Question

in a circle of radius 26cm l, ab and cd are two parellel chords on opposite sides of the centre. if AB = 48cm and CD = 20cm, find the distance between the chords

please explain.
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Answer 1

Answer:

ans = OP⊥AB at M and OQ⊥CD at N.

To find out - 

If the length of PQ=? 

Solution- 

We join OC and OA. 

So, OC=OA=25 cm, since OC and OA are radii 

ΔOAP and ΔOCQ are right ones, since OP⊥AB at P and OQ⊥CD at Q.

Now AP=21AB=21×14 cm =7 cm and

CQ=21CD=21×48 cm =24 cm

Since the perpendicular from the centre of a circle to a chord bisects the latter.

So, in ΔOAP, by Pythagoras theorem, we have

OP=OA2−AP2=252−72 cm =24 cm

Again in ΔOCQ, by Pythagoras theorem, we have

OQ=OC2−CQ2=252−242 cm =7 cm

∴PQ=OP−OQ=(24−7) cm =17 cm

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