Question Number 103318 by I want to learn more last updated on 14/Jul/20

Commented bysom(math1967) last updated on 14/Jul/20

((AT)/(AB))=cos∠BAT=cos30° AT=15×((√3)/2)cm ∴BP=QD=(15−((15(√3))/2))cm perimeter of shaded part ={(1/6)×2π×((15(√3))/2)+2(15−((15(√3))/2))+2×15}cm

Commented byI want to learn more last updated on 14/Jul/20

Thanks sir, i appreciate.

Commented byTawa11 last updated on 15/Sep/21

nice

Answered by 1549442205 last updated on 14/Jul/20

The radius of the circle equal to the altitude of equilateral ABD ,so R=((15(√3))/2).The length of the arc PQ is l=((Rπ)/3)=((15π(√3))/6)=((5π(√3))/2).QD=PB=15−R=15−((15(√3))/2) =((15(2−(√3))/2)⇒PB+QD=15(2−(√3)) The perimeter of shaded part is BC+CD+QD+arc(PQ)+PB= 15×2+15(2−(√3))+((5π(√3))/2) =((5(24−6(√3)+𝛑(√3)))/2)

Commented byI want to learn more last updated on 14/Jul/20

Thanks sir, i appreciate.

Commented byI want to learn more last updated on 14/Jul/20

Sirs, what of the Area of the shaded. Am i going to say: Area of shaded = l^2 sin(θ) − Area of sector?

Commented bysom(math1967) last updated on 14/Jul/20

Area of rhombus−Area of sector

Commented byI want to learn more last updated on 14/Jul/20

Thanks sir

Commented by1549442205 last updated on 14/Jul/20

The area of rhombs is S=((AC×BD)/2)=((15(√3)×15(√3))/4)=((675)/4) The area of the secror is S_0 =((πR^2 )/6)=(π/6)×(((15(√3))/2))^2 =((675π)/(24)) The area of shaded part is: S_(sh) =S−S_0 =((675)/4)−((675𝛑)/(24))=((675(6−𝛑))/(24))

Commented byI want to learn more last updated on 14/Jul/20

Thanks sir. I apreciate.