The day before yesterday we discussed whether or not a certain rectangular prism with a piece missing would have the same surface area as one that is whole. We came to the conclusion that it would.
When looking at them in one view/one face at a time it will be the same.
The next shape we looked at was this:
We were asked if this figure was any different. Once we really thought about it, we found that it was. 2 small square units would be missed if we were to simply use the method for the last two figures.
With that in mind we easily solved the surface area for the figure:
The next figure we inspected looked like this:
Once again, we figured out that the surface area for this one would also have a different surface area in comparison to one that is whole.
SA= 2(Top/bottom)+2(side)+2(front/back- missing units)+(inside)
Afterwards we discussed finding the surface area of a cylinder.
We were asked to make the net of a cylinder.
The circumference of one of the circles will give you the width/length of the rectangle that makes up the cylinder.
You can find the surface area of a cylinder using these formulas
We then were given a cylinder to solve the surface area for. To make things different we were also asked to solve in pi. This means that we do not multiply anything by pi, but to simply leave it in to solve for later.
= (170)π u^2 (solve by pi)
= 534.07 u^2 (solve to two decimals)
- number 22, 23: textbook
- 1.3 in the homework book and extra practice sheets
That's right, Kim. I went there.