# Surface area of a triangular prism...

• Feb 3rd 2010, 07:36 PM
Sherpa
Surface area of a triangular prism...
Hello everyone. I'm hoping someone can give me a little bit of help. Here's the deal, I helped a friend with her homework - calculating the surface area of triangular prisms - a couple of nights ago. It looks like I've done something wrong, however, because when she returned home she hadn't gotten a single correct answer. Anyway, here's what I've got, maybe someone can tell me if my answers are correct. This is the formula that I used: ab+ (s1+s1+s3)h

1: 144 square inches
6x8 + (10 + 6 + 8)4 = 144

2: 216 square feet

3: 48 square km

4: 60 square cm

5: 212 square yards

6: 84 square inches

http://img46.imageshack.us/img46/8247/prisms.jpg
• Feb 3rd 2010, 08:03 PM
Soroban
Hello, Sherpa!

Code:

        *       /    *       *        *       |  *        *       |      *        *     a |        c *        *       |              *        *       |                  *    / h       * - - - - - - - - - - - *                   b

The triangular faces are all right triangles.

. . The area of a triangle is: .$\displaystyle \tfrac{1}{2}ab\text{ units}^2$

. . Hence, the area of two triangles is: .$\displaystyle ab\text{ units}^2$

There are three rectangluar panels.
. . Their areas are: .$\displaystyle ah,\;bh,\;ch\text{ units}^2$

Therefore, the total surace area is: .$\displaystyle S \;=\; ab + (a+b+c)h$

• Feb 3rd 2010, 08:19 PM
Sherpa
It looks like the teacher has his formulas wrong! I guess I should explain that to him.