Answer:
[tex]\%A_F=77.335\%[/tex]
Explanation:
Given:
The triangle ORQ is symmetric about OS as it comes from center O on the cord QR at S.
[tex]RO=QO=7\ ft[/tex] (∵ radius of the cylinder)
[tex]RS=\sqrt{RO^2-OS^2}[/tex]
[tex]RS=3.6055\ ft[/tex]
Now the area of triangle ORQ:
[tex]A_t=\frac{1}{2}\times QR\times OS[/tex]
[tex]A_t=RS\times OS[/tex]
[tex]A_t=3.6055\times 6[/tex]
[tex]A_t=21.6333\ ft^2[/tex]
Now the angle ROS:
[tex]\cos\theta=\frac{OS}{OR}[/tex]
[tex]\theta=\cos^{-1}(\frac{6}{7} )[/tex]
[tex]\theta=31.0027^{\circ}[/tex]
Therefore the reflex angle ROQ:
[tex]\rangle ROQ=360^{\circ}-\theta^{\circ}[/tex]
[tex]\rangle ROQ=360^{\circ}-31.0027^{\circ}[/tex]
[tex]\rangle ROQ=328.9973^{\circ}[/tex]
Now the area of sector ROQPR:
We have the area of full circle, [tex]A=\pi.r^2[/tex]
where:
r = radius of the circle
hence for sector:
[tex]A_S=\pi\times 7^2\times \frac{328.9973}{360}[/tex]
[tex]A_S=140.6811\ ft^2[/tex]
Now the cross sectional area filled with water:
[tex]A_F=A_S+A_t[/tex]
[tex]A_F=140.6811-21.6333[/tex]
[tex]A_F=119.0478\ ft^2[/tex]
Total cross sectional area of tank:
[tex]A=\pi.r^2[/tex]
[tex]A=\pi\times 7^2[/tex]
[tex]A=153.9380\ ft^2[/tex]
Now the percentage of total capacity used:
[tex]\%A_F=\frac{A_F}{A}\times 100\%[/tex]
[tex]\%A_F=\frac{119.0478}{153.9380} \times 100[/tex]
[tex]\%A_F=77.335\%[/tex]
