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 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?

Guest May 17, 2017
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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 

 

 

cool cool cool

CPhill  May 17, 2017

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