Have you ever gotten a list of school supplies to buy for the year? Suppose, on that list, is a \(12\) inch ruler. Your parents are convinced that if you don’t get a pencil case to put it in, you will lose that ruler. If your pencil case from last year is a rectangular prism \(1\) in x \(6\) in x \(10\) in, will you need to buy a new pencil case to fit your ruler?
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In this target, we will focus on finding missing lengths given surface area or volume of a polyhedron. To be able to master this target, it is imperative that you focus on your skills in solving linear equations. It is also important to recognize when two terms are considered like terms.
Example 1:
If a cylinder whose base has a radius of \(7.5\) inches has a total surface area of \(1202\) square inches, what is the height of the cylinder?
Solution: First sketch a picture of this situation. Then, you can follow the steps in the video to find the solution. Compare your sketch.
Example 1:
If a cylinder whose base has a radius of \(7.5\) inches has a total surface area of \(1202\) square inches, what is the height of the cylinder?
Solution: First sketch a picture of this situation. Then, you can follow the steps in the video to find the solution. Compare your sketch.
Example 2:
In the following diagram, the lateral area is \(156\) square feet. Find the value of \(x\).
In the following diagram, the lateral area is \(156\) square feet. Find the value of \(x\).
Solution: Since this is a triangular prism. We will use the lateral surface area formula for a prism to solve this problem.
Here’s what we know from the diagram:
\(LA = 156\) ft\(^2\)
Height of prism = \(x\) ft (unknown)
Perimeter of base = \(3\cdot 4 = 12\) ft
So let’s substitute the values we know and don’t know into the \(LA\) equation:
\(LA\) = \(PH\)
\(156 = 12(x)\) Substitute \(156\) for \(LA\), \(12\) for \(P\), perimeter of base, \(x\) for \(H\), height of the prism.
\(x = 13\) ft Divide both sides of the equation by \(12\) to solve for \(x\).
Example 3:
Find the length of the diagonal shown (\(\overline{AG}\)) in the rectangular prism below if \(GH = 2\) cm, \(FD = 6\) cm, and \(AB = 8\) cm. Please note: the diagonal of a rectangular prism is a segment that connects opposite vertices from each base of the prism.
Here’s what we know from the diagram:
\(LA = 156\) ft\(^2\)
Height of prism = \(x\) ft (unknown)
Perimeter of base = \(3\cdot 4 = 12\) ft
So let’s substitute the values we know and don’t know into the \(LA\) equation:
\(LA\) = \(PH\)
\(156 = 12(x)\) Substitute \(156\) for \(LA\), \(12\) for \(P\), perimeter of base, \(x\) for \(H\), height of the prism.
\(x = 13\) ft Divide both sides of the equation by \(12\) to solve for \(x\).
Example 3:
Find the length of the diagonal shown (\(\overline{AG}\)) in the rectangular prism below if \(GH = 2\) cm, \(FD = 6\) cm, and \(AB = 8\) cm. Please note: the diagonal of a rectangular prism is a segment that connects opposite vertices from each base of the prism.
To find the length of the diagonal in a rectangular prism, you can use several right triangles and the Pythagorean theorem. Or, you can use the formula below:
Length of diagonal = \(\sqrt{x^2 + y^2 + z^2}\) where \(x, y,\) and \(z\) are the dimensions of the rectangular prism.
Solution:
Option 1: Use Right Triangles
Let's start by drawing in the diagonal \(\overline{AC}\), focus on \(\triangle{ABC}\), and find \(AC\).
Length of diagonal = \(\sqrt{x^2 + y^2 + z^2}\) where \(x, y,\) and \(z\) are the dimensions of the rectangular prism.
Solution:
Option 1: Use Right Triangles
Let's start by drawing in the diagonal \(\overline{AC}\), focus on \(\triangle{ABC}\), and find \(AC\).
Let's focus on \(\triangle{ACG}\) and find \(AG\).
\((AC)^2 + (CG)^2 = (AG)^2\)
\((\sqrt{68})^2 + 6^2 = (AG)^2\)
\(68 + 36 = (AG)^2\)
\(104 = (AG)^2\)
\(\pm\sqrt{104}=AG\)
\(AG = \sqrt{104} = 2\sqrt{26}\)
Our answer is
\(AG = \sqrt{104} cm = 2\sqrt{26}\) cm \(\approx 10.2\) cm
Option 2: Use the diagonal formula that you discovered in the Investigation.
Since \(x, y,\) and \(z\) represent the dimensions of the rectangular prism, we will substitute \(2\) cm, \(6\) cm, and \(8\) cm into the formula:
\(AG = \sqrt{x^2 + y^2 + z^2}\)
\(AG = \sqrt{2^2+6^2+8^2}\)
\(AG =\sqrt{4+36+64}\)
\(AG = \sqrt{104} cm = 2\sqrt{26}\) cm \(\approx 10.2\) cm
Example 4:
If the volume of a regular hexagonal prism is \(540\sqrt3\) m\(^3\), what is the length of its height?
\((\sqrt{68})^2 + 6^2 = (AG)^2\)
\(68 + 36 = (AG)^2\)
\(104 = (AG)^2\)
\(\pm\sqrt{104}=AG\)
\(AG = \sqrt{104} = 2\sqrt{26}\)
Our answer is
\(AG = \sqrt{104} cm = 2\sqrt{26}\) cm \(\approx 10.2\) cm
Option 2: Use the diagonal formula that you discovered in the Investigation.
Since \(x, y,\) and \(z\) represent the dimensions of the rectangular prism, we will substitute \(2\) cm, \(6\) cm, and \(8\) cm into the formula:
\(AG = \sqrt{x^2 + y^2 + z^2}\)
\(AG = \sqrt{2^2+6^2+8^2}\)
\(AG =\sqrt{4+36+64}\)
\(AG = \sqrt{104} cm = 2\sqrt{26}\) cm \(\approx 10.2\) cm
Example 4:
If the volume of a regular hexagonal prism is \(540\sqrt3\) m\(^3\), what is the length of its height?
Solution:
Let’s sketch a picture of the problem. Since this problem gives you the volume of the prism, we will start with that formula, and substitute values that we know. \(V = BH\) \(540\sqrt3 = BH\) Notice we have two variables in this equation, so we cannot solve it as is. We must first find the area of the hexagon base before we can continue with this equation. Let’s draw the hexagon and find the area of its base. Forget how to do that? Go back to this target to review. |
Now we can go back to where we left off and substitute what we found for the area of the hexagon base:
\(540\sqrt3 = BH\)
\(540\sqrt3 = 54\sqrt3(H)\)
\(H = 10\) m
Example 5:
Given the volume of a sphere is \(972\pi\) cubic meters, what is the length of its radius?
Solution: Since this problem gives you the volume of the sphere, we will use the volume formula of a sphere:
\(V = \Large\frac{4r^3\pi}{3}\) Volume formula of a sphere
\(972\pi = \Large\frac{4r^3\pi}{3}\) Substitute \(972\pi\) for volume
\(\left(\Large\frac{3}{4\pi}\right)\normalsize (972\pi)\) = \(\Large\frac{4r^3\pi}{3}\left(\Large\frac{3}{4\pi}\right)\) Multiply both sides of the equation by the reciprocal \(\Large\frac{3}{4\pi}\).
\(729 = r^3\) Simplify.
To solve this equation for \(r\), we can do one of two things:
Option 1: Take the cube root of both sides of the equation.
\(729 = r^3\)
\(\sqrt[3]{729}= \sqrt[3]{r^3}\)
\(9 = r\) Thus, \(r = 9\) meters
Option 2: Raise both sides of the equation to the \(\frac{1}{3}\) power
\(729 = r^3\)
\(729^{(\frac{1}{3})} = r^{(3)(\frac{1}{3})}\)
Thus, \(r = 9\) meters.
\(540\sqrt3 = BH\)
\(540\sqrt3 = 54\sqrt3(H)\)
\(H = 10\) m
Example 5:
Given the volume of a sphere is \(972\pi\) cubic meters, what is the length of its radius?
Solution: Since this problem gives you the volume of the sphere, we will use the volume formula of a sphere:
\(V = \Large\frac{4r^3\pi}{3}\) Volume formula of a sphere
\(972\pi = \Large\frac{4r^3\pi}{3}\) Substitute \(972\pi\) for volume
\(\left(\Large\frac{3}{4\pi}\right)\normalsize (972\pi)\) = \(\Large\frac{4r^3\pi}{3}\left(\Large\frac{3}{4\pi}\right)\) Multiply both sides of the equation by the reciprocal \(\Large\frac{3}{4\pi}\).
\(729 = r^3\) Simplify.
To solve this equation for \(r\), we can do one of two things:
Option 1: Take the cube root of both sides of the equation.
\(729 = r^3\)
\(\sqrt[3]{729}= \sqrt[3]{r^3}\)
\(9 = r\) Thus, \(r = 9\) meters
Option 2: Raise both sides of the equation to the \(\frac{1}{3}\) power
\(729 = r^3\)
\(729^{(\frac{1}{3})} = r^{(3)(\frac{1}{3})}\)
Thus, \(r = 9\) meters.
Quick Check
1) Given a cone whose base has a radius of \(5\) inches and total surface area is \(219.9\) in\(^2\), find the height of the cone.
1) Given a cone whose base has a radius of \(5\) inches and total surface area is \(219.9\) in\(^2\), find the height of the cone.
2) Find the length of the diagonal in a rectangular prism whose dimensions are \(1\) ft, \(6\) ft, and \(9\) ft.
3) Find the length of the radius of a hemisphere whose total surface area is \(147\) square centimeters.
Quick Check Solutions
3) Find the length of the radius of a hemisphere whose total surface area is \(147\) square centimeters.
Quick Check Solutions