NicoRobin
Answered

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Mack designed a water fountain with a square pyramid flowing into a cube. The edges of the bases of the pyramid and the cube have the same length and the heights of the pyramid and the cube are the same. Describe the relationship between the volume of the cube and the volume of the pyramid.

Please help me on this question. There are six questions in total and I did the rest (finding the volume/surface area) but this one I just don't understand. I'm willing to give up 60 of the 70 points I have for help (help that tells me the answer and explains it). Maybe even all 70. Please help if you understand. Thank you, minna.



Sagot :

tadvisohil886
volume of pyramid = 1/3 b2h
volume of  cube = b3

now , provided pyramid and cube have same edges and same heights.

thus, volume of pyramid = 1/3 b3
volume of cube = b3.

thus, ratio = 1/3b3/ b3 = 1/3.

Thus, volume of pyramid = 1/3volume of cube
kate200468
[tex]V_{cube}=a^3\ \ \ and\ \ \ V_{pyramid}= \frac{1}{3} \cdot a^2\cdot h=\frac{1}{3}a^3\\ \\ \frac{V_{pyramid}}{V_{cube}} = \frac{\frac{1}{3}a^3}{a^3}= \frac{1}{3}\\ \\ \\A_{cube}=6a^2\\ \\h^2=a^2+( \frac{1}{2} a)^2\ \ \ \Rightarrow\ \ \ h^2= \frac{5a^2}{4}\ \ \ \Rightarrow\ \ \ h= \frac{a \sqrt{5} }{2}\\\\A_{pyramid}=a^2+4\cdot \frac{1}{2}ah=a^2+2a\cdot \frac{a \sqrt{5} }{2} =a^2(1+ \sqrt{5}) \\ \\ \frac{A_{pyramid}}{A_{cube}}= \frac{a^2(1+ \sqrt{5}) }{6a^2} = \frac{1+ \sqrt{5}}{6}[/tex]
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