Surface Area of a pyramid

Subsection 2.7.3 Surface Area of a pyramid

Activity 2.7.5. Net of a triangle based Pyramid.

🔗

Paper Folding to form a Pyramid.

🔗

Materials Needed.

🔗

Colored paper or cardstocks
🔗

🔗
Scissors
🔗

🔗
Rulers
🔗

🔗
Glue or tape
🔗

🔗
Markers
🔗

🔗

🔗

Steps.
🔗

♦ Observe the net of the pyramid shown above.
🔗

♦ Draw on the Colored paper or cardstocks the net of a pyramid that is, four triangles. Note the sides of the triangles and height should be equal.
🔗

♦ Cut off the extra paper to remain with the pyramid net.
🔗

♦ Fold along the edges to form a pyramid. Hold the triangles in position using a tape.
🔗

♦ Using a ruler, measure the base height and slant height of the triangular faces.
🔗

🔗
Find the volume of the pyramid you have constructed.
🔗

🔗

🔗

Activity 2.7.6. Observing pyramidal objects in the surrounding environment.

🔗

♦ Identify pyramid-shaped objects around school/home e.g., tent(Shown alongside), roof, food container, toys etc.
🔗

♦ Measure dimensions or use estimated values.
🔗

♦ Find their surface area using the formula and compare with classmates.
🔗

🔗

🔗

A $\text{pyramid}$ is a geometric solid object that has a polygon as its base and faces that converge at a point called the apex. In other words the faces are not perpendicular to the base

🔗

The polygons that can act as the face of the pyramids include:

🔗

🔹 Triangle: Thus called triangular pyramid.

🔗

🔹 Squares: Thus called square pyramid.

🔗

🔹The two pyramids take their names after the shape of their base.

🔗

A right pyramid’s line between the apex and the centre of the base is perpendicular to the base.

🔗

Example 2.7.6.

Find the surface area of the following triangular pyramid (correct to 1 d.p)

🔗

Solution.

The surface area of a triangular pyramid is the sum of the areas of all $\textbf{four triangular faces}$

🔗

In finding the surface area we need to recall the definations of a net and a prism.

🔗

The net of a pyramid consists of a base triangle $ABC\text{,}$ which is a polygon that represents the base of the pyramid, and triangular faces$\angle \, ABD,\, \angle \, BCD, \, \angle ACD$ that represent the sides of the pyramid.

🔗

Each triangle’s area will be found using Heron’s Formula:

\begin{align*} A =\amp \sqrt{S(S-a)(S-b)(S-c)} \end{align*}

Where $S$ is the semi perimeter.

\begin{align*} S = \amp \frac{a+b+c}{2} \end{align*}

🔗

Find the Area of the Base Triangle $ABC\text{.}$ Given sides $AB = 4\, \text{cm}, \, BC = 5\, \text{cm} , \,AC = 3\, \text{cm}$

🔗

Calculate the semi-perimeter:

\begin{align*} S = \amp \frac{4\, \text{cm}+5\, \text{cm}+3\, \text{cm}}{2} = 6 \, \text{cm} \end{align*}

Apply Heron’s Formula:

\begin{align*} A_{ABC} = \amp \sqrt{6(6-4)(6-5)(6-3)} \\ = \amp \sqrt{6 \times(2)\times(1)\times(3)} \\ = \amp \sqrt{36} \\ = \amp 6 \, \text{cm}^2 \end{align*}

Find the Area of Side Triangle$ABD$ Given sides $AB = 4\, \text{cm}, \, AD = 6\, \text{cm} , \, BD = 7\, \text{cm} $

\begin{align*} S = \amp \frac{4 \,\text{cm}+6\, \text{cm}+7\, \text{cm}}{2} = 8.5 \, \text{cm} \\ A_{ABD} = \amp \sqrt{8.5(8.5-4)(8.5-6)(8.5-7)} \\ = \amp \sqrt{8.5 \times(4.5)\times(2.5)\times(1.5)} \\ = \amp \sqrt{143.44} \\ = \amp 11.97 \, \text{cm}^2 \end{align*}

Find the Area of Side Triangle$BCD$ Given sides $BC = 5, \, \text{cm} \,CD = 6 , \, \text{cm} \, BD = 7 \text{cm}$

\begin{align*} S = \amp \frac{5 \, \text{cm}+6 \, \text{cm}+7 \, \text{cm}}{2} = 9 \, \text{cm} \\ A_{BCD} = \amp \sqrt{9(9-4)(9-6)(9-7)} \\ = \amp \sqrt{9 \times(4)\times(2)\times(3)} \\ = \amp \sqrt{216} \\ = \amp 14.7 \, \text{cm}^2 \end{align*}

Find the Area of Side Triangle$ACD$ Given sides $AC = 3, \, AD = 6 , \, CD = 6 $

\begin{align*} S = \amp \frac{3 \, \text{cm}+6 \, \text{cm}+6 \, \text{cm}}{2} = 7.5 \, \text{cm} \\ A_{ACD} = \amp \sqrt{7.5(7.5-4)(7.5-6)(7.5-7)} \\ = \amp \sqrt{7.5 \times(4.5)\times(1.5)\times(1.5)} \\ = \amp \sqrt{75.94} \\ = \amp 8.71\, \text{cm}^2 \end{align*}

Find the Total Surface Area. Surface Area = $A_{ABC} + A_{ABD} +A_{BCD} + A_{ACD}$

\begin{align*} = \amp 6 \, \text{cm}^2 + 11.97 \, \text{cm}^2 + 14.7 \, \text{cm}^2 + 8.71 \, \text{cm}^2 \\ = \amp 41.38 \, \text{cm}^2 \end{align*}

🔗

Example 2.7.7.

Draw the net of the pyramid.

🔗

$Net of square pyramid$

$Square pyramid$

Solution.

The net of the pyramid is shown below.

🔗

The surface area of a pyramid is the total area of the lateral faces and the base of the pyramid.

🔗

Example 2.7.8.

Calculate the surface area of a pyramid whose base is rectangular and slant height $6$$\text{cm}$ as shown below.

🔗

Solution.

$\text{First we form the net of the pyramid.}$

🔗

The area of the rectangular base

\begin{align*} = \amp \text{length} \times \text{width} \\ = \amp5 \, \text{cm} \times 3 \, \text{cm} \\ = \amp 15 \, \text{cm}^2 \end{align*}

The height of triangles with base of $5 \, \text{cm}$

\begin{align*} = \amp\sqrt{36-6.25} \\ = \amp\sqrt{29.75} \\ = \amp 5.45 \, \text{cm} \, (2 \text{d.p}) \end{align*}

The area of a triangle with base of 5 cm

\begin{align*} = \amp\frac{1}{2}\times \text{base} \times \text{height} \\ = \amp\frac{1}{2} \times 5 \times 5.45 \\ = \amp13.625 \, \text{cm}^2 \end{align*}

Since this triangles are two,Therefore total area is:

\begin{align*} = \amp 2 \times 13.625\\ = \amp27.25 \, \text{cm}^2 \end{align*}

The height of triangles with base of 3 cm

\begin{align*} = \amp\sqrt{6^2-1.5^2} \\ = \amp\sqrt{36-2.25} \\ = \amp\sqrt{33.75} \\ = \amp5.81 \, \text{cm}^2 (2 \text{d.p}) \end{align*}

The area of a triangle with base of 3 cm

\begin{align*} = \amp\frac{1}{2}\times \text{base} \times \text{height} \\ = \amp\frac{1}{2} \times 3 \times 5.81 \, \text{cm}^2\\ = \amp 8.715 \, \text{cm}^2 \end{align*}

Since this triangles are two,Therefore total area

\begin{align*} = \amp 2 \times 8.715\\ = \amp17.43 \, \text{cm}^2 \end{align*}

Therefore, the surface area of the pyramid is,

\begin{align*} = \amp15 \, \text{cm}^2 + 27.25 \, \text{cm}^2 + 17.43 \, \text{cm}^2 \\ = \amp59.68 \, \text{cm}^2 \end{align*}

🔗

Example 2.7.9.

Find the surface area of a square pyramid with a height of 6 cm and a side length of 4cm.

🔗

Solution.

Select the correct formula and substitute with the given values.

🔗

We are given b = 2 by 2 and H = 4, therefore

\begin{align*} \text{Surface Area} = \amp \text{Base area} + 4\text{(area of triangles)} \\ \text{Base Area} = 2 \, \text{cm} \times 2 \, \text{cm}\amp \\ = \amp 4 \, \text{cm}^2 \\ \text{Area of one triangle} = \amp \frac{1}{2} b \times h \\ = \amp \frac{1}{2} \times 2 \, \text{cm} \times 4 \, \text{cm} \\ = \amp 4 \, \text{cm}^2 \\ \text{Area of all triangles}= \amp 4 \, \text{cm}^2 \times 4\, \text{triangles} \\ = \amp 16 \, \text{cm}^2\\ \text{Surface area}= \amp 16 \, \text{cm}^2 + 4 \, \text{cm}^2\\ = \amp 20 \, \text{cm}^2 \end{align*}

🔗

The surface area for the square pyramid is $20 \, \text{cm}^2\text{.}$

🔗

$\textbf{Exercise}$

🔗

$Square pyramid$

1. A square-based pyramid has a base with sides measuring 10 cm each, while its triangular faces have a slant height of 15 cm. Determine the total surface area of this pyramid, including the base and all four triangular faces.

🔗

2. A pyramid with a square base has a total surface area of 400 cm², and its base side measures 8 cm. Using the formula for surface area, calculate the slant height of the pyramid.

🔗

3. A square pyramid has a base with sides of 24 cm each, and the height of the triangular face is 18 cm. Find the total surface area of the pyramid.

🔗

4. A miniature paper pyramid is being designed with a square base of 12 cm by 12 cm and a slant height of 20 cm. How much paper is required to construct the entire pyramid?

🔗

5. The roof of a small storage building is in the shape of a rectangular pyramid with a base side length of 5 m and a slant height of 6 m. If the entire roof needs to be covered with wooden shingles, calculate the total area that needs to be covered.

🔗

$Net of square pyramid$

Checkpoint 2.7.10.

🔗

Checkpoint 2.7.11.

🔗

Prev Top Next