Topics
7. 3D Geometry
topic badge
United States of AmericaVirginia
Grade 8

7.04 Problem solving with volume and surface area

Lesson
Practice

Problem solving with volume and surface area

Recall that the volume of a three-dimensional shape is the amount of space that the shape takes up. The surface area is the total area of the two-dimensional faces that make up the figure. When considering whether a situation represents volume or surface area, examples of some helpful terms to keep in mind include

VolumeSurface Area
containaround
interiorexterior
fillcover
capacitywrap
cubic unitssquare units

Surface area measures the outside of a three-dimensional figure and is measured in \text{units}^2. Some examples of real-life scenarios that require calculating surface area are:

A cone shaped wooden block.

\\\\ \,\\\\\\ \,\\\\\\ \,\\Determining how much paint is required to paint a cone shaped block.

A pyramid shaped tent.

\\\\ \,\\\\\\ \,\\\\\\ \,\\ Determining how much material is required to make a pyramid shaped tent.

Volume measures how much space is inside of a three-dimensional figure, or the capacity that the figure can hold and is measured in \text{units}^3. Some examples of real-life scenarios that require calculating volume are:

Ice cream cone.

\\\\ \,\\\\\\ \,\\\\\\ \,\\Determining how much ice cream can fit in a cone.

A pyramid shaped jar

\\\\ \,\\\\\\ \,\\\\\\ \,\\Determining how much sand would be needed to fill a pyramid shaped jar.

Examples

Example 1

A podium is formed by sawing off the top of a cone. How much space is inside the podium? Round your answer to two decimal places.

An 8m-podium formed by sawing off the top of a cone with radius of 8 m and 4 m. Ask you teacher for more information.
Worked Solution
Create a strategy

Because the question is asking about the space inside the podium, we must calculate the volume of the figure. The volume of podium is the difference of the volume of the large cone and the volume of the small cone at the top.

Apply the idea
Two right triangles formed with a common side length. Ask you teacher for more information.

Based on the diagram, the radius of the small cone is r_1=2 \operatorname{m} and the radius of the large cone is r_{2}=4 \operatorname{m}.

The slant height of the large cone is 16\operatorname{ cm} and the slant height of the smaller cone is 8\operatorname{ cm}.

To find the perpendicular heights of the cones we can use the Pythagorean theorem.

Let's work with the smaller cone first. Let a be the perpendicular height:

A Right triangle. hypotenuse of 8, shorter leg is 2 and longer leg is 'a'.
\displaystyle a^{2}+b^{2}\displaystyle =\displaystyle c^{2}The Pythagorean theorem
\displaystyle a^{2}+2^{2}\displaystyle =\displaystyle 8^{2}Substitute b=2 and c=8
\displaystyle a^{2}+4\displaystyle =\displaystyle 64Evaluate the exponents
\displaystyle a^{2}\displaystyle =\displaystyle 60Subtract 4 from both sides
\displaystyle a\displaystyle =\displaystyle \sqrt{60}Take the square root of both sides

We will leave a=\sqrt{60} in exact form for now because the longer we wait to round, the more precise the answer will be.

Now that we know the perpendicular height of the smaller cone, we can calculate its volume.

\displaystyle V_\text{small}\displaystyle =\displaystyle \dfrac{1}{3}\pi r^{2} hFormula for volume of a cone
\displaystyle V_\text{small}\displaystyle =\displaystyle \dfrac{1}{3}\pi \cdot 2^{2} \cdot \sqrt{60}Substitute r=2 and h =\sqrt{60}
\displaystyle V_\text{small}\displaystyle =\displaystyle \dfrac{1}{3}\pi \cdot 4 \cdot \sqrt{60}Evaluate the exponent
\displaystyle =\displaystyle 32.45 \operatorname{ m}^{3}Use a calculator to evaluate to two decimal places

Now let's look at the large cone. Let b be the perpendicular height:

A Right triangle. hypotenuse of 16, shorter leg is 4 and longer leg is 'b'.
\displaystyle a^{2}+b^{2}\displaystyle =\displaystyle c^{2}The Pythagorean theorem
\displaystyle 4^{2}+b^{2}\displaystyle =\displaystyle 16^{2}Substitute a=4 and c=16
\displaystyle 16+b^{2}\displaystyle =\displaystyle 256Evaluate the exponents
\displaystyle b^{2}\displaystyle =\displaystyle 240Subtract 16 from both sides
\displaystyle b\displaystyle =\displaystyle \sqrt{240}Take the square root of both sides

We will also leave b=\sqrt{240} in exact form.

Now that we know the perpendicular height of the larger cone, we can calculate its volume.

\displaystyle V_{\text{large}}\displaystyle =\displaystyle \frac{1}{3} \pi r^{2} hFormula for volume of a cone
\displaystyle V_{\text{large}}\displaystyle =\displaystyle \frac{1}{3} \pi \cdot 4^{2} \cdot \sqrt{240}Substitute r=4 and h=\sqrt{240}
\displaystyle V_{\text{large}}\displaystyle =\displaystyle \frac{1}{3} \pi \cdot 16 \cdot \sqrt{240}Evaluate the exponent
\displaystyle V_{\text{large}}\displaystyle =\displaystyle 259.57 \operatorname{ m}^{3}Use a calculator to evaluate to two decimal places

We can now calculate the volume of the podium.

\displaystyle \text{Volume of podium }\displaystyle =\displaystyle 259.57 \operatorname{ m}^{3}-32.45 \operatorname{ m}^{3}Subtract the volume of the small cone from the large cone
\displaystyle =\displaystyle 227.12\operatorname{ m}^3 Evaluate

Example 2

You are making pyramid-shaped terrarium displays for an art exhibit. Each terrarium has a square base with a width of 9 inches and the slant height of the terrarium is 14 inches. How many square inches of glass do you need to buy to make a terrarium?

Worked Solution
Create a strategy

The problem is asking to find the total area of the flat faces making up the pyramid, so the surface area formula S.A.=\dfrac{1}{2}lp + B will be used to find the surface area of the square-based pyramid.

Apply the idea

The slant height, l, is given as 14 inches, and a side of the square base being 9 inches.

The perimeter, p, of the square base is 4 \cdot 9, or 36.

The area of the base, B, is 9^{2}, or 81.

With all values, we can substitute for the variables in the formula and simplify.

\displaystyle \dfrac{1}{2}lp+B\displaystyle =\displaystyle \dfrac{1}{2}(14)(36)+(81)Substitute l=14, p=36, and B=81
\displaystyle =\displaystyle 252+81Evaluate the multiplication
\displaystyle =\displaystyle 333Evaluate the addition

The terrarium needs 333 \operatorname{ in}^{2} of glass.

Reflect and check

The net of the pyramid-shaped terrarium with a square base would look like this:

A square-based pyramid shaped terrarium and its net. base side width is 9 inches and slant height is 14 inches.

We can use the net to determine that the area of the base would be 81 \operatorname{ in}^{2}.

We can use the formula for area of a triangle, \dfrac{1}{2}bh to determine the area of each of the triangular sides to be 63 \operatorname{ in}^{2}. There are four triangular bases, so the area of all of the faces would be 252 \operatorname{ in}^{2}.

If we add together the area of the square base and the area of the four triangular bases, we get that the surface area is 252 \operatorname{ in}^{2}+81 \text{ in}^2=333 \operatorname{ in}^{2}. So the surface area of the terrarium is 333 \text{ in}^{2}.

Idea summary

We can use the concept of surface area and volume in real-world problems.

The surface area is the sum of the areas of all surfaces of a figure.

The volume of a three dimensional shape is the amount of space that is taken up by the shape, or the amount it can hold.

Outcomes

8.MG.2

The student will investigate and determine the surface area of square-based pyramids and the volume of cones and square-based pyramids.

8.MG.2d

Solve problems in context involving volume of cones and square-based pyramids and the surface area of square-based pyramids.

What is Mathspace

About Mathspace