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mbma526

  • one year ago

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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  1. mbma526
    • one year ago
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    @Thats_Kyy

  2. Jacob902
    • one year ago
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    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?

  3. Jacob902
    • one year ago
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    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

  4. mbma526
    • one year ago
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    thank you so much @Jacob902

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