Question

12. Two chords AB and CD of lengths 5 cm and 11 cm
respectively of a circle are parallel to each other and are on the same side of its centre. If the distance between AB and CD is 3 cm, find the radius of the circle.
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Answer 1

Answer:

answer is 5.6 cm

Step-by-step explanation:

Join OA and OC.

Let the radius of the circle be r cm and O be the centre

Draw OP⊥AB and OQ⊥CD.

We know, OQ⊥CD, OP⊥AB and AB∥CD.

Therefore, points P,O and Q are collinear. So, PQ=6 cm.

Let OP=x.

Then, OQ=(6–x) cm.

And OA=OC=r.

Also, AP=PB=2.5 cm and CQ=QD=5.5 cm.

(Perpendicular from the centre to a chord of the circle bisects the chord.)

In right triangles QAP and OCQ, we have

OA

2

=OP

2

+AP

2

and OC

2

=OQ

2

+CQ

2

∴r

2

=x

2

+(2.5)

2

..... (1)

and r

2

=(6−x)

2

+(5.5)

2

..... (2)

⇒x

2

+(2.5)

2

=(6−x)

2

+(5.5)

2

⇒x

2

+6.25=36−12x+x

2

+30.25

12x=60

∴x=5

Putting x=5 in (1), we get

r

2

=52+(2.5)

2

=25+6.25=31.25

⇒r

2

=31.25⇒r=5.6

Hence, the radius of the circle is 5.6 cm