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- mbma526

will fan and medal
A triangular prism has a surface area of 475 square feet, a length of 15 feet, a height of 10 feet, and sides of 5 feet. Find the width of the base of the triangular prism.

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- Jacob902

example A triangular prism has a surface area of 350 square feet, a length of 15 feet, a height of 10 feet, and a side?

- Jacob902

One vertical face is 150 ft^2.
Another vertical faces is 50 ft^2.
That leaves 10W for a third vertical face,
in addition to which, one has the two triangular bases.
If the bases were right triangles,
their width would be sqrt(15^2-10^2) = sqrt(125)
= 5 sqrt(5) = about 11.7 ft,
but this doesn't "work" because the area of the two bases
would be 58.5 ft^2 and 117 ft^2 for the 3rd vertical face,
for a total surface area of about 375 ft^2.
Hence, the angle between the "side" and the "width" is larger than 90 degrees.
However, I'm assigning the name "theta" to the angle between "side" and "length".
W^2 = S^2 + L^2 - 2 LS cos(theta)
LS sin(theta) + WH = 150 ft^2
The only two unknowns in these two equations are theta and W,
so the number of solutions should be 0,1, or 2.
W^2 = 250 - 150 cos(theta)
75 sin(theta) + WH = 150
W = 15 - 7.5 sin(theta)
I feed to "fooplot.com" the two equations
y = sqrt(250 - 150 cos(x)) and
y = 15 - 7.5 sin(x)
and find an intersection point at theta = 0.5529 radians,
W = 11.06 ft

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- mbma526

thank you so much @Jacob902

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