Height of the cone

1) You will need the two formulas. What are they?

2) Is there actually a solution?

3) It is kind of an odd question, with mixed up units, so I have to wonder if you understand what it is asking.
 
Scdmathelp said:
Its asking for Height. Im not sure if the surface area is significant to it
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Yes .... you got it. ???

You'll need to use Pythagoras's theorem to calculate height from the given data. You don't need surface area.
 
You have a right triangle given to you! The height is unknown and the other two sides are known. What do you do next?
 
tkhunny said:
2) Is there actually a solution?
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As tkhunny suggests, there is no solution to this question.
The facts in the question, contradict the facts presented in the diagram. (There's no point in squaring anything)!

Of course, for any cone, you can choose a unit of measurement of length so that the surface area is numerically equal to the volume.
However then, [MATH]\hspace2ex l=\hspace1ex \mathrel{\raise{8pt}{r}} \hspace{-3pt} \left(\frac{r^2+9}{r^2-9}\right)[/MATH] [MATH](r>3)[/MATH] when measured in those units.
So, measured in those units, r must be greater than 3.
It also means if you want the sides of the triangle to be integer, there is only one solution, namely a 6, 8, 10 triangle [MATH](r, h, l)[/MATH].
 
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The problem is asking for the length of one leg of a right triangle where the other leg has length 5 cm and the hypotenuse has length 21 cm. Pythagorean triangle!
 
HallsofIvy said:
The problem is asking for the length of one leg of a right triangle where the other leg has length 5 cm and the hypotenuse has length 21 cm. Pythagorean triangle!
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No, there are two conditions in the question.
One is:
The value of the surface area (in square centimeters) of the cone is equal to the value of the volume (in centimeters) of the cone. (Find the height of the cone).
The second is:
The height is one side of a right-angled triangle, with hypotenuse 21 and other side 5.

If you use Pythagoras' Theorem you will find that your proposed solution is not a solution to the problem because it does not satisfy the first condition.
In fact there is no solution, as I pointed out in post #6.

It is like being asked to find [MATH]x[/MATH]:
[MATH]x^2>16\\ x^2-2x-3=0[/MATH]and answering, 'I know how to solve a quadratic equation'
Ans: [MATH]x=3[/MATH] or [MATH]x=-1[/MATH](In fact there is no solution to this problem either).
 
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