Question Number 209847 by SonGoku last updated on 23/Jul/24
Answered by mr W last updated on 23/Jul/24
$${depth}\:={h} \\ $$$${h}=\sqrt{\mathrm{23}^{\mathrm{2}} −\left(\frac{\mathrm{53}−\mathrm{41}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{493}}=\mathrm{22}.\mathrm{20}{m} \\ $$$${cross}\:{section}\:{area}\:={A} \\ $$$${A}=\frac{\mathrm{53}+\mathrm{41}}{\mathrm{2}}×{h}=\mathrm{1043}.\mathrm{57}\:{m}^{\mathrm{2}} \\ $$$${wetted}\:{perimeter}\:={P} \\ $$$${P}=\mathrm{23}+\mathrm{41}+\mathrm{23}=\mathrm{87}\:{m} \\ $$$${hydraulic}\:{radius}\:={R} \\ $$$${R}=\frac{{A}}{{P}}=\frac{\mathrm{1043}.\mathrm{57}}{\mathrm{87}}=\mathrm{12}\:{m} \\ $$
Commented by SonGoku last updated on 23/Jul/24
$$\mathrm{Thank}\:\mathrm{you}! \\ $$