Question 163654
You have the formula for the surface area of the triangular prism:
{{{S = Ph+2B}}} where: P = perimeter of the base, h = the height of the prism, and B = the area of the base.
You are given:
{{{h = 12}}}m.
You can find P from knowing the lengths of the sides of the right triangular base in which the two legs are 3m and 4m, so the hypotenuse must be 5m which you can show by using the Pythagorean theorem:
{{{c^2 = a^2+b^2}}} c is the hypotenuse, a and b are the two legs (3m and 4m), so...
{{{c^2 = 3^2+4^2}}}
{{{c^2 = 9+16}}}
{{{c^2 = 25}}} Take the square root of both sides.
{{{c = 5}}} Ignoring the negative solution.
The perimeter is:
{{{P = 3+4+5}}}
{{{P = 12}}}m
The area of the triangular base (B) is:
{{{B = (1/2)bh}}}
{{{B = (1/2)(3*4))}}}
{{{B = 6}}}sq.m
Now we have all we need to calculate the surface area (S) of the right-triangular prism.
{{{S = Ph+2B}}} Making the appropriate substitutions, we get:
{{{S = 12*12+2*6}}}
{{{S = 144+12}}}
{{{P = 156}}}sq.m