In a circle with centre O and radius 5cm, AB and CD are two parallel chords on the same side of centre. if AB=6cm and CD=8cm. find the difference of AB and CD from the centre.
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Answer:
Given- A B=6 cm and CD=8 cm are the chords of a circle of radius 5 cm with centre at O.
OP⊥AB at M and OQ⊥CD at N.
To find out - the length of P Q=?
Solution-
We join OC and O A.
So, OC=O A=5 cm, since OC and O A are radii.
ΔOAP and ΔOCQ are right ones, since OP⊥AB at P and OQ⊥CD at Q.
Now A P=
2
1
A B=
2
1
×6 cm =3 cm and C Q=
2
1
CD=
2
1
×8 cm =4 cm, since the perpendicular from the centre of a circle to a chord bisects the latter.
So, in ΔOAP, by Pythagoras theorem, we have
O P=
O A
2
−A P
2
=
5
2
−4
2
cm =3 cm
Again in ΔOCQ, by Pythagoras theorem, we have
The Q=
OC
2
−C Q
2
=
5
2
−3
2
cm =4 cm.
∴P Q=O P−The Q=(4−3) cm =1 cm
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