(c) In a circle of radius 17cm, two II chords of length 30cm and 16cm are
drawn.
Find the distance between the chords if both chords are (l) on the
opposite side of the centre (ii) on the same side of
the centre.
Answers
Answer:
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the center.
AB=30cm and CD=16cm [ Given ]
Draw OL⊥AB and OM⊥CD.
Join OA and OC.
OA=OC=17cm [ Radius of a circle ]
The perpendicular from the center of a circle to a chord bisects the chord.
∴ AL=
2
AB
=
2
30
=15cm
Now, in right angled △OLA,
∴ (OA)
2
=(AL)
2
+(LO)
2
[ By Pythagoras theorem ]
⇒ (LO)
2
=(OA)
2
−(AL)
2
⇒ (LO)
2
=(17)
2
−(15)
2
⇒ (LO)
2
=289−225
⇒ (LO)
2
=64
⇒LO=8
Similarly,
In right angled △CMO,
⇒ (OC)
2
=(CM)
2
+(MO)
2
⇒ (MO)
2
=(OC)
2
−(CM)
2
⇒ (MO)
2
=(17)
2
−(8)
2
⇒ (MO)
2
=289−64
⇒ (MO)
2
=225
∴ MO=15cm
Hence, distance between the chords =(LO+MO)=(8+15)cm=23cm
Answer:
thanks for free points