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McGraw Hill Integrated II, 2012 View details

5. Volumes of Pyramids and Cones

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Exercise 28 Page 845

Divide the composite solid into a cone and a cylinder.

About 7698.5 cubic centimeters

Practice makes perfect

Let's analyze the given composite solid. Since the diameter of the base is 26 cm, its radius is 262=13 cm.

It is made of two solids.

A cone with a radius of r= 13 cm and a height of h_1= 12 cm,
A cylinder with a radius of r= 13 cm and a height of h_2= 10.5 cm.

Now, let's use the formulas for the volume of a cone and for the volume of a cylinder.

Solid	Cone	Cylinder
Radius	r= 13	r= 13
Height	h_1= 12	h_2= 10.5
Volume	V_\text{cone}=\dfrac{1}{3}\pi {\color{#0000FF}{r}}^2{\color{#009600}{h_1}}	V_\text{cylinder}=\pi {\color{#0000FF}{r}}^2{\color{#009600}{h_2}}
Volume	\textcolor{darkorange}{V_\text{cone}}=\dfrac{1}{3}\pi({\color{#0000FF}{13}})^2({\color{#009600}{12}})=\textcolor{darkorange}{676\pi}	\textcolor{darkviolet}{V_\text{cylinder}}=\pi ({\color{#0000FF}{13}})^2({\color{#009600}{10.5}})=\textcolor{darkviolet}{1774.5\pi}

Now, to find the volume of the composite solid we will add the volume of the cone and the volume of the cylinder. Next, we will round our answer to the nearest tenth.

V_\text{solid}=\textcolor{darkorange}{V_\text{cone}}+\textcolor{darkviolet}{V_\text{cylinder}}

▼

Substitute values and evaluate

SubstituteII

\textcolor{darkorange}{V_\text{cone}}={\color{#0000FF}{\textcolor{darkorange}{676\pi}}}, \textcolor{darkviolet}{V_\text{cylinder}}={\color{#009600}{\textcolor{darkviolet}{1774.5\pi}}}

V_\text{solid}=\textcolor{darkorange}{676\pi}+\textcolor{darkviolet}{1774.5\pi}

AddTerms

Add terms

V_\text{solid}=2450.5\pi

π ≈ 3.1416

V_\text{solid}\approx 2450.5({\color{#0000FF}{3.1416}})

Multiply

V_\text{solid}\approx 7698.4908

RoundDec

Round to 1 decimal place(s)

V_\text{solid}\approx 7698.5

Finally, we find that the volume of the given solid is about 7698.5 cubic centimeters.