Circular pool

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

Correct answer:

x =  6102 hl

Step-by-step explanation:

$$\begin{gathered} r=10\text{m} \\ t=10\text{m} \\ h=2\text{m} \\ a=t/2=10/2=5\text{m} \\ \alpha =2\cdot \arcsin (\frac{a}{r})=2\cdot \arcsin (\frac{5}{10})\doteq 1.0472\text{rad} \\ {\alpha }_2=\alpha \to \text{°}=\alpha \cdot \frac{180}{\pi }\text{°}=1.0472\cdot \frac{180}{\pi }\text{°}=60\text{°} \\ {S}_1=\pi \cdot r^2\cdot \frac{2\pi -\alpha }{2\pi }=3.1416\cdot 10^2\cdot \frac{2\cdot 3.1416-1.0472}{2\cdot 3.1416}\doteq 261.7994{\text{m}}^2 \\ b=\sqrt{r^2-a^2}=\sqrt{10^2-5^2}=5\sqrt{3}\text{m}\doteq 8.6603\text{m} \\ {S}_2=\frac{a\cdot b}{2}=\frac{5\cdot 8.6603}{2}\doteq 21.6506{\text{m}}^2 \\ S=S_1+2\cdot S_2=261.7994+2\cdot 21.6506\doteq 305.1007{\text{m}}^2 \\ V=h\cdot S=2\cdot 305.1007\doteq 610.2013{\text{m}}^3 \\ x=V\to \text{hl}=V\cdot 10\text{hl}=610.2013\cdot 10\text{hl}=6102\text{hl}=6102\text{hl} \end{gathered}$$

Try another example