Question

qno.29 plzzzz its urgent

Answer 1

In ∆ AOB

tan theta = AB / OA
OA = r

hence ,
Tan theta = AB/r
r tan theta = AB ......(1)

∆ AOB

sec theta = OB / OA

OA=r

r sec theta = OB

r sec theta = OP + BP
r sec theta - OP = BP .....(2)


length of arc AP = theta /360 × 2πr
= theta πr/180 ......(3)


perimeter of shaded region=
AB + BP + AP
from eq. 1,2 and 3
= r tan theta + r sec theta - OP + theta × π r/180
= r( than theta + sec theta + theta π /180 - OP)
= r ( tan theta + sec theta + theta π /180 - 1 ). ••• ( OP = AO = r)

Answer 2

Here is the required answer