Question

an elastic belt is placed round tha rim of a pulley of radius 5 cm .one point on tha belt is pulled directly away from tha center o of tha pulley untill it is at p 10 cm from o . Find tha arm of tha pulley .Also find tha shaded region
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Answer 1

Step-by-step explanation:

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

Hope it helps! Please mark the brainliest!!