#1)
Using the similarity statement, we can find \(YR\) by generating a proportion.
\(\frac{DZ}{LZ}=\frac{MR}{YR}\) | There are a few options here. This proportion is the one I chose to set up for this particular problem. Substitute in the known side lengths and solve for the missing one. |
\(\frac{18}{21}=\frac{28.8}{YR}\) | It is generally wise to simplify any fractions completely before one cross multiplies. Taking this precaution beforehand can ensure that the computation does not get out of hand. |
\(\frac{6}{7}=\frac{28.8}{YR}\) | Now we can cross multiply. |
\(6YR=201.6\) | |
\(YR=33.6\text{in}\) | Do not forget the units! |
\(YR=34\text{in}\) | The question asks that the answer is rounded to the nearest whole inch, so I complied. |
#2)
This question becomes simple once you know the formula for the volume of a pyramid: \(V_{\text{pyramid}}=\frac{1}{3} lwh\)
\(V_{\text{pyramid}}=\frac{1}{3}lw h\) | Of course, we must look at the given information; the figure is a square pyramid, so the side length of the bases is equivalent. |
\(V_{\text{pyramid}}=\frac{1}{3}*183*183*110\) | 183 is divisible by 3, so we can reduce that portion now. |
\(V_{\text{pyramid}}=61*183*110\) | |
\(V_{\text{pyramid}}=1227930\text{m}^3\) | |
#3)
The fill-in-the-blank questions are really just testing one's knowledge of the individual formulas.
\(V_{\text{cylinder}}=\hspace{3mm}\pi r^2 h\\ V_{\text{cone}}\hspace{5mm}=\frac{1}{3} \pi r^2 h\)
When you place the formulas side by side, basic observation shows that a cone's formula is 1/3 of the volume of a cylinder with the same base and height. Of course, I already revealed what the formula is, so the volume of a cylinder is \(\pi r^2 h\) , and the formula for a cone is \(\frac{1}{3} \pi r^2 h\)
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