Question 1158208
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Your error in the first problem is that you must have taken the word "similar" in the statement of the problem in the vernacular sense.  You need to consider it in the plane geometry sense; that is, if two triangles are similar, then the measures of their corresponding angles are the same, and the measures of the corresponding sides are in proportion.  For your problem, which said the shorter side of the new, larger triangle is 2 cm longer than the short side of the original triangle, you have the following proportions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{13}{11}\ =\frac{b}{12}\ =\ \frac{c}{\sqrt{121+144}}]


where *[tex \Large b] and *[tex \Large c] are the measures of the long leg and hypotenuse of the new triangle respectively.


First *[tex \Large b],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{13}{11}\ =\ \frac{b}{12}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 11b\ =\ 132]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  b\ \approx\ 14.2]


Then you can either use Pythagoras, *[tex \Large c\ \approx\ \sqrt{169\ +\ 201.6}] or the proportion *[tex \Large \frac{13}{11}\ \approx \ \frac{c}{16.3}] to find the hypotenuse of the new triangular base.


I have no idea what you did wrong on the problem with the volume of a cone.  I tried making every common mistake and could not come up with your answer.  Looks like you just pulled a number out of your backside.


The volume of a cone is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{cone}\ =\ \frac{1}{3}\pi{r^2}h]


Since your diameter is 10 cm, your radius is 5 cm, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{cone}\ =\ \frac{1}{3}(\pi{25}\cdot{15})]


But since you want the volume of three candles, eliminate the factor of one-third:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{cone}\ =\ {3.14}\cdot{25}\cdot{15}]


You can do your own arithmetic.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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