Question

Try to solve this frnds.. ​

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 x 5
=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
=25 x 1.73 -26.17
=43.25-26.17
=17.08 cm²

Hope this helps