Allday Education

Arcs & Chords

A chord is a segment with both endpoints on a circle, and every chord cuts off an arc. This lesson is about how chords and their arcs control each other. There are three theorems, and each one is an if-and-only-if — it works in both directions.

The three facts: congruent chords cut off congruent arcs, a diameter perpendicular to a chord bisects the chord and its arc, and chords the same distance from the center are congruent. Almost every problem in this lesson is one of these three plus a little arithmetic.

Congruent chords, congruent arcs

In the same circle (or in congruent circles), two chords are congruent if and only if their arcs are congruent. Equal chords pull off equal pieces of the circle, and equal pieces of the circle are held by equal chords.

So if a problem tells you JKLM\overline{JK} \cong \overline{LM} and gives you the measure of arc JKJK, the measure of arc LMLM is the same number — no computation needed.

A diameter perpendicular to a chord

If a diameter (or radius) is perpendicular to a chord, it bisects the chord and its arc. The perpendicular from the center cuts the chord exactly in half.

This sets up a right triangle you'll use constantly: the radius rr to a chord endpoint is the hypotenuse, and the legs are the distance dd from the center to the chord and half the chord. The Pythagorean theorem ties them together: r2=d2+(chord2)2r^2 = d^2 + \left(\dfrac{\text{chord}}{2}\right)^2.

Given any two of the three lengths — radius, distance to the chord, half the chord — you can find the third. The figure below shows that right triangle: the radius rr is the hypotenuse, the distance dd from the center is one leg, and half the chord is the other.

dd
rr
OO
AA
BB
MM

Chords equidistant from the center

Two chords in the same circle are congruent if and only if they are equidistant from the center. The closer a chord sits to the center, the longer it is; the diameter, passing through the center at distance zero, is the longest chord of all.

So if two chords are marked as the same distance from the center, their lengths are equal — and if two chords are congruent, the perpendicular distances from the center to each chord match.

Worked examples

Example 1: congruent chords

In circle PP, ABCD\overline{AB} \cong \overline{CD} and arc ABAB measures 9898^\circ. Find the measure of arc CDCD.

Congruent chords cut off congruent arcsABCD\overline{AB} \cong \overline{CD}
Set the arcs equalmCD=mABm\overset{\frown}{CD} = m\overset{\frown}{AB}
SubstitutemCD=98m\overset{\frown}{CD} = 98^\circ

Answer: mCD=98m\overset{\frown}{CD} = 98^\circ

Example 2: chord length from the right triangle

The radius of circle PP is 1010, and the perpendicular distance from PP to chord AB\overline{AB} is 66. Find ABAB.

Radius, distance, and half the chord form a right triangler2=d2+(AB2)2r^2 = d^2 + \left(\dfrac{AB}{2}\right)^2
Substitute102=62+(AB2)210^2 = 6^2 + \left(\dfrac{AB}{2}\right)^2
Solve for the squared leg(AB2)2=10036=64\left(\dfrac{AB}{2}\right)^2 = 100 - 36 = 64
Take the square rootAB2=8\dfrac{AB}{2} = 8
Double it for the full chordAB=16AB = 16

Answer: AB=16AB = 16

Example 3: equidistant chords

In circle OO, chords FG\overline{FG} and HJ\overline{HJ} are each 55 units from the center, and FG=24FG = 24. Find HJHJ.

Chords equidistant from the center are congruentHJ=FGHJ = FG
SubstituteHJ=24HJ = 24

Answer: HJ=24HJ = 24

Try one yourself

OO
JJ
KK
LL
MM
8282^\circ
??

Common questions

Does the perpendicular-diameter theorem work for any chord?

Yes, as long as the diameter (or radius) is actually perpendicular to the chord. If it crosses the chord at some other angle, it does not bisect it, and the right-triangle setup falls apart.

Why is half the chord in the Pythagorean formula, not the whole chord?

The perpendicular from the center bisects the chord, so the right triangle's leg only reaches the midpoint of the chord. Solve for the half first, then double it — forgetting to double is the most common mistake in this lesson.

Which chord is longer, one near the center or one near the edge?

The one near the center. Chord length shrinks as the distance from the center grows. The diameter, at distance zero, is the longest possible chord.

Do these theorems work across two different circles?

Only if the circles are congruent (same radius). In circles of different sizes, equal chords do not cut off equal arc measures.

Want the video version?

Allday Everyday Math has video lessons, practice, and an AI tutor for every topic, Pre-Algebra through Algebra 2.

Try it for $1