In a circle with centre O, chord AB=6cm and CD=5cm, If OM is perpendicular to AB and ON is perpendicular to CD. Then ___________
Answers
Answer:
Step-by-step explanation:
Given- AB=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 PQ=?
Solution-
We join OC and OA.
So, OC=OA=5 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=
2
1
AB=
2
1
×6 cm =3 cm and CQ=
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
OP=
OA
2
−AP
2
=
5
2
−4
2
cm =3 cm
Again in ΔOCQ, by Pythagoras theorem, we have
OQ=
OC
2
−CQ
2
=
5
2
−3
2
cm =4 cm.
∴PQ=OP−OQ=(4−3) cm =1 cm
Answer:
Step-by-step explanation:
Step-by-step explanation given in the pdf attached