This question requires knowledge of multiple math genres.
What is the surface area of a triangular prism with a base of the dimensions 18, 24, and 36and a height of 2. There is enough information giving.

Sorry, 36and should be 30 and

Have you tried using the sketch, and 'open' the structure of the prism?

1. The area of the base: You need to find the angle, using cosine rule:

a^2 = b^2+c^2 - 2bc cosA

A = cos^{-1}(\frac{a^2-b^2-c^2}{-2bc})

Let a = 30, b = 18, c = 24;

A = cos^{-1}(\frac{(30)^2-(18)^2-(24)^2}{-2(18)(24)})

A = cos^{-1}(0) = 90

Ok, so you have a right angled triangular base. Since A = 90 and a = 30, that makes 'a' a hypotenuse.

Area = \frac12 (18)(24) = 216

Now, you have two cross sectional areas, that make a total of (216*2) = 432 sq units.

Then, you have three rectangles at the sides, with dimensions 2*18, 2*30 and 2*24. Find the area of these rectangles, and add them up with the area of the two triangular cross sections.

I hope it helped! :)

Wow, I figure it out by finding that the base is a right triangle, so if a circle is drawn connecting all points, so 30 is the hypotonuse and diameter so the height is the midpoint of 30 (midpoint of circle), to the opposite point which is 30/2=15, and 30*15 divided by two times two plus the dimensions' sum times 2.

I didn't understand your last post as from "so the height is the midpoint..."

Isn't the height given as 2?

Total Surface Area = area of two triangles + area of 3 rectangles.

TSA = 432 + (2*18 + 2*30 + 2*24) = 576

Sorry, the height of the base (triangle)