The volume of a right cone is one-third of the product of the area of the base B and the
Since the area of the base and the height is already given, substitute the values in to the equation and then simplify.
We may also find the exact value by substituting the given values into the equation below.
where V is the volume, r is the radius, and h is the height.
To obtain the exact value of the height, we use the Pythagorean theorem.
![\begin{gathered} a^2+b^2=c^2 \\ a^2=c^2-b^2 \\ a=\sqrt[]{c^2-b^2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%5E2%2Bb%5E2%3Dc%5E2%20%5C%5C%20a%5E2%3Dc%5E2-b%5E2%20%5C%5C%20a%3D%5Csqrt%5B%5D%7Bc%5E2-b%5E2%7D%20%5Cend%7Bgathered%7D)
where a and b are the legs of the formed right triangle while c is its hypothenuse.
From the figure, the slant height is the hypothenuse while the radius, 9 ft, serve as one of the legs. Thus, the other leg, which is the height of the cone can be solved as follows.
![\begin{gathered} a=\sqrt[]{19^2-9^2} \\ =\sqrt[]{361-81} \\ =\sqrt[]{280} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%3D%5Csqrt%5B%5D%7B19%5E2-9%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B361-81%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B280%7D%20%5Cend%7Bgathered%7D)
Therefore, the height of the cone is as stated above.
Substitute the obtained height and the radius into the second formula that was stated.
![\begin{gathered} V=\frac{\pi r^2h}{3} \\ V=\frac{\pi(9^2)(\sqrt[]{280})}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%3D%5Cfrac%7B%5Cpi%20r%5E2h%7D%7B3%7D%20%5C%5C%20V%3D%5Cfrac%7B%5Cpi%289%5E2%29%28%5Csqrt%5B%5D%7B280%7D%29%7D%7B3%7D%20%5Cend%7Bgathered%7D)
Simplify the right side of the equation. Evaluate the exponential expression.
![V=\frac{\pi(81)(\sqrt[]{280})}{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B%5Cpi%2881%29%28%5Csqrt%5B%5D%7B280%7D%29%7D%7B3%7D)
Multiply the numerator and then divide it by 3.
![\begin{gathered} V=\frac{\pi(81)(\sqrt[]{280})}{3} \\ \approx\frac{(3.1416)(81)(16.7332)}{3} \\ \approx\frac{4258.0907}{3} \\ \approx1419.36 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%3D%5Cfrac%7B%5Cpi%2881%29%28%5Csqrt%5B%5D%7B280%7D%29%7D%7B3%7D%20%5C%5C%20%5Capprox%5Cfrac%7B%283.1416%29%2881%29%2816.7332%29%7D%7B3%7D%20%5C%5C%20%5Capprox%5Cfrac%7B4258.0907%7D%7B3%7D%20%5C%5C%20%5Capprox1419.36%20%5Cend%7Bgathered%7D)
Note that the value that we obtain at first, 1415,603, is slightly different from the one that we obtained, which is 1419.36, since there are values that we already rounded off before substituting the values in the equation.
Therefore, to be more exact, it is best to indicate that the volume of the given cone is approximately 1419.36 ft³.
