Question

Find the perimeter of the shaded region in the given figure

Answer 1

Answer

$\sf r(tan\theta+\frac{1- cos\theta}{cos\theta}+\frac{\theta \pi}{180^{\circ}})$

Explanation

Given, radius of the circle = r

Angle formed by sector = ∅

To find, the perimeter of the shaded region

Clearly, the perimeter of the figure is AB + BC + arc AC

We will find the algebraic values of them and add them to find the perimeter of the shaded region is

We know that OA is the radius and AB the tangent, hence

∠OAB = 90°

Hence, we can use trigonometry in the triangle

So, tan∅ = AB/OA

⇒ tan∅ = AB/r

⇒ AB = r tan∅.................(1)

Again,

cos∅ = OA/OB

⇒ cos∅ = r/(OC + BC)

⇒ cos∅ = r/(r + BC)

⇒ (r + BC)cos∅ = r

⇒ BC cos∅ = r - r cos∅

⇒ BC = $\sf \frac{r(1 - cos\theta)}{cos\theta}$..................(2)

And, length of arc AC = ∅/360° × 2πr

= ∅πr/180°....................(3)

So, perimeter of shaded region = AB + BC + length of arc AC

From (1), (2) and (3)

Perimeter = r tan∅ + $\sf \frac{r(1 - cos\theta)}{cos\theta}$ + ∅πr/180°

Taking r common, we get perimeter

$\sf r(tan\theta+\frac{1- cos\theta}{cos\theta}+\frac{\theta \pi}{180^{\circ}})$

Answer 2

Answer:

Perimeter of shaded region = BC + AB + arc (AC)

Finding the perimeter of triangle:

Perimeter = AB + BO + OA

=> AB + BO + r

Since AB is tangent, /_BAO = 90°

=> tan Ø = BA/OA

=> sec Ø = BO/OA

If we add tan Ø, sec Ø and 1:

=> BA/OA + BO/OA + 1 = (BA + BO + OA)/OA

=> Perimeter/r

If we need only perimeter in terms of trigonometric ratios:

Perimeter/r × r = Perimeter

Perimeter of triangle = r(tan Ø + sec Ø + 1)

Length of arc = Ø × $\sf{\frac{2 \pi r}{360}}$

Arc = Ø$\sf{\frac{\pi r}{180}}$

Shaded region = Perimeter of triangle - 2r + Arc

=> r(tan Ø + sec Ø + 1) - 2r + Ø$\sf{\frac{\pi r}{180}}$

=> tan Ø r + sec Ø r + r - 2r + Ø$\sf{\frac{\pi r}{180}}$

=> tan Ø r + sec Ø r - r + Ø$\sf{\frac{\pi r}{180}}$

=> r(tan Ø + sec Ø - 1 + Ø$\sf{\frac{\pi}{180}}$)

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