Hello!
The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³
A 12 cm cone with a dome on top of it that has an 8 cm diameter
Data: (Cone)
h (height) = 12 cm
r (radius) = 4 cm (The diameter is 8 being twice the radius)
Adopting: [tex]\pi \approx 3.14[/tex]
V (volume) = ?
Solving: (Cone volume)
[tex]V = \dfrac{ \pi *r^2*h}{3}[/tex]
[tex]V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}[/tex]
[tex]V = 3.14*16*4[/tex]
[tex]\boxed{V = 200.96\:cm^3}[/tex]
Note: Now, let's find the volume of a hemisphere.
Data: (hemisphere volume)
V (volume) = ?
r (radius) = 4 cm
Adopting: [tex]\pi \approx 3.14[/tex]
If: We know that the volume of a sphere is [tex]V = 4* \pi * \dfrac{r^3}{3}[/tex] , but we have a hemisphere, so the formula will be half the volume of the hemisphere [tex]V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}[/tex]
Formula: (Volume of the hemisphere)
[tex]V = 2* \pi * \dfrac{r^3}{3}[/tex]
Solving:
[tex]V = 2* \pi * \dfrac{r^3}{3}[/tex]
[tex]V = 2*3.14 * \dfrac{4^3}{3}[/tex]
[tex]V = 2*3.14 * \dfrac{64}{3}[/tex]
[tex]V = \dfrac{401.92}{3}[/tex]
[tex]\boxed{ V_{hemisphere} \approx 133.97\:cm^3}[/tex]
Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)
Volume of the figure = cone volume + hemisphere volume
Volume of the figure = 200.96 cm³ + 133.97 cm³
[tex]\boxed{\boxed{\boxed{Volume\:of\:the\:figure = 334.93\:cm^3}}}\end{array}}\qquad\quad\checkmark[/tex]
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I Hope this helps, greetings ... Dexteright02! =)
