Circular pool

The base of the pool is a circle with a radius r = 10 m, excluding a circular segment that determines the chord length 10 meters. The pool depth is h = 2m. How many hectoliters of water can fit into the pool?

Result

x =  6102 hl

Solution:

r=10 m t=10 m h=2 m a=t/2=10/2=5 m  α=2 arcsin(ar)=2 arcsin(510)1.0472 rad α2=180πα=rad21.0472=60   S1=π r2 2πα2π=3.1416 102 2 3.14161.04722 3.1416261.7994 m2  b=r2a2=102525 3 m8.6603 m  S2=a b2=5 8.6603221.6506 m2  S=S1+2 S2=261.7994+2 21.6506305.1007 m2  V=h S=2 305.1007610.2013 m3  x=Vhl=V 10 hl=6102.01316 hl=6102 hlr=10 \ \text{m} \ \\ t=10 \ \text{m} \ \\ h=2 \ \text{m} \ \\ a=t/2=10/2=5 \ \text{m} \ \\ \ \\ α=2 \cdot \ \arcsin ( \dfrac{ a }{ r } )=2 \cdot \ \arcsin ( \dfrac{ 5 }{ 10 } ) \doteq 1.0472 \ \text{rad} \ \\ α_{2}=\dfrac{ 180^\circ }{ \pi } \cdot α=rad_{21}.0472^\circ =60 \ ^\circ \ \\ \ \\ S_{1}=\pi \cdot \ r^2 \cdot \ \dfrac{ 2 \pi - α }{ 2 \pi }=3.1416 \cdot \ 10^2 \cdot \ \dfrac{ 2 \cdot \ 3.1416 - 1.0472 }{ 2 \cdot \ 3.1416 } \doteq 261.7994 \ \text{m}^2 \ \\ \ \\ b=\sqrt{ r^2-a^2 }=\sqrt{ 10^2-5^2 } \doteq 5 \ \sqrt{ 3 } \ \text{m} \doteq 8.6603 \ \text{m} \ \\ \ \\ S_{2}=\dfrac{ a \cdot \ b }{ 2 }=\dfrac{ 5 \cdot \ 8.6603 }{ 2 } \doteq 21.6506 \ \text{m}^2 \ \\ \ \\ S=S_{1} + 2 \cdot \ S_{2}=261.7994 + 2 \cdot \ 21.6506 \doteq 305.1007 \ \text{m}^2 \ \\ \ \\ V=h \cdot \ S=2 \cdot \ 305.1007 \doteq 610.2013 \ \text{m}^3 \ \\ \ \\ x=V \rightarrow hl=V \cdot \ 10 \ hl=6102.01316 \ hl=6102 \ \text{hl}



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