is called bisecting the chord.

Equal Chords, Equal Distance

Chords that are the same length sit the same distance from the center. Chords closer to the center are longer; chords farther from the center are shorter.

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KEY TAKEAWAY

Think of a circle like a pizza. Every slice from the center to the crust (the radius) is the same length. A chord is like a straight cut across the pizza that doesn't have to go through the center. The cut that does go through the center — the biggest possible cut — is the diameter. Because every radius is equal, you can use that fact to find missing lengths in many problems.

SECTION 3

Seeing the Parts of a Circle

A picture makes these ideas much easier to understand. The diagram below shows a circle with its center, a radius, a chord, and a diameter all labeled.

The circle has center O. The purple line is a radius. The pink line AB is the diameter (the longest chord). The cyan line CD is a chord that does not pass through the center. The dashed yellow line shows the perpendicular distance from the center to chord CD.

Notice that every point on the circle is the same distance from the center. That distance is the radius. The diameter stretches all the way across through the center, so it's always twice the radius. The chord CD is shorter than the diameter because it doesn't pass through the center.

SECTION 4

Key Formulas and Relationships

You don't need complicated algebra to work with circles on the SSAT. Instead, you need a few simple relationships and the ability to spot right triangles hiding inside the circle.

DIAMETER AND RADIUS

d = 2 × r

Here d is the diameter and r is the radius. If you know one, you can always find the other.

PERPENDICULAR BISECTOR RULE

If OM ⊥ chord AB, then AM = MB

When a line from center O hits chord AB at a right angle (⊥ means perpendicular) at point M, it splits the chord into two equal halves. AM = MB.

RIGHT TRIANGLE IN A CIRCLE

r² = d² + (half-chord)²

When you draw a radius to the end of a chord and a perpendicular from the center to the chord, you get a right triangle. Here r is the radius (hypotenuse), d is the distance from the center to the chord, and half-chord is half the chord's length. This comes from the Pythagorean theorem.

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Pythagorean Theorem Reminder

The Pythagorean theorem says that in a right triangle, a² + b² = c², where c is the longest side (the hypotenuse). In circle problems, the radius is usually the hypotenuse.

SECTION 5

The Hidden Right Triangle

Here's the most powerful trick for circle problems: whenever you see a chord, draw a line from the center perpendicular to that chord. That line, together with half the chord and a radius, makes a right triangle. Let's see this in a diagram.

A chord AB with the perpendicular from center O meeting the chord at point M. The pink line OB is the radius (r = 10). The yellow line OM is the perpendicular distance (d = 6). Each half of the cyan chord is 8. Together they form a 6-8-10 right triangle.

In this example, the radius is 10, and the distance from the center to the chord is 6. Because OM is perpendicular to AB, point M splits the chord in half. The right triangle OMA has legs of 6 and 8, and the hypotenuse (the radius) is 10. This is a well-known 6-8-10 Pythagorean triple (which is just the 3-4-5 triple doubled). On the SSAT, look for triples like 3-4-5, 5-12-13, and 6-8-10.

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Common Pythagorean Triples

Memorize these sets of three whole numbers that satisfy the Pythagorean theorem: 3-4-5, 5-12-13, and 8-15-17. Multiples of these (like 6-8-10 or 10-24-26) appear often in circle chord problems.

SECTION 6

Worked Example: Finding a Chord Length

Let's walk through a full problem step by step. This is the kind of question you might see on the SSAT.

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Problem

A circle has a radius of 13 cm. A chord is drawn 5 cm from the center. What is the length of the chord?

Step-by-Step Solution

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Step 1 — Draw a Picture and Identify What You Know

Imagine a circle with center O and radius 13. A chord AB is drawn inside the circle, and the perpendicular distance from O to the chord is 5. Call the point where the perpendicular meets the chord M.

r = 13, d = 5, find chord AB
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Step 2 — Recognize the Right Triangle

The perpendicular from O to the chord creates a right triangle OMA. The hypotenuse is the radius OA = 13. One leg is OM = 5. The other leg is MA (half the chord).

Triangle OMA: hypotenuse = 13, one leg = 5
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Step 3 — Apply the Pythagorean Theorem

Use r² = d² + (half-chord)². Plug in the values: 13² = 5² + (half-chord)². That gives us 169 = 25 + (half-chord)².

169 = 25 + (half-chord)²
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Step 4 — Solve for the Half-Chord

Subtract 25 from both sides: (half-chord)² = 169 − 25 = 144. Take the square root: half-chord = √144 = 12.

Half-chord = 12 cm
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Step 5 — Find the Full Chord

The perpendicular bisects (cuts in half) the chord, so the full chord is 2 × 12 = 24.

Chord AB = 24 cm

Notice that we used a 5-12-13 Pythagorean triple here. If you memorize common triples, you can solve problems like this very quickly without even needing to calculate square roots!

SECTION 7

Common Mistakes and How to Avoid Them

Circle problems aren't too hard once you know the rules, but there are some traps that students fall into. Let's look at the most common mistakes and how to avoid them.

Common Mistake	Why It Happens	How to Fix It
Mixing up radius and diameter	The problem gives the diameter but you use it as the radius (or vice versa)	Always write down whether the number is r or d. Remember d = 2r.
Forgetting to double the half-chord	You find the half-chord correctly but forget to multiply by 2	Circle the question and ask yourself: does it want the whole chord or half?
Using the wrong side as the hypotenuse	You put the distance or half-chord as the longest side instead of the radius	The radius is ALWAYS the hypotenuse in the chord-distance triangle.
Thinking all chords are diameters	You assume every chord passes through the center	Only a chord that goes through the center is a diameter. Most chords do not.

Common mistakes in circle chord problems

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KEY TAKEAWAY

Think of the radius as the VIP of every circle problem. It's like the team captain — it's always the longest side of the triangle you form inside the circle, and all the other measurements depend on it. When in doubt, start by finding the radius.

SECTION 8

Connecting to Bigger Ideas

The properties of radii and chords are just the beginning. As you learn more math, these same ideas will connect to bigger topics. Here's a preview of where this leads.

What You Know Now	What Comes Next
All radii in a circle are equal	The equation of a circle uses the radius: (x − h)² + (y − k)² = r²
A perpendicular from the center bisects a chord	In geometry proofs, this becomes a key theorem for proving triangles are congruent
The diameter is the longest chord	In advanced math, the diameter helps define the concept of a "metric" — how distances work
Pythagorean theorem with chords	Trigonometry uses the same right-triangle idea, connecting circles to sine and cosine

From current knowledge to future topics

You don't need to worry about these advanced topics for the SSAT. But it's good to know that mastering radii and chords now gives you a strong foundation for high school geometry and beyond.

SECTION 9

Practice Problems

Try these five problems on your own before reading the answers. They go from easier to harder, just like questions on the real SSAT.

Question 1: CONCEPTUAL

A circle has a radius of 7 inches. What is the length of the diameter? (A) 3.5 inches (B) 7 inches (C) 14 inches (D) 21 inches (E) 49 inches

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The correct answer is (C) 14 inches. The diameter is always twice the radius, so d = 2 × 7 = 14. Choice (A) is the radius divided by 2, which is a common mistake. Choice (B) is the radius itself, not the diameter. Choice (D) is 3 × 7, which confuses the diameter formula with something else. Choice (E) is 7², which would relate to area calculations, not diameter.

Question 2: BASIC CALCULATION

A circle has a diameter of 26 cm. A chord is drawn 5 cm from the center. What is the length of half of that chord? (A) 5 cm (B) 8 cm (C) 12 cm (D) 13 cm (E) 24 cm

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The correct answer is (C) 12 cm. First find the radius: r = 26 ÷ 2 = 13. Now use the Pythagorean theorem: 13² = 5² + (half-chord)², so 169 = 25 + (half-chord)², giving (half-chord)² = 144, and half-chord = 12. Choice (A) is the distance from center, not the half-chord. Choice (B) would come from a 5-8-13 guess, but 5² + 8² ≠ 13². Choice (D) is the radius, not the half-chord. Choice (E) is the full chord, but the question asked for half.

Question 3: INTERMEDIATE

A chord in a circle is 24 cm long. The radius of the circle is 13 cm. How far is the chord from the center of the circle? (A) 1 cm (B) 5 cm (C) 7 cm (D) 11 cm (E) 12 cm

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The correct answer is (B) 5 cm. The half-chord is 24 ÷ 2 = 12. Using the Pythagorean theorem: 13² = d² + 12², so 169 = d² + 144, giving d² = 25, and d = 5. Choice (A) is 13 − 12, which incorrectly subtracts instead of using the Pythagorean theorem. Choice (C) results from an arithmetic error. Choice (D) comes from incorrectly using the full chord (24) instead of the half-chord (12). Choice (E) is the half-chord itself, not the distance.

Question 4: APPLIED

A circular garden has a radius of 10 feet. A straight fence (a chord) is built across part of the garden, 6 feet from the center. How long is the fence? (A) 4 feet (B) 8 feet (C) 12 feet (D) 16 feet (E) 20 feet

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The correct answer is (D) 16 feet. The radius is 10, and the distance from the center to the chord is 6. Use the Pythagorean theorem: 10² = 6² + (half-chord)², so 100 = 36 + (half-chord)², giving (half-chord)² = 64, and half-chord = 8. The full fence length is 2 × 8 = 16 feet. Choice (A) is 10 − 6, a common subtraction error. Choice (B) is just the half-chord. Choice (C) is 2 × 6, which doubles the wrong value. Choice (E) is the diameter, not the chord length.

Question 5: CRITICAL THINKING

Two chords are drawn in the same circle. Chord PQ is 8 cm from the center, and chord RS is 6 cm from the center. The radius of the circle is 10 cm. Which statement is true? (A) Chord PQ is longer than chord RS (B) Chord RS is longer than chord PQ (C) Both chords have the same length (D) Chord RS is the diameter (E) There is not enough information to compare them

Show answer

The correct answer is (B). Use the Pythagorean theorem for each chord. For PQ: 10² = 8² + (half-PQ)², so half-PQ = √(100 − 64) = √36 = 6, and PQ = 12 cm. For RS: 10² = 6² + (half-RS)², so half-RS = √(100 − 36) = √64 = 8, and RS = 16 cm. Since 16 > 12, chord RS is longer. This makes sense because chords closer to the center are longer. Choice (A) gets the relationship backwards. Choice (C) is wrong because different distances from the center mean different lengths. Choice (D) is wrong because the diameter would be 20, not 16. Choice (E) is wrong because we have enough information to compute both lengths.

SUMMARY

Review: Radii and Chords

Summary

Every circle is defined by its center and its radius. All radii in the same circle are equal. A chord is a segment connecting two points on the circle, and the diameter is the longest chord, always equal to 2 × radius.

The most useful tool is the perpendicular from the center to a chord, which always bisects (cuts in half) the chord. This creates a right triangle where the radius is the hypotenuse. Use the Pythagorean theorem (r² = d² + half-chord²) to find any missing length. Remember: chords closer to the center are longer, and chords farther away are shorter. Memorize common Pythagorean triples like 3-4-5 and 5-12-13 to save time on the SSAT.