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.