Question

In Fig. is 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 shaded region is :
$r(tan\theta+sec\theta+\dfrac{\pi\theta}{180}-1)$

Answer 1

Answer:

QED

Step-by-step explanation:

using x instead of theta

tan x = AB/r

AB = r tanx

PB = OB - OP

PB = rsecx - r    ( as cosx = r/OB so OB = r/cosx = rsecx)

PB = r (secx -1)

Arc PA = (x/360)(2*pie*r)

Arc PA = (pie x / 180)r

Perimeter of shaded region

= AB + PB + ArcPA

= rtanx + r(secx -1) + (pie x / 180)r

= r ( tanx + secx -1 + (pie x / 180))

= r ( tanx + secx  + (pie x / 180)  - 1 )

QED

Answer 2

Step-by-step explanation:

From figure:

In ΔAOB,

(i)

Cos θ = OA/OB

⇒ OB cos θ = r

⇒ OB = r/cosθ

⇒ OB = r sec θ

∴ PB = OB - r

        = r secθ - r        

(ii)

tanθ = AB/OA

⇒ AB = OA tanθ

⇒ AB = r tanθ

∴ Length of the arc = (θ/360°) * 2πr

                                = πrθ/180

∴ Perimeter of the shaded region = AB + PB + Length of arc AP

= r tanθ + r secθ - r + πrθ/180

= r tanθ + r secθ + πrθ/180 - r

= r(tanθ + secθ + πθ/180 - 1)

Hope it helps!