Question

An elastic belt is placed around the rim of a pulley of radius 5 cm. From one point on the belt 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 = 3.14 and 3 = 173) ​

Answer 1

Given

Radius (R) = 5cm

length of OP = 10cm

Now, In ∆AOP and ∆BOP

AP = BP {The length of the tangents drawn from an external point to a circle are equal}

OP = OP {common}

∠OAP = ∠OBP {Each 90°, tangents are ⏊ to radius through the point of contact}

∴ ΔAOP ≡ ΔBOP {by SAS congruency rule}

So, Area of (∆AOP) = Area of (∆BOP) ---> (1)

AOP is a triangle, right angled at A

By applying applications of trigonometry

$\mathsf{\implies\: Cos\theta = \frac{OA}{OP}}$

$\mathsf{\implies\:Cos\theta = \frac{5}{10}}$

$\mathsf{\implies\:Cos\theta = \frac{1}{2}}$

$\mathsf{\implies\:\cancel{Cos}\theta = \cancel{Cos} 60°}$

$\mathsf{\implies\:\theta = 60°}$

∴ ∠POB = 60°

∠AOB = θ + ∠POB

∠AOB = 60° + 60°

∠AOB = 120°

Reflex ∠AOB = 360° - 120°

Reflex ∠AOB = 240°

Length of the belt that is still in contact with the pulley = $\mathsf{\frac{∠AOB}{360} × 2 \pi R }$

Taking π = 3.14

= $\mathsf{\frac{240}{360} × 2 × 3.14 × 5 }$

= $\mathsf{\frac{2}{3} × 2 × 3.14 × 5 }$

= $\mathsf{\frac{62.8}{3} }$

= $\fbox{\mathsf{20.93 cm}}$

Now, finding the length of AP

OP² = OA² + AP²

AP = $\mathsf{\sqrt{OP^2 - OA^2}}$

AP = $\mathsf{\sqrt{100 - 25}}$

AP = $\mathsf{\sqrt{75}}$

AP = 5√3 cm

Area of the shaded part = Area of (∆AOP) + Area of (∆BOP) - Area of the sector (AOCB)

= Area of (∆AOP) + Area of (∆AOP) - Area of the sector (AOCB) {from (1)}

= 2 × Area of (∆AOP) - Area of the sector (AOCB)

= $\mathsf{2 × \frac{1}{2} × 5 × 5\sqrt{3} - \frac{∠AOB}{360} × \pi × R^2}$

Taking √3 = 1.73

= $\mathsf{2 × \frac{1}{2} × 25 × 1.73- \frac{120}{360} × 3.14× 25}$

= $\mathsf{ \frac{86.5}{2} - \frac{78.5}{3}}$

= $\mathsf{ \frac{259.5 - 157}{6}}$

= $\mathsf{ \frac{102.5}{6}}$

= $\fbox{\mathsf{17.083\: cm^2}}$

Answer 2

Answer:

refer the image