Thread: Triangular prism

I believe or hope that i have solved the two cones with formula using
S.A =PIr(s+r)
=(3.14)(3)(6+3)
=3.14(3)(9)
=84.8yards

I can't seem to be able to solve for the rectangular portion of the problem.

Any help is greatly appreciated

To what question are you referring?

My psychic powers are a little off today, would you please post a picture or description of the shape you are trying to evaluate the surface area of?

Its not allowing me to transmit my attachment. I have no problems with it the other day

MG_0003_math prism.jpg - Upload of file failed.

i keep getting this message

The file was to large, so i needed to compress it.

I only need to answer this question then i have completed lessons 1-10. In wishing to complete this course to return to college. I'm not asking for someone to do my work. I'm really asking is just for some direction, some ideas. Below is what i believe i have been able to solve so far. I'm not even sure if I'm using the right formula.

S.A =PIr(s+r)
=(3.14)(3)(6+3)
=3.14(3)(9)
=84.8yards

I'm slightly confused: how did you get that value for s? The way I see it, the radius of the half-cone is , but the height is , making by Pythagoras. Once you know this, you can calculate the surface area of the three rectangles and the surface area of the cone and put them together.

The sides was just an estimated guess of 6. I don't know how to figure the value for the sides. So, the sides would actually be 5.8 ?

Originally Posted by hotboxdriver

The sides was just an estimated guess of 6. I don't know how to figure the value for the sides. So, the sides would actually be 5.8 ?

No, they would be . Try to keep things in an exact form until the final stages of your working. Do you understand how I got to that conclusion?

yes,
solving for the sides using Hypotenuse rule X2=5^2+3^2=25+9
=√34