1
o is the centre of circle op = 41
units...pm =40 units then om using activity
opsq. =
omsq. +Pmsq.
Answers
Answered by
0
Answer:
Area of ΔOPQ=
2
1
r
2
sin(θ)=
2
1
⋅225⋅
2
1
=
4
225
Area of sector = π
360
o
θ
⋅r
2
=
12
225
π=
4
75
π
Area between the chord and arc = Area of sector - Area of triangle =
4
75
π−
4
225
=
4
75
(π−3)
Answered by
5
Given that,
6
Radius of circle (OA) = 41cm
Chord (AB) = 80cm
Draw OC⊥AB
We know that
The perpendicular from centre to chord bisects the chord
∴AC=BC=802=40cm
Now in ΔOCA, by Pythagoras theorem
AC2 + OC2 = OA2
=>402 + OC2 = 412
=>1600 + OC2 = 1681
=>OC2 = 81
=> OC2 = 81
=>OC = 81−−√
=>OC = 9cm
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