Question

In the given figure, shown a sector OAP of a circle with centre O, containing angle theta. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of the shaded region is :
$r( \tan(theta) + \sec(theta) + \frac{\pi \: theta}{180} - 1)$
​

Answer 1

Given

The radius of circle = r

Angle of sector OAP = ∅

To prove

Perimeter of shaded region = r(tan∅ + sec∅ + $\sf\frac{\pi\theta}{180}$ - 1)

Proof

Perimeter of the shaded region = AB + PB + arc PA

We know that radius is perpendicular to tangent

→ angle OAB = 90°

So, we can apply trigonometry in the triangle OAB

→ tan∅ = AB/OA

→ tan∅ = AB/r (since OA is the radius)

→ AB = r tan∅

Again,

sec∅ = OB/OA

→ sec∅ = (OP + PB)/OA

→ sec∅ = (r + PB)/r

→ r + PB = r sec∅

→ PB = r sec∅ - r

→ PB = r(sec∅ - 1)

Now, arc length of PA would be

∅/360° × 2πr

→ PA = ∅πr/180

So, perimeter of shaded region = AB + PB + arc PA

→ r tan∅ + r(sec∅ - 1) + ∅πr/180

Taking r common,

r(tan∅ + sec∅ - 1 + $\sf\frac{\pi\theta}{180})$

Hence Proved.