Chord Properties of a Circle — Multiple Choice Questions

If a chord is at a distance of 8 cm from the centre of the circle of radius 17 cm, then the length of the chord will be :

1. 15 cm

2. 20 cm

3. 30 cm

4. 45 cm

Answer

Let the chord be AB and the perpendicular from the center O meet the chord at M.

In the right-angled triangle △OMA:

Using the Pythagorean theorem:

OA² = AM² + OM²

17² = AM² + 8²

289 = AM² + 64

AM² = 289 - 64

AM² = 225

AM = $\sqrt{225}$ = 15 cm

Since the perpendicular from the center bisects the chord.

Length of chord = 2 × AM = 30 cm.

Hence, option 3 is the correct option.

A chord of length 40 cm is drawn at a distance of 15 cm from the centre of a circle, then the radius of the circle will be :

1. 12 cm

2. 15 cm

3. 17 cm

4. 25 cm

Answer

Let the chord be AB and the perpendicular from the center O meet the chord at M.

Since the perpendicular from the center bisects the chord AB.

AM = $\dfrac{AB}{2} = \dfrac{40}{2}$ = 20 cm

In the right-angled triangle △OMA:

Using the Pythagorean theorem:

OA² = AM² + OM²

OA² = 20² + 15²

OA² = 225 + 400

OA² = 625

OA = $\sqrt{625}$ = 25 cm.

Hence, option 4 is the correct option.

In the figure, O is the centre of the circle and AC = BC = 3.5 cm. ∠ACO will be :

1. 60°

2. 80°

3. 90°

4. 100°

Answer

The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Since,

AC = BC [C is the midpoint of AB]

O is the centre.

Thus, OC is perpendicular to AB.

∠ACO = 90°.

Hence, option 3 is the correct option.

In the figure, O is the centre of the circle and OP = OQ and CD = 6 cm. The length of AB is :

1. 6 cm

2. 12 cm

3. 3 cm

4. 5 cm

Answer

Given,

OP = OQ

CD = 6 cm

Chords equidistant from the centre of a circle are equal.

AB = CD

∴ AB = 6 cm

Hence, option 1 is the correct option.

A chord of length 70 cm is drawn in a circle of radius 37 cm. The distance of the chord from the centre of the circle is :

1. 20 cm

2. 15 cm

3. 14 cm

4. 12 cm

Answer

Let the chord be AB and the perpendicular from the center O meet the chord at M.

Since the perpendicular from the center bisects the chord AB.

AM = $\dfrac{70}{2}$ = 35 cm

In the right-angled triangle △OMA:

Using the Pythagorean theorem:

OA² = AM² + OM²

37² = 35² + OM²

OM² = 1369 - 1225

OM² = 144

OM = $\sqrt{144}$ = 12 cm.

Hence, option 4 is the correct option.