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HELP ME PLEASE!!!!!!!!!!!!!

HELP ME PLEASE!!!!!!!!!!!!!-example-1
User Wole
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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:
\pi \approx 3.14

V (volume) = ?

Solving:(Cone volume)


V = ( \pi *r^2*h)/(3)


V = ( 3.14 *4^2*\diagup\!\!\!\!\!12^4)/(\diagup\!\!\!\!3)


V = 3.14*16*4


\boxed{V = 200.96\:cm^3}

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting:
\pi \approx 3.14

If: We know that the volume of a sphere is
V = 4* \pi * (r^3)/(3) , but we have a hemisphere, so the formula will be half the volume of the hemisphere
V = (1)/(2)* 4* \pi * (r^3)/(3) \to \boxed{V = 2* \pi * (r^3)/(3)}

Formula: (Volume of the hemisphere)


V = 2* \pi * (r^3)/(3)

Solving:


V = 2* \pi * (r^3)/(3)


V = 2*3.14 * (4^3)/(3)


V = 2*3.14 * (64)/(3)


V = (401.92)/(3)


\boxed{ V_(hemisphere) \approx 133.97\:cm^3}

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


\boxed{\boxed{\boxed{Volume\:of\:the\:figure = 334.93\:cm^3}}}\end{array}}\qquad\quad\checkmark

________________________

I Hope this helps, greetings ... Dexteright02! =)

User Biraj Zalavadia
by
7.0k points
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