Question

An elastic belt is placed around the rim of a pulley of radius 5 cm. (fig. 10) from one point c on 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.

Answer 1

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

Answer 2

please refer all the photos
:-)