solve this question i will mark you brainlist
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Hi rohit,
Solution :
HINT : perimeter of shaded region = AB + PB + length of arc PA
in triangle OAB,
tanФ = AB/r
AB = rtanФ
arcPA = 2πr * (Ф/360)
arcPA = πrФ/180
PB = OB - OP
= OB - r
in triangle OAB,
CosФ = r/OB
OB = r/CosФ
OB = r SecФ
PB = OB - r
PB = rSecФ - r
PB = r(SecФ -1)
so shaded perimeter
= AB + PB + length of arc PA
= rtanФ + r(SecФ -1) + πrФ/180
= r ( tanФ + SecФ + πФ/180 -1)
Hence proved
Hope it helped.
Let me know if any doubts.
Cheers !!!
Solution :
HINT : perimeter of shaded region = AB + PB + length of arc PA
in triangle OAB,
tanФ = AB/r
AB = rtanФ
arcPA = 2πr * (Ф/360)
arcPA = πrФ/180
PB = OB - OP
= OB - r
in triangle OAB,
CosФ = r/OB
OB = r/CosФ
OB = r SecФ
PB = OB - r
PB = rSecФ - r
PB = r(SecФ -1)
so shaded perimeter
= AB + PB + length of arc PA
= rtanФ + r(SecФ -1) + πrФ/180
= r ( tanФ + SecФ + πФ/180 -1)
Hence proved
Hope it helped.
Let me know if any doubts.
Cheers !!!
jish4you:
perimeter* not areas
done...pls check the solution
thanks
thank you for asking this question...loved solving it
:) keep going :)
-_-thanks loved you
haha welcome....n plsss do mark it as brainliest if possible
if someone answer this question so it possible
ok u can mark it then ...but i had fun solving it ...thnx again :))
later*
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