Question

AB (= 20 cm) is diameter of the given circle and AP (= 16 cm). The distance of chord AP from centre O is ______.

Options

• 12 cm

• 18 cm

• 9 cm

• 6 cm

Answer 1

AB (= 20 cm) is diameter of the given circle and AP (= 16 cm). The distance of chord AP from centre O is 6 cm.

Explaantion:

Since AB is the diameter of the circle with length 20 cm, the radius r of the circle is:

r = `20/2` = 10 cm

The angle subtended by a diameter at any point on the circumference is a right angle (90°). Therefore, ∠APB = 90°, which means ΔAPB is a right-angled triangle with hypotenuse AB = 20 cm and side AP = 16 cm.

Using the Pythagorean theorem in ΔAPB to find the length of the other chord segment PB:

AP² + PB² = AB²

16² + PB² = 20²

256 + PB² = 400

PB² = 400 − 256 = 144

∴ PB = 12 cm

To find the distance of chord AP from the centre O, let us drop a perpendicular from centre O to chord AP, meeting it at point M. Alternatively, consider the mid-point theorem/line joining the centre to the mid-point of a chord. In ΔAPB, the line segment joining the centre O (which is the mid-point of diameter AB) and the mid-point of chord AP is parallel to PB and its length is half the length of PB:

Distance from O to AP = `\frac{1}{2} \times PB = \frac{1}{2} \times 12 = 6\text{ cm}`

Thus, the distance of chord AP from the centre O is 6 cm.