In a circle, a chord is away from the centre. If the radius of the circle is , what is the length of the chord?

Understand the Question The perpendicular drawn from the centre of a circle to a chord bisects the chord. Connecting the centre to an endpoint of the chord forms a right-angled triangle where: The hypotenuse is the radius of the circle (). One leg is the distance from the centre to the chord (). The other leg is half the length of the chord. Using the Pythagoras theorem, we find half the chord's length and double it to get the total length of the chord. Step 1 · Find Half the Chord Length Let the circle have centre and chord . Draw meeting at .Given: In right-angled , using Pythagoras theorem: Step 2 · Calculate the Total Length of the Chord The perpendicular from the centre of a circle to a chord bisects the chord: Answer > Common Mistakes Forgetting to Double: Stopping after finding , which is only half the chord's length. Hypotenuse Confusion: Confusing the perpendicular distance () with the hypotenuse. The radius () is always the hypotenuse in .

Circles and Geometric Shapes | EOT

Question 1

In a circle, a chord is $5\text{ cm}$ away from the centre. If the radius of the circle is $13\text{ cm}$ , what is the length of the chord?

Solution

Understand the Question

The perpendicular drawn from the centre of a circle to a chord bisects the chord.
Connecting the centre to an endpoint of the chord forms a right-angled triangle where:
- The hypotenuse is the radius of the circle ( $r = 13\text{ cm}$ ).
- One leg is the distance from the centre to the chord ( $d = 5\text{ cm}$ ).
- The other leg is half the length of the chord.
Using the Pythagoras theorem, we find half the chord's length and double it to get the total length of the chord.

Step 1 · Find Half the Chord Length

Let the circle have centre $\text{O}$ and chord $\text{AB}$ . Draw $\text{OM} \perp \text{AB}$ meeting $\text{AB}$ at $\text{M}$ .Given: $\text{Radius } (\text{OA}) = 13\text{ cm}$ $\text{Distance from centre } (\text{OM}) = 5\text{ cm}$

In right-angled $\Delta \text{OMA}$ , using Pythagoras theorem:

\begin{aligned} \text{OA}^2 &= \text{OM}^2 + \text{AM}^2 \\ 13^2 &= 5^2 + \text{AM}^2 \\ 169 &= 25 + \text{AM}^2 \\ \text{AM}^2 &= 169 - 25 \\ \text{AM}^2 &= 144 \\ \text{AM} &= \sqrt{144} = 12\text{ cm} \end{aligned}

Step 2 · Calculate the Total Length of the Chord

The perpendicular from the centre of a circle to a chord bisects the chord: $\text{AB} = 2 \times \text{AM}$

\begin{aligned} \text{AB} &= 2 \times 12 \\ &= 24\text{ cm} \end{aligned}

Answer

$24\text{ cm}$

Common Mistakes

Forgetting to Double: Stopping after finding $\text{AM} = 12\text{ cm}$ , which is only half the chord's length.
Hypotenuse Confusion: Confusing the perpendicular distance ( $5\text{ cm}$ ) with the hypotenuse. The radius ( $13\text{ cm}$ ) is always the hypotenuse in $\Delta \text{OMA}$ .

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Q18

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Q20

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Q23

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Q24

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