An elastic belt is placed around the rim of a pulley of radius 5 cm. From one point Con
the belt, the elastic belt is pulled directly away from the centre o of the pulley until it is at P, 10
cm from the point O. Find the length of the belt that is still in contact with the pulley. Also find
the shaded area, (user = 3.14 and V3 = 1.73)
5 cm
B
Answers
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:
Wilcox doc fix Eric Eric Eric Eric Don rob