Two parallel chords of a circle whose radius is root 117 cm are on the same side of the centre of the distance between the chords is 3cm find the length of both the chords...pls give revalant answer
Answers
Answer:
In a circle, the perpendicular bisector from the centre to the mid point of the chord, forms a right angled triangle as shown in the figure.
Now, the perpendicular distance between the centre to the chord PQ is calculated as
=>OP
2
=OM
2
+MP
2
=>15
2
=OM
2
+9
2
(Since, MP is half of the chord PQ )
=>OM=12cm
Similarly, the perpendicular distance between the centre to the chord CD is calculated as
=>OC
2
=OE
2
+EC
2
=>15
2
=OE
2
+9
2
(
=>OE=12cm
Since the chords are on either side of the centre, distance between the chords =OM+OE=12+12=24cm
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= 2AB
= 230
=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