Question

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.


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Answer 1

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

solution

Answer 2

Answer:

Here, OM ⊥ AB and ON ⊥ CD. is drawn and OB and OD are joined.

As we know, AB bisects BM as the perpendicular from the centre bisects the chord.

Since AB = 5 so,

BM = AB/2

Similarly, ND = CD/2 = 11/2

Now, let ON be x.

So, OM = 6− x.

Consider ΔMOB,

OB2 = OM2 + MB2

Or,

OB2 = 36 + x2 – 12x + 25/4 ……(1)

Consider ΔNOD,

OD2 = ON2 + ND2

Or,

OD2 = x2+121/4 ……….(2)

We know, OB = OD (radii)

From eq. (1) and eq. (2) we have;

36 + x2 -12x + 25/4 = x2 + 121/4

12x = 36 + 25/4 – 121/4

12x = (144 + 25 -121)/4

12x = 48/4 = 12

x = 1

Now, from eq. (2) we have,

OD2 = 11 + (121/4)

Or OD = (5/2) × √5