Every Straight Cut Across a Circle Has the Same Hidden Rule
Slice a round cake anywhere with a straight knife, not necessarily through the middle, and the cut you leave on the top is a chord. Make the cut nearer the edge and it is short; slide it toward the centre and it grows. Push it right through the middle and you get the longest cut the cake allows. Behind that simple picture sits a single right-triangle relationship that fixes the length of every chord from just two numbers.
Once you can see the radius, half the chord, and that perpendicular forming a right triangle, every chord-length problem becomes the Pythagorean theorem in disguise.
What Is a Chord of a Circle?
A chord of a circle is a straight line segment whose two endpoints both lie on the circle's boundary. Join any two points on the circumference with a straight line, and that segment is a chord.
The diameter is a special chord: it is the chord that passes through the centre, and it is the longest chord any circle has. Every other chord is shorter, because no straight line across a circle can be longer than the one that goes through the middle. A chord also splits the circle into two pieces, called segments, a larger major segment and a smaller minor segment.
It helps to place the chord alongside the other circle parts. The radius runs from the centre to the boundary; a chord runs from boundary to boundary; the diameter is the chord that happens to pass through the centre. A short menu of these terms sits in parts of a circle, but the one feature that defines a chord is simple: both ends touch the circle, and it need not go through the centre.
The Chord Length Formula — Two Ways to Find It
The length of a chord can be found two ways, depending on what the problem gives you. Define the variables first: L is the chord length, r the radius of the circle, d the perpendicular distance from the centre to the chord, and θ the angle the chord subtends at the centre.
From the perpendicular distance. Drop a perpendicular from the centre to the chord. It hits the chord at its midpoint (that is a theorem we prove below), so it splits the chord into two equal halves. The radius, one half-chord, and the perpendicular form a right triangle, with the radius as the hypotenuse. By the Pythagorean theorem, the half-chord is $\sqrt{r^2 - d^2}$, and the full chord is twice that:
$$L = 2\sqrt{r^2 - d^2}.$$
From the central angle. If instead you know the angle θ the chord subtends at the centre, the chord is the base of an isosceles triangle with two radius sides. Splitting that triangle down its line of symmetry gives a right triangle whose opposite side is half the chord, so the half-chord is $r \sin!\left(\tfrac{\theta}{2}\right)$, and the full chord is:
$$L = 2r\sin!\left(\frac{\theta}{2}\right).$$
Both formulas describe the same chord; you pick whichever matches the information you are handed. If the angle is given, use the second; if a perpendicular distance is given, use the first.
The Chord Theorems — and Why They Hold
A handful of theorems about chords come up again and again, and most school problems are an application of one of them. Here are the three that matter most.
Theorem 1: The perpendicular from the centre bisects the chord
If you drop a perpendicular from the centre of a circle to a chord, it cuts the chord exactly in half. The proof is short: join the centre to both ends of the chord, giving two radii of equal length; the perpendicular is shared by the two triangles formed, and each has a right angle where it meets the chord. The two triangles are congruent (right angle, hypotenuse, side), so the two half-chords are equal. This is the theorem the length formula quietly relies on.
Theorem 2: Equal chords are equidistant from the centre, and chords equidistant from the centre are equal
Two chords of the same length sit the same perpendicular distance from the centre, and the rule works in reverse too. This follows straight from the length formula: $L = 2\sqrt{r^2 - d^2}$ depends only on $r$ and $d$, so for a fixed radius, equal $L$ forces equal $d$.
Theorem 3: The larger of two unequal chords is closer to the centre
Look at the formula again: as the perpendicular distance $d$ shrinks toward zero, the chord length $L$ grows. When $d = 0$ the chord passes through the centre and becomes the diameter, the longest chord. So the closer a chord sits to the centre, the longer it is.
These three are not separate facts to memorise; they are all the single relationship $L = 2\sqrt{r^2 - d^2}$ read in different directions. Equal chords, equidistant chords, the longest chord, each is what the formula says when you hold one quantity fixed and vary another.
Examples of the Chord of a Circle
With the formulas and theorems in place, here is the chord doing real work. The problems move from a one-step length calculation up to a proof-style application of the bisection theorem.
Example 1: A chord lies 4 cm from the centre of a circle whose radius is 5 cm. Find the length of the chord
Use the perpendicular-distance formula with $r = 5$ and $d = 4$:
$$L = 2\sqrt{r^2 - d^2} = 2\sqrt{25 - 16} = 2\sqrt{9} = 2 \times 3 = 6 \text{ cm}.$$
Final answer: L = 6 cm.
Example 2: A chord subtends an angle of $60°$ at the centre of a circle of radius 10 cm. Find the chord length
A first instinct is to forget the half-angle and write $L = 2r\sin\theta = 2(10)\sin 60° \approx 17.3$ cm. Check that against the picture. The chord is the straight line across, and it can never be longer than the diameter, which here is 20 cm; 17.3 cm is plausible in size but comes from the wrong formula. The two radii to the chord's endpoints form an isosceles triangle, and splitting it uses half the central angle, not the whole angle.
The correct formula uses $\tfrac{\theta}{2}$:
$$L = 2r\sin!\left(\frac{\theta}{2}\right) = 2(10)\sin 30° = 2(10)(0.5) = 10 \text{ cm}.$$
Final answer: L = 10 cm. (Here $\sin 30° = 0.5$; the sine ratio is the opposite side over the hypotenuse in a right triangle.) In Bhanzu's Grade 10 cohort at the McKinney TX center, dropping the half-angle is the most common first-attempt error on this formula, appearing in close to half of first tries until students sketch the isosceles triangle before substituting.
Example 3: The perpendicular distance from the centre to a chord is 8 cm, and the chord is 12 cm long. Find the radius of the circle
Rearrange the length formula. With $L = 12$, the half-chord is 6, and $d = 8$:
$$r^2 = \left(\frac{L}{2}\right)^2 + d^2 = 6^2 + 8^2 = 36 + 64 = 100, \qquad r = \sqrt{100} = 10 \text{ cm}.$$
Final answer: r = 10 cm.
Example 4: Two parallel chords of a circle of radius 13 cm have lengths 10 cm and 24 cm. Find the distance between them if they lie on the same side of the centre
Find each chord's distance from the centre using $d = \sqrt{r^2 - (L/2)^2}$:
$$d_1 = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 \text{ cm}, \qquad d_2 = \sqrt{13^2 - 12^2} = \sqrt{25} = 5 \text{ cm}.$$
On the same side of the centre, the gap between them is the difference: $12 - 5 = 7$ cm.
Final answer: 7 cm.
Example 5: A chord of a circle is equal in length to the radius. Find the angle it subtends at the centre
If the chord equals the radius, the two radii to its endpoints and the chord itself are all the same length, so the triangle is equilateral. Every angle in an equilateral triangle is $60°$, so the central angle is $60°$.
Final answer: $60°$.
Example 6: Prove that the perpendicular from the centre of a circle to a chord bisects the chord, using the chord AB and centre O
Join OA and OB; both are radii, so $OA = OB$. Let the perpendicular from O meet AB at M, so $\angle OMA = \angle OMB = 90°$. Triangles OMA and OMB share the side OM and have equal hypotenuses $OA = OB$, so by the right-angle, hypotenuse, side rule they are congruent. Congruent triangles give $AM = MB$, so M is the midpoint and the perpendicular bisects the chord.
Final answer: the perpendicular from the centre bisects the chord (Theorem 1 proved).
Where Chords Show Up
Chords matter because they turn a curved boundary into a straight, measurable line, which is exactly what engineers and designers need to work with a circle.
Bridges and arches. A circular arch is described by its span (a chord) and its rise. Builders specify an arch by its chord and the perpendicular height, then back out the radius the stone must be cut to.
Machining and inspection. When a machinist measures a round part across two points without reaching the centre, that measurement is a chord; the chord-length formula recovers the true radius from it, which is how circular parts are checked against spec.
Astronomy and surveying. The straight-line distance across a circular arc, a chord, is how surveyors lay out curved roads and how the angular size of a distant object converts to a real distance.
Music and acoustics. A curved concert-hall wall reflects sound along chord-like straight paths between points on the curve, which is why the geometry of chords shows up in acoustic design.
For a Class 9 or 10 student, chords are the gateway into the whole circle-theorems chapter: master the single relationship $L = 2\sqrt{r^2 - d^2}$ and most of the theorems stop being separate rules and become one idea seen from different angles.
Where Students Trip Up on Chords
Mistake 1: Using the full central angle instead of half
Where it slips in: A problem gives the central angle and the student writes $L = 2r\sin\theta$ with the whole angle.
Don't do this: Plug the full angle straight into the sine.
The correct way: Splitting the isosceles triangle of two radii uses half the central angle, so the formula is $L = 2r\sin!\left(\tfrac{\theta}{2}\right)$. The memorizer who recalls "$2r\sin$ something" without the diagram is the one who drops the half.
Mistake 2: Forgetting that the perpendicular bisects the chord
Where it slips in: A student uses the full chord as a triangle side instead of the half-chord.
Don't do this: Set up the right triangle with $L$ as a leg.
The correct way: The perpendicular from the centre cuts the chord in half, so the leg of the right triangle is $\tfrac{L}{2}$, not $L$. Always halve the chord before applying the Pythagorean theorem.
Mistake 3: Confusing the perpendicular distance with the radius
Where it slips in: A student assumes the radius reaches the chord's midpoint, mixing up the perpendicular distance $d$ with the radius $r$.
Don't do this: Substitute the radius where the perpendicular distance belongs, or the reverse.
The correct way: Keep the right triangle clear: $r$ is the hypotenuse (centre to endpoint), $d$ is one leg (centre to chord midpoint), and $\tfrac{L}{2}$ is the other leg. Label them on a sketch before substituting.
Key Takeaways
A chord of a circle is a straight segment joining two points on the boundary; the diameter is the longest chord.
Chord length is found two ways: $L = 2\sqrt{r^2 - d^2}$ from the perpendicular distance, and $L = 2r\sin!\left(\tfrac{\theta}{2}\right)$ from the central angle.
The perpendicular from the centre bisects the chord, which is the right triangle behind the first formula.
Equal chords are equidistant from the centre, and the closer a chord is to the centre, the longer it is.
The most common mistake is using the full central angle instead of half it in the sine formula.
Practice These Problems to Solidify Your Understanding
A chord is 6 cm from the centre of a circle of radius 10 cm. Find the chord length.
A chord subtends a $90°$ angle at the centre of a circle of radius 8 cm. Find the chord length.
A 16 cm chord lies in a circle of radius 10 cm. Find its perpendicular distance from the centre.
Answer to Question 1: L = 16 cm. Answer to Question 2: L = $8\sqrt{2} \approx 11.3$ cm. Answer to Question 3: d = 6 cm. If Question 2 gave you 16, you used the full angle instead of half it (see Mistake 1).
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