Question

The parallel chords of a circle with centre at C
are such that PQ = 8 cm and RS = 16 cm. The
chords lie on one side of centre and the distance
between them is 4 cm. The radius of circle is
(a) 3√2 cm
(b) 3√5 cm
(c) 4√5 cm
(d) 5√5 cm​

Answer 1

Answer:

Step-by-step explanation:

Given, chords PQ=8 cm and RS=16 cm and AB=4 cm

Let perpendicular from the centre meet the chord RS at B and PQ at A.

Since, the perpendicular from the centre of the circle to the chord bisects the chord,

RB=BS=8 cm and PA=AQ=4 cm.

Let OB=x cm, then OA=OB+AB=(x+4) cm.

In right triangle OBS,

OS

2

=OB

2

+BS

2

=OB

2

+64          (i)

In right triangle OAQ,

OQ

2

=OA

2

+AQ

2

=OA

2

+16          (ii)

∴ OS and OQ are the radii of the circle

∴ OB

2

+64=OA

2

+16

⇒ x

2

+64=(x+4)

2

+16

⇒ x

2

+64=x

2

+8x+16+16

⇒ 8x=32⇒x=4

∴ From eqn (i)

Radius of the circle(OS)

2

=16+64=80

⇒ OS=4

5

​

 cm

I HOPE IT HELPS YOU