Surface Area of Composite solids

Subsection 2.7.7 Surface Area of Composite solids

Activity 2.7.10.

\(\textbf{Tree Model Surface Area}\)

Theme: Modeling a tree using a cylinder \\(textbf{trunk}\) and hemisphere \(\textbf{tree top}\) or cone \(\textbf{pine tree top}\)

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We will model a tree trunk as a composite solid, then calculate the total surface area, excluding the part where the top and trunk connect.

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\(\displaystyle \textbf{Materials needed:}\)
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Nets or templates for:
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Cylinder (tree trunk)
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Cone or hemisphere (tree top)
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Scissors, glue or tape
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Rulers and string
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Step-by-Step Instructions.
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1. Create the Model
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Each student or group builds a tree model using
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A cylinder for the trunk
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Either a cone (pine tree) or a hemisphere (bushy tree) for the top
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For example, a pine tree would be a cone on top of a cylinder.
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2. Label Dimensions
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Measure and label your dimensions (or give them pre-set values). For example:
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♦ Radius of trunk = 3 cm
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♦ Height of trunk = 10 cm
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♦ Radius of cone = 3 cm
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♦ Slant height of cone = 5 cm
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3. Surface Area Calculation
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Surface area of cylinder: ♦ Lateral area: \(2 \pi r h\) ♦ Bottom circle: \(\pi r2\) ♦ Do NOT count the top circle — it is covered by the cone
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Surface area of cone:
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♦ Lateral area: \(2 \pi r \ell\)
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♦ Do NOT count the base of the cone — it is attached to the trunk
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♦ Add all visible surfaces:
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♦ \(\text{S.A}_{\text{total}} = \textbf{Lateral area of cone} + \textbf{Lateral area of cylinder} + \textbf{Base of cylinder}\)
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Sample Calculation:
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With the above values:
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Cylinder lateral: \(2\pi \times (3 \, \text{cm})\times(10 \, \text{cm}) = 60 \, \text{cm} \times 3.14 =188.40 \, \text{cm}^2\)
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Cylinder base: \(\pi \times (3 \, \text{cm})^2 =9 \, \text{cm} \times 3.14 = 28.26\, \text{cm}^2\)
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Cone lateral: \(\pi (3 \, \text{cm}) \times (5 \, \text{cm}) = 15 \, \text{cm} \times 3.14 = 47.10 \,\text{cm}^2\)
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\(S.A = 188.40 \, \text{cm}^2 + 28.26\, \text{cm}^2 + 47.10 \,\text{cm}^2 = 263.76 \, \text{cm}^2\)
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\(\textbf{Study Questions}\)
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Why don’t we count the base of the cone or the top of the cylinder?
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What happens if the cone is bigger than the cylinder?
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How would the surface area change if the tree had branches modeled as small cylinders?
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A \(\textbf{Solid}\) is a three dimensional shape. \(\textbf{Solids}\) are objects with three dimensions i.e Width, Length and Height and they have surface area and volumes..

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\(\textbf{Area of composite solids} \)

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When two or more different solids are placed together, the result is composite solids. The surface area of a composite solids can be found by adding areas of the parts of the solids.

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Example 2.7.25.

Find the surface area of the figure alongside. Leave your answer in \(\text{m}^2\)

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Solution.

Surface Area of a cone.

\begin{align*} \text{CSA_{cone}} = \amp \pi r \times ( \ell = \sqrt{r^2 +h^2} )\\ = \amp 3.14 \times (40 \,\text{cm}) \times \sqrt{(40)^2 \, \text{cm} + (30)^2 \, \text{cm}} \\ = \amp 3.14 \times 40 \, \text{cm} \times 50 \, \text{cm}\\ = \amp 125.60 \times 50 \\ = \amp 6,280 \text{cm}^2 \end{align*}

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Curved Surface Area of the Cylinder

\begin{align*} \text{CSA_{cylinder}} = \amp 2\pi r \times h\\ = \amp 2 \times 3.14 \times (40 \,\text{cm}) \times 50 \, \text{ cm} \\ = \amp 2 \times 3.14 \times 40 \, \text{cm} \times 50 \, \text{cm}\\ = \amp 251.20\, \text{cm}^2\times 50 \, \text{cm} \\ = \amp 12, 560 \,\text{cm}^2 \end{align*}

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Base Area of the Cylinder (since only the bottom is exposed)

\begin{align*} \text{Base Area} = \amp \pi r^2 \\ = \amp \pi \times (40)^2 \, \text{cm} \\ = \amp 5024 \, \text{cm}^2 \end{align*}

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Total Surface Area \(= \text{CSA_{cylinder}} + \text{CSA_{cone}} + \,\text{Base Area} \)

\begin{align*} = \amp 12, 560\, \text{cm}^2 + 6,280 \, \text{cm}^2 + 5024 \, \text{cm}^2 \\ = \amp 23,864 \, \text{cm}^2 \end{align*}

Converting to \(\text{m}^2 \)

\begin{align*} 1 \text{m} = \amp 100 \, \text{cm}\\ 1 \text{m}^2 = \amp 10,000\, \text{cm}^2\\ ? \amp 23,864\, \text{cm}^2 \\ = \amp 1\, \text{m} \times \frac{23,864\, \text{cm}^2} {10,000\, \text{cm}^2} \\ = \amp 2.3864 \, \text{m}^2 \\ = \amp 2.39\, \text{m}^2 \,\ \text{(to two d.p)} \end{align*}

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\(\textbf{Exercise}\)

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1. The solid below is made of a cube and a square pyramid whose height is 33cm and cube sides measures 14cm. Answer the following

Find the surface area of the solid shown. Give your answers to two decimal places.
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Now determine the volume of the composite solids.
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2. Calculate the volume and surface area of the solid alongside.

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3. A right circular icecream cone with a radius of 3 cm and a height of 12 cm holds a half scoop of ice cream in the shape of a hemisphere on top. If the ice cream melts completely, will it fit inside the cone? Show all calculations to justify your answer.

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4. A lampshade is in the shape of a frustum of a cone. The top and bottom circular openings have diameters of 12 cm and 20 cm, respectively. If the slant height is 15 cm, find the lateral surface area of the lampshade.

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5. Mogaka a grade 10 student was trying to sketch an image of a ice cream cone container with icecream. Find the surface area of the sketched image alongside.

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6.A birthday cake has a cylindrical base of radius 10 cm and height 15 cm. The top is shaped like a hemisphere with the same radius. Find the total volume of the cake.

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7.A goblet consists of a hemisphere on top of a cylindrical base. The hemisphere has a radius of 5 cm, and the cylinder has the same radius with a height of 12 cm. Find the surface of the goblet and determine how much liquid it can hold.

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Checkpoint 2.7.26.

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Checkpoint 2.7.27.

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