A circle has radius $15\text{ cm}$. The perpendicular distance of a chord from the centre is $9\text{ cm}$. What is the length of the chord?

Step-by-step solution

1. Let the chord be $AB$ and $M$ be the midpoint of $AB$; then $OM\perp AB$ and $OM=9\text{ cm}$.
2. In right triangle $△OMA$, $OA$ is the radius $15\text{ cm}$ and $OM=9\text{ cm}$.
3. Apply Pythagoras: $O{A}^2=O{M}^2+A{M}^2$.
4. So $15^2=9^2+A{M}^2\Rightarrow A{M}^2=225-81=144$.
5. Hence $AM=\sqrt{144}=12\text{ cm}$.
6. Since $M$ is the midpoint, $AB=2\times AM=2\times 12=24\text{ cm}$.

Concept tested: Chord length from its distance from centre

Published 25 August 2026 · How this solution was prepared and checked

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