Two triangular prisms, each 41 in.41 in. long, have bases that are the same size isosceles right triangles. The sides of each triangle are 19 in.19 in., 19 in.19 in., and 26.9 in.26.9 in. What is the surface area of each triangular prism? Imagine you glue the lateral faces with the 26.926.9-in. sides together. What is the surface area of the resulting prism?
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Using Heron's formula to find the area of one of the triangle bases ....
The semi-perimeter, s, = [ 19 + 19 + 26.9] / 2 = 32.45
And the area of one base = sqrt [ ( 32.45) (32.45 - 19)^2 ( 32.45 - 26.9) ] ≈ 180.5 in^2
So.....the surface area of one prism =
Area of the bases + Area of the sides =
2 (180.5) + 41 [ 19 + 19 + 26.9] = 3021.9 in ^2
And doubling this gives the total surface area of both prisms combined ≈ 6043.8 in^2
If we glued the 26.9 in sides together we would have a rectangular prism with square bases of 19 inches on each side
So.......the surface area of such a prism =
Area of both bases + Area of the sides
2(19)^2 + 41 ( 19 * 4 ) = 3838 in^2
