FWT wrote:find the volume of a bowl with the measurements of 16 ft diameter at the bottom, 20 ft diameter at the top and the bowl is 2 ft high. the slope of the sides of the bowl (their straight, not curved) if you were looking at a cross section of it is 45 degrees.
Draw the cross-sectional view, but upside-down. Continue the sides (as dashed lines, if you like) until they meet at the top.
You should have a triangle, being the side view of a cone. The whole cone is the actual bowl part, plus the dashed-in extra part on top. The top, dashed-in bit is extra.
The bottom part is a "truncated" cone (because it's had its top chopped off), and represents the bowl. This part has a height of h = 2 and a base radius of r = 10.
Draw the height line from the top of the cone down to the base. Label the portion inside the bowl portion as "2".
This height line splits the isosceles triangle of your side-view picture into two halves. Each of these halves contains nested similar right triangles. Label the height of the triangle for the top part as "H". The base is clearly "8". You already know that the base of the larger triangle is 10.
You have similar triangles with heights H and H + 2, and bases 8 and 10, respectively. You can either use the similarity of the triangles to solve for H, or else use what you know about forty-five-degree right triangles to find the two heights. (Hint: What is always
true about the two legs of a 45-45-90 triangle?)
Once you have this information, you can find the volume of the whole cone and the smaller (top) cone. Subtract the volume of the top cone from the volume of the whole cone to find the volume of the truncated cone, and thus the volume of the bowl.
If you get stuck, please reply showing how far you have gotten. Thank you!