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 (use *π* = 3.14 and `sqrt3=1.73)`

#### Solution

Given:

OA = 5 cm

OP = 10 cm

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, ∆OAP is a right-angled triangle.

⇒∠OAP=90°

Now,

OP^{2 }= OA^{2} + AP^{2 }

⇒ 10^{2} = 5^{2} + AP^{2}

⇒ AP^{2} = 75

⇒ AP = 53`sqrt3 cm`

Also,

`costheta=(OA)/(OP) = 5/10`

`=>costheta=1/2`

⇒θ=60°

Now,

∠AOP=∠BOP=60° (∵∆OAP≅∆OBP)

⇒∠AOB=120°

Length of the belt still in contact with the pulley = Circumference of the circle − Length of the arc ACB

`=2xx3.14xx5-120^@/360^@xx2xx3.14xx5`

`=2xx3.14xx5xx(1-1/3)`

`=2xx3.14xx5xx2/3`

=20.93 cm (Approx.)

Now

Area of ∆OAP =

`1/2xxAPxxOA=1/2xx5sqrt3xx5=(25sqrt3)/2 cm^2`

Similarly,

Area of ∆OBP = `(25sqrt3)/2cm^2`

∴ Area of ∆OAP + Area of ∆OBP = 25`sqrt3`cm^{2}=25×1.73=43.25 cm^{2}

Area of sector OACB = `120^@/360^@xx3.14xx(5^2)=1/3xx3.14xx25=26.17cm^2`

∴ Area of the shaded region = (Area of ∆OAP + Area of ∆OBP) − Area of the sector OACB

= 43.25 cm^{2 }− 26.17 cm^{2}

= 17.08 cm^{2} (Approx.)