Circles and Geometric Shapes | Exercise 5.5

Question 1

Find the length of the chord of a circle where the radius is 7 cm7\text{ cm} and perpendicular distance is 6 cm6\text{ cm}.

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Solution
Understand the Question
  • The perpendicular drawn from the centre of a circle to a chord bisects the chord.
  • The radius (extOA ext{OA}), perpendicular distance (extOM ext{OM}), and half-chord (extAM ext{AM}) form a right-angled triangle ΔOMA\Delta \text{OMA} with the radius as the hypotenuse.
  • We use the Pythagoras theorem to calculate the half-chord length and then multiply by 22 to obtain the total chord length.

Step 1 · Find Half the Chord Length

Let AB\text{AB} be the chord of the circle with centre O\text{O}, and let OM⊥AB\text{OM} \perp \text{AB}.Diagram 1

Given radius OA=7 cm\text{OA} = 7\text{ cm} and perpendicular distance OM=6 cm\text{OM} = 6\text{ cm}.

In right-angled triangle ΔOMA\Delta \text{OMA}, by Pythagoras theorem

OA2=OM2+AM272=62+AM249=36+AM2AM2=49−36AM2=13AM=13 cm\begin{aligned} \text{OA}^2 &= \text{OM}^2 + \text{AM}^2 \\ 7^2 &= 6^2 + \text{AM}^2 \\ 49 &= 36 + \text{AM}^2 \\ \text{AM}^2 &= 49 - 36 \\ \text{AM}^2 &= 13 \\ \text{AM} &= \sqrt{13} \text{ cm} \end{aligned}

Step 2 · Calculate Full Chord Length

Since the perpendicular from the centre to a chord bisects the chord, AB=2×AM\text{AB} = 2 \times \text{AM}.

AB=2×AM=2×13=213 cm\begin{aligned} \text{AB} &= 2 \times \text{AM} \\[0.6em] &= 2 \times \sqrt{13} \\[0.6em] &= 2\sqrt{13} \text{ cm} \end{aligned}
Answer

213 cm2\sqrt{13}\text{ cm}

Common Mistakes
  • Forgetting to Double the Length: Stopping after finding AM=13 cm\text{AM} = \sqrt{13}\text{ cm}, which is only half the chord length.
  • Hypotenuse Confusion: Confusing the perpendicular distance with the hypotenuse. The radius of the circle is always the hypotenuse in ΔOMA\Delta \text{OMA}.

More questions in Exercise 5.5

Q1

Find the length of the chord of a circle where the radius is 7 cm7\text{ cm} and perpendicular distance is 6 cm6\text{ cm}.

Q2

Explain why the following statement is true: If the perpendicular distance of a chord from the centre is dd and the radius is rr, then the chord length is 2r2−d22\sqrt{r^2 - d^2}.

Q3

In a circle, if the distance of chord ABAB from the centre is twice the distance of another chord CDCD from the centre, then can we conclude that CD=2ABCD = 2AB? Give reasons for your answer.

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