. An elastic belt is placed around the rim of a pulley
of radius 5 cm. From one point on the belt, elastic
belt is pulled directly away from the centre o of
the pulley until it is at P, 10 cm from the point 0.
Find the length of the belt that is still in contact
with the pulley. Also find the shaded area.
(Use π= 3.14 and √3 = 1.73)
Answers
Answer:
ANSWER
We know that, tangent of a circle is perpendicular to the radius of the circle through the point of contact.
⇒ ∠OAP=90
o
⇒ cosθ=
OP
OA
⇒ cosθ=
10
5
⇒ cosθ=
2
1
⇒ θ=cos
−1
(
2
1
)
∴ θ=60
o
⇒ ∠AOB=60
o
+60
o
=120
o
⇒ Reflex ∠AOB=360
o
−∠AOB
=∠360
o
−120
o
=240
o
∴ α=240
o
We know that,Length of belt in contact with pulley =ADB
⇒ Length of major arc =
360
o
α
×2π×r
=
360
o
240
o
×2×3.14×5
=
3
62.8
=20.93cm
Now, in right angled triangle OAP we have:
sin60
o
=
OP
AP
2
3
=
10
AP
⇒ AP=5
3
cm
⇒ Area of triangle =
2
1
×Base×Height
=
2
1
×AP×OA
=
2
1
×5
3
×5
=
2
25
5
cm
2
∴ Area of triangle OAP= Area of triangle OBP=
2
25
3
cm
2
⇒ Area of minor sector OACB=
360
o
θ
×πr
2
=
360
o
120
o
×3.14×5×5
=
3
78.5
=26.17cm
2
∴ Area of shaded region = Area of triangle OAP+ Area of triangle OBP− Area of minor sector OABC
=
2
25
3
+
2
25
3
−26.17
=25
3
−26.17
=25×1.73−26.17
=17.08cm
2
Hence, area of the shaded region will be 17.08cm
2
Answer:
We know that, tangent of a circle is perpendicular to the radius of the circle through the point of contact.
⇒ ∠OAP=90
o
⇒ cosθ=
OP
OA
⇒ cosθ=
10
5
⇒ cosθ=
2
1
⇒ θ=cos
−1
(
2
1
)
∴ θ=60
o
⇒ ∠AOB=60
o
+60
o
=120
o
⇒ Reflex ∠AOB=360
o
−∠AOB
=∠360
o
−120
o
=240
o
∴ α=240
o
We know that,Length of belt in contact with pulley =ADB
⇒ Length of major arc =
360
o
α
×2π×r
=
360
o
240
o
×2×3.14×5
=
3
62.8
=20.93cm
Now, in right angled triangle OAP we have:
sin60
o
=
OP
AP
2
3
=
10
AP
⇒ AP=5
3
cm
⇒ Area of triangle =
2
1
×Base×Height
=
2
1
×AP×OA
=
2
1
×5
3
×5
=
2
25
5
cm
2
∴ Area of triangle OAP= Area of triangle OBP=
2
25
3
cm
2
⇒ Area of minor sector OACB=
360
o
θ
×πr
2
=
360
o
120
o
×3.14×5×5
=
3
78.5
=26.17cm
2
∴ Area of shaded region = Area of triangle OAP+ Area of triangle OBP− Area of minor sector OABC
=
2
25
3
+
2
25
3
−26.17
=25
3
−26.17
=25×1.73−26.17
=17.08cm
2
Hence, area of the shaded region will be 17.08cm