An elastic belt is placed round the rim of a pulley of radius 5 cm on point on the belt is directly pulled away from the centre O of the pulley until it is at point P 10 cm away from the centre O . Find the length of the belt that is in contact with the rim of the pulley . Find the area of the shaded part also
Answers
Answer:
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
Given :
∠OAP=90
cosθ=OA/Op
=5/10
=1/2
=60 °
Therefore θ=60
∠ AOB=60+60
=120
Reflex∠ AOB=360-∠AOB=360-120
α=240°
Length of belt in contact with pulley =Length of major arc=(α /360) 2π r
=(240/360 ) 2x 3.14 x 5
=62.8/3
=20.9333cm
In right angled Δ OAP,sin60=AP/OP√3/2=AP/102AP=10√ 3AP=5√3cm
Area of traingleOAP=1/2 basex height
=1/2 x Ap X OA
=1/2 5√3 x5
=25√3/2 cm²
Ar(Δ OAP)=ar(ΔOBP)=25√3/2 cm2
Area of minor sector= OACB=(θ/360 )π r²
=(120/360) x3.14 x 5x5
=78.5/3=26.17cm2
Shaded area=arc(Δ OAP)+ ar(ΔOBP)- area of minor sector OACB
=25√3/2+25√3/2- 26.17
=25√3-26.17
=25x1.73 -26.17
=43.25-26.17
=17.08 cm²
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