Question

A chord of length 24 cm is at a perpendicular distance of 5 cm from the centre of the

circle. Find the length of the chord of the same circle which is at a perpendicular distance

of 12cm from the centre. ​

Answer 1

Hope this helps you!!

Answer 2

Answer:-

Given:

• A circle with center O.

• Length of chord AB = 24cm

• Distance of chord of from the centre OC = 5cm

To Find:

• Radius of circle

• Length of chord PQ which is at the distance of 12 cm from the centre

Theoream used:

• The line perpendicular from the center to the chord bisects the chord.

Solution:

➥ Construction-

Join OA(radius)

➣ In ∆OAC

OC = 5cm

AC = ½ AB = ½×24 = 12cm

(The line perpendicular from the center to the chord bisects the chord)

OC⊥AC

So, it is a right angle triangle.

By pythagoras theoream

h² = b² + p²

➟ (OA)² = (AC)² + (OC)²

➟ (OA)² = 12² + 5²

➟ (OA)² = 144 + 25

➟ OA = √169

➟ OA = √13×13

∴ OA = radius = 13cm

➥ Construction

Join OP(radius)

➣ In ∆PRO

OP = radius = 13cm

OR = 12 cm (Given)

OR⊥PQ[Given]

So, it is a right angle triangle.

By pythagoras theoream

h² = b² + p²

➟ (OP)² = (PR)² + (OR)²

➟ (13)² = (PR)² + 12²

➟ 169 = PR² + 144

➟ PR = √169-144

➟ PR = √5×5

∴ PR = 5cm

As we know,

The line perpendicular from the center to the chord bisects the chord.

PQ = 2×PR = 2×5 = 10cm

∴ Measure of chord which is at the distance of 12 cm is 10cm.