Content code
m1504
Slug (identifier)
the-volume-of-truncated-solids
Grades
Secondary III
Topic
Mathematics
Tags
solid
formula for the volume of solids
truncated solid
volume of truncated solids
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A truncated solid is a solid that has been cut by a plane, and only a portion remains. The plane can be parallel or not parallel to the base.

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Any solid can be truncated. Sometimes, the resulting solid is similar to the initial solid, and the volume can be calculated easily. However, for truncated pyramids and cones, it is necessary to use subtraction. Here are two examples.

Title (level 2)
The Volume of Truncated Pyramids
Title slug (identifier)
truncated-pyramid
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A contractor insulates the roof of a house using polyurethane foam applied with a spray gun. Once the foam has hardened, it will look like a truncated rectangular-based pyramid.

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Example of a truncated pyramid where the volume is sought.
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If the product costs |$4\ | for |9\ \text{dm}^3,| how much will it cost to insulate the roof?

  1. Identify the solids
    It is a truncated rectangular-based pyramid. To make the calculations easier, find the initial truncated pyramid and determine its dimensions as follows:

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Trunk of pyramid, complete pyramid, and truncated portion of the initial pyramid
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In a truncated pyramid, like in a truncated cone, the corresponding measurements are proportional. ||\dfrac{\text{Length}_2}{\text{Length}_3} = \dfrac{\text{Width}_2}{\text{Width}_3} = \dfrac{\text{Height}_2}{\text{Height}_3}||In order to calculate the height of the removed pyramid (pyramid 3), the proportion must be solved first.||\begin{align} \dfrac{\color{#EC0000}{\text{Width}_2}}{\color{#FF55C3}{\text{Width}_3}} &= \dfrac{\text{Height}_2}{\text{Height}_3} \\\\ \dfrac{\color{#EC0000}{38}}{\color{#FF55C3}{28{.}5}} &= \dfrac{h_3+\color{#EFC807}{9{.}5}}{h_3} \\\\ 38h_3 &= 28{.}5(h_3+9{.}5) \\ 38h_3 &= 28{.}5h_3 + 270{.}75 \\ 9{.}5h_3 &= 270{.}75 \\ h_3 &= 28{.}5\ \text{dm} \end{align}||Next, determine the height of the full pyramid (pyramid 2).||\begin{align} h_2 &= h_3 +\color{#EFC807}{9{.}5} \\ &= 28{.}5 + \color{#EFC807}{9{.}5} \\ &= 38 \ \text{dm} \end{align}||

  1. Calculate the volume
    The volume of the roof (image 1) is obtained by subtracting the volume of pyramids 2 and 3.||\begin{align} V_{1} &= V_{2} - V_{3}\\ &= \dfrac{A_{\text{base}_2}\times h_2}{3} - \dfrac{A_{\text{base}_{3}}\times h_3}{3} \\ &= \dfrac{(47{.}5 \times 38) \times 38}{3} - \dfrac{(35{.}72 \times 28{,}5)\times 28{.}5}{3} \\ &= \dfrac{1 \ 805 \times 38}{3} - \frac{1 \ 018{.}02 \times 28{.}5}{3} \\ &\approx 22\ 863{.}33 - 9 \ 671{.}19 \\ &\approx 13 \ 192{.}14 \ \text{dm}^3 \end{align}||

  2. Interpret the answer
    Create a proportion and solve it using cross-multiplication. ||\begin{align}\dfrac{$4\ }{?} &= \dfrac{9 \ \text{dm}^3}{13 \ 192{.}14\ \text{dm}^3} \\\\? &= \dfrac{4 \times 13 \ 192{.}14}{9} \\? &\approx $5 \ 863{.}17\ \end{align}|| ​​Therefore, the cost of the insulation product is approximately |$5\ 863{.}17.\\|

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Be careful not to confuse truncated pyramids with trapezoidal-based prisms. The lateral faces of truncated pyramids are trapezoids inclined towards the interior of the solids, whereas the lateral faces of trapezoidal-based prisms are rectangles and the bases are congruent trapezoids.

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Truncated pyramid and trapezoidal-based prism.
Title (level 2)
The Volume of a Truncated Cone
Title slug (identifier)
truncated-cone
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A horticulturist plants tomato seeds in pots shaped like inverted truncated cones to produce a bountiful vegetable harvest.

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Example of an inverted cone where the volume of the upper portion is sought.
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The horticulturist must have 14 plants to produce enough tomatoes. If she fills her pots to capacity, how much soil will she need to plant all the seeds?

  1. Identify the solids
    It is an inverted truncated cone. Consider the initial cone to make the calculations easier.

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Trunk of cone, full cone, and truncated portion of initial cone.
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  1. Calculate the volume
    Obtain the volume of the pot (image 1) by subtracting the volumes of cones 2 and 3. ||\begin{align} V_{1} &= V_{2} - V_{3}\\ &= \frac{A_{\text{base}_2}\times h_2}{3} - \frac{A_{\text{base}_{3}}\times h_3}{3}\end{align}||Since the solids are cones, each base is a circle. Thus, use the following formula:||A_{\text{base}} =\pi r^2||Next, find the radii measurements using the given diameters. ||r_2 = \dfrac{31}{2} =15{.}5\ \text{cm}\\r_3 = \dfrac{18{.}4}{2} =9{.}2\ \text{cm}||It is also necessary to calculate the height of cone 3. ||h_3 = 63{.}7 - 25{.}89 = 37{.}81\ \text{cm}|| ||\begin{align} V_{1} &= V_{2} - V_{3}\\ &= \dfrac{A_{\text{base}_2}\times h_2}{3} - \dfrac{A_{\text{base}_{3}}\times h_3}{3} \\ &= \dfrac{(\pi \times 15{.}5^2) \times 63{.}7}{3} - \dfrac{(\pi \times 9{.}2^2)\times 37{.}81}{3} \\ &= \dfrac{240{.}25\pi \times 63{.}7}{3} - \dfrac{84{.}64\pi \times 37{.}81}{3} \\ &\approx 16\ 026{.}23 - 3 \ 351{.}28 \\ &\approx 12\ 674{.}95 \ \text{cm}^3 \end{align}||

  2. Interpret the answer
    The answer above is the volume of soil required for 1 pot. Since there are 14 pots to fill, perform this calculation: ||​14\ \text{pots} \times 12 \ 674{.}95\ \text{cm}^3/\text{pot}=177 \ 449{.}3 \ \text{cm}^3|| Thus, the amount of soil needed is approximately |177 \ 449{.}3\ \text{cm}^3.|

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It is possible to use formulas instead of completing the previous steps. However, the formulas will vary depending on the nature of the truncated solid.

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Volume formulas associated with truncated solids
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  1. The volume of a truncated cone

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2 columns
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50% / 50%
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Truncated cone with its different dimensions.  The radii of the two bases and the height are shown.
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||V = ​\dfrac{\pi\color{#ec0000}{h}}{3}\Big(\color{#3B87CD}{R}^2+ \color{#3B87CD}{R} \color{#3A9A38}{r} + \color{#3A9A38}{r}^2 \Big)\\ \text{​where}\\ \begin{align} \color{#ec0000}{h} &: \text{Height of the truncated cone} \\ \color{#3a9a38}{r} &: \text{Radius of the small base} \\ \color{#3B87CD}{R} &: \text{Radius of the large base} \end{align}||

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  1. The volume of a regular truncated pyramid

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2 columns
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Truncated pyramid with its different dimensions. The area of the two bases and the height are shown.
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||V​ = \dfrac{\color{#ec0000}{h}}{3}\Big(\color{#3B87CD}{B} + \sqrt{\color{#3B87CD}{B}\times \color{#3a9a38}{b}}+ \color{#3a9a38}{b} \Big)\\\text{​where}\\\begin{align} \color{#ec0000}{h} &: \text{Height of the truncated pyramid} \\ \color{#3a9a38}{b} &: \text{Area of the small base} \\ \color{#3B87CD}{B} &: \text{Area of the large base} \end{align}||

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The solutions shown in the previous examples are more effective than formulas because they can be reused when looking for missing measurements in the solids.

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see-also
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