- Two parallel chords of length 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. Find the distance between the chords.
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
Step-by-step explanation:
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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 = AB =30 = 15
2 2
Now, in right angled △OLA,
=(OA)² =(OA)² − (AL)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=15cm
Hence, distance between the chords =(LO+MO)=(8+15)cm=23cmMO)² = 225
=MO=15cm
Hence, distance between the chords =(LO+MO)=(8+15)cm=23cm
