Explain why the following statement is true: If the perpendicular distance of a chord from the centre is and the radius is , then the chord length is .

Question

Homi · Accepted Answer

Understand the Question

The perpendicular drawn from the centre of a circle to a chord bisects the chord.
Joining the centre to an endpoint of the chord creates a right-angled triangle where:
- The hypotenuse is the radius $r$ .
- One leg is the perpendicular distance $d$ .
- The other leg is half the chord length.
Applying the Pythagoras theorem gives half the chord length as $\sqrt{r^2 - d^2}$ , making the total chord length $2\sqrt{r^2 - d^2}$ .

Step 1 · Set Up the Geometry

Let $O$ be the centre of the circle, $AB$ be the chord, and $OM \perp AB$ where $M$ lies on $AB$ . Diagram 1

Given:

Radius $OA = r$
Perpendicular distance $OM = d$

Since the perpendicular from the centre to a chord bisects the chord, $M$ is the midpoint of $AB$ : $AM = MB = \dfrac{AB}{2} \implies AB = 2 \times AM$

Step 2 · Apply Pythagoras Theorem

In right-angled triangle $\triangle OMA$ (with $\angle OMA = 90^\circ$ ):

\begin{aligned} OA^2 &= OM^2 + AM^2 \\[0.6em] r^2 &= d^2 + AM^2 \\[0.6em] AM^2 &= r^2 - d^2 \\[0.6em] AM &= \sqrt{r^2 - d^2} \end{aligned}

Therefore, the full chord length is:

\begin{aligned} AB &= 2 \times AM \\[0.6em] &= 2\sqrt{r^2 - d^2} \end{aligned}

Answer

Hence proved, the chord length is $2\sqrt{r^2 - d^2}$ .

Common Mistakes

Forgetting to Double the Segment: Finding $AM = \sqrt{r^2 - d^2}$ and forgetting that $AM$ is only half the chord, so total chord length requires multiplying by $2$ .
Misidentifying the Hypotenuse: Treating the perpendicular distance $d$ or half-chord $AM$ as the hypotenuse instead of the radius $r$ , which incorrectly leads to $r = \sqrt{d^2 - AM^2}$ .

Question 2

More questions in Exercise 5.5