Circles and Geometric Shapes | Exercise 5.5

Question 3

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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Solution
Understand the Question
  • The length of a chord at a perpendicular distance dd from the centre of a circle with radius rr is given by 2r2−d22\sqrt{r^2 - d^2}.
  • Because the chord length depends on the square root of the difference of squares, the relationship between chord length and its distance from the centre is non-linear.
  • Doubling the distance from the centre does not simply halve the chord length, so we cannot conclude that CD=2ABCD = 2AB.

Step 1 · Compare Chord Lengths Algebraically

Let the radius of the circle be rr, and the perpendicular distance of chord CDCD from the centre be dd. Then the distance of chord ABAB from the centre is 2d2d.Diagram 1

The formula for the length of a chord at distance xx from the centre is 2r2−x22\sqrt{r^2 - x^2}.

Length of chord CDCD: CD=2r2−d2CD = 2\sqrt{r^2 - d^2}

Length of chord ABAB:

AB=2r2−(2d)2=2r2−4d2\begin{aligned} AB &= 2\sqrt{r^2 - (2d)^2} \\[0.6em] &= 2\sqrt{r^2 - 4d^2} \end{aligned}

Now, calculate 2AB2AB:

2AB=2×2r2−4d2=4r2−4d2\begin{aligned} 2AB &= 2 \times 2\sqrt{r^2 - 4d^2} \\[0.6em] &= 4\sqrt{r^2 - 4d^2} \end{aligned}

Since 2r2−d2≠4r2−4d22\sqrt{r^2 - d^2} \neq 4\sqrt{r^2 - 4d^2} in general, we cannot conclude that CD=2ABCD = 2AB.

Step 2 · Verify with a Numerical Counterexample

Let radius r=5 cmr = 5\text{ cm}, distance of CDCD from centre d=2 cmd = 2\text{ cm}, and distance of ABAB from centre 2d=4 cm2d = 4\text{ cm}.

Length of chord CDCD:

CD=252−22=225−4=221≈9.16 cm\begin{aligned} CD &= 2\sqrt{5^2 - 2^2} \\[0.6em] &= 2\sqrt{25 - 4} \\[0.6em] &= 2\sqrt{21} \approx 9.16\text{ cm} \end{aligned}

Length of chord ABAB:

AB=252−42=225−16=29=2×3=6 cm\begin{aligned} AB &= 2\sqrt{5^2 - 4^2} \\[0.6em] &= 2\sqrt{25 - 16} \\[0.6em] &= 2\sqrt{9} \\[0.6em] &= 2 \times 3 = 6\text{ cm} \end{aligned}

Calculating 2AB2AB:

2AB=2×6=12 cm\begin{aligned} 2AB &= 2 \times 6 = 12\text{ cm} \end{aligned}

Since 221≠122\sqrt{21} \neq 12, CD≠2ABCD \neq 2AB.

Answer

No, we cannot conclude that CD=2ABCD = 2AB because the chord length 2r2−d22\sqrt{r^2 - d^2} has a non-linear relationship with distance dd.

Common Mistakes
  • Assuming Linear Proportionality: Assuming that chord length is directly or inversely proportional to distance from the centre (i.e., doubling the distance halves the chord length).
  • Forgetting the Square Root: Neglecting the Pythagoras relation L=2r2−d2L = 2\sqrt{r^2 - d^2}, which shows chord length depends on the difference of squares under a radical.

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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