Finding the approximate value of the volume

[aburchett]

Find the approximate value of the volume of the right circular cone with a circular base shown below. Approximate your solution to the nearest hundredth.

[attachment=0:1lqbz2g6]10.png[/attachment:1lqbz2g6]

My possible choices are
320 m3
4021.24 m3
4557.40 m3
1005.31 m3
1139.35 m3

My work:
D=16 R=8

V=1/3 pi r^2 h
V=1/3 (3.14)(8)^2(17)
V=1/3(3.14)(64)(17)
V=1/3 (3416.32)
V= 1138.77 m^3

This is not any of the answers that I can choose from, can someone please tell me where I went wrong?

 

[soroban]

Hello, aburchett!

The "17" is not the height of the cone . . . it is the "slant height".

Find the approximate value of the volume of the right circular cone with a circular base shown below.
Approximate your solution to the nearest hundredth.

Possible choices:. . \(\displaystyle (a)\;320\;m^3 \qquad (b)\; 4021.24\;m^3 \qquad (c)\;4557.40\;m^3 \qquad (d)\;1005.31\;m^3 \qquad (e)\;1139.35\;m^3\)

            *
           /|\
          / | \
         /  |  \ 17
        /   |h  \
       /    |    \
      * - - + - - *
      :  8  :  8  :

\(\displaystyle \text{Pythagorus says: }\;h^2 +\,8^2 \:=\:17^2 \quad\Rightarrow\quad h^2 +\,64 \:=\:289 \quad\Rightarrow\quad h^2 \:=\:225 \quad\Rightarrow\quad h \:=\:15\)


\(\displaystyle \text{Then: }\:V \;=\;\frac{\pi}{3}r^2h \;=\;\frac{\pi}{3}(8^2)(15) \;=\;320\pi \;=\;1005.309649 \;\approx\;1005.31\:m^3\) .... edited

 

[lookagain]

\(\displaystyle . \ . \ .= 320\pi \;=\; >> 105.309649 <<\;\approx\;1005.31\:m^3\)

\(\displaystyle 320\pi \ = \ 1005.3096... \ \approx \ 1005.31 \ \longrightarrow\)


\(\displaystyle The \ approximate \ volume \ is \ 1005.31 m^3.\)