Q1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.

Answer:

Given: Radius (r) = 7 cm, distance from centre to chord (d) = 6 cm.

Using the formula: length of chord = \( 2\sqrt{r^2 - d^2} \)

= \( 2\sqrt{7^2 - 6^2} = 2\sqrt{49 - 36} = 2\sqrt{13} \) cm

Therefore, the length of the chord is \( 2\sqrt{13} \) cm.

Q2. Explain why the following statement is true: If the perpendicular distance of a chord from the centre is \( d \) and the radius is \( r \), then the chord length is \( 2\sqrt{r^2 - d^2} \).

Answer:

Let the chord be AB and its midpoint be M. The line from the centre O to M is perpendicular to AB. In right triangle OMA:

• OA = r (radius)
• OM = d (distance from centre to chord)
• AM = half the chord length (since M is the midpoint)

Using the Baudhayana-Pythagoras theorem:

\( OA^2 = OM^2 + AM^2 \)
\( r^2 = d^2 + AM^2 \)
\( AM^2 = r^2 - d^2 \)
\( AM = \sqrt{r^2 - d^2} \)

Since AB = 2 × AM, we get:

Length of chord AB = \( 2\sqrt{r^2 - d^2} \)

Q3. In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that \( \mathrm{CD} = 2 \mathrm{AB} \)? Give reasons for your answer.

Answer:

No, we cannot conclude that CD = 2AB.

Let the radius of the circle be r. Let the distances from the centre to AB and CD be d₁ and d₂ respectively.

Given: d₁ = 2d₂.

Using the chord length formula:

AB = \( 2\sqrt{r^2 - d_1^2} \) and CD = \( 2\sqrt{r^2 - d_2^2} \)

If d₁ = 2d₂, then AB = \( 2\sqrt{r^2 - 4d_2^2} \), while CD = \( 2\sqrt{r^2 - d_2^2} \).

For CD to be twice AB, we would need \( \sqrt{r^2 - d_2^2} = 2\sqrt{r^2 - 4d_2^2} \), which is not generally true. The relationship between chord lengths and their distances from the centre is not linear.