hydraulic radius (Rh) of a trapezoidal section?

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saraxo

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I'm trying to calculate the hydraulic radius but I think the solution in my textbook may be wrong. But I wanted to see if anyone here might be able to point out my mistake, if any. 

A trapezoidal channel has a bottom width of 3 ft and side slopes of 2 ft vertical and 1 ft horizontal. Assume a depth of 2 ft. 

SInce the depth is 2 ft, the "base" of the triangles should be 1 ft as per the side slopes given. To find the length of the side, I did 1^2 + 2^2 = c^2.

c was calculated to be 2.236

This will give a wetted perimeter of 2.236(2) + 3 = 7.472

Area of triangle: 1/2 x 1 ft x 2ft x 2 triangles  = 2 ft

Area of rectangle: 3 ft x 2 ft = 6 ft

So the hydraulic area = 6 ft + 2 ft = 8

Hydraulic radius = A/P = 8/7.472 = 1.07 ft. 

My textbook says Rh should be 1.72 ft. Am I doing something wrong?

 
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