Question

Two chords AB and CD of lengths 8 cm and 10 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 10cm , find the radius
of the circle.

The one who answer correct I'll follow him

I'll mark him brainliest
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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=10cm.

Let OP=x.

Then, OQ=(10–x) cm.

And OA=OC=r.

Also, AP=PB=4cm and CQ=QD= 5cm.

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

In right triangles QAP and OCQ,

we have

OA²=OP² +AP² and OC²=OQ²+CQ²

∴r² =x² +(4)² ......{1}

and r²=(10−x)² +(5)² ..... {2}

⇒x²+(4)² =(10−x)² +(5)²

⇒x²+16=100−20x+x²+25

⇒x²-x²+16=100-20x+25

⇒20x=100+25-16

⇒20x=109

⇒ x=109/20

∴ x =5.45

Putting x=5.45 in (1), we get

r²=5.45²+(4)²

=29.70+16= 45.70

⇒r²=45.70⇒ r=22.85.

Hence, the radius of the circle is 22.85 cm