Question

Solve the ques....with full explanation...ques.is here

Answer 1

perimeter= AB+BP+ARC AP
Put BP=BO-PO
then take r as common
then put name of trigonometric ratio
you will find the answer
try it
agar phir bhi na hoa to btana I will send the solution

Answer 2

In triangle AOB ,

tan¥ = AB / OA

=> AB = tan ¥× OA
=> AB = r × tan¥ -----(i) ( bcz OA is radius )

& .... Sec¥ = OB / OA
=> OB = Sec¥ × OA
=> OB = r × Sec¥ ------(ii)

Now perimeter of AOB ,

AO + AB +OB

rSec¥ + r + rtan ¥ = r ( sec¥+tan¥+1)

perimeter of shaded region
= perimeter of triangle ABC - perimeter of sector AOC

= r ( sec¥ + tan¥ + 1) - ( OA+ OC + AC )

= r ( sec¥ + tan¥ + 1) - (r + r + ¥/180×πr)

= r ( sec¥ + tan¥ + 1) - ( 2r + ¥/180×πr)

= r ( sec¥ + tan¥ + 1) - 2r - π¥r/180

= rsec¥ + rtan¥ + r - 2r - π¥r/180

= rsec¥ + rtan¥ - r - π¥r/180

= r ( sec¥ + tan¥ - π¥/180 -1 )

hope it helped ...!!