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7. 3D Geometry
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Grade 8

7.01 Surface area of square-based pyramids

Lesson
Practice

Surface area of square-based pyramids

A pyramid is a polyhedron formed from a polygonal base and a set of triangular faces. The triangular faces connect to one side of the base and all join together at the apex.

Polyhedron

A solid figure whose faces area all polygons.

A square-based pyramid with labels apex, face, and base.

A square-based pyramid is a pyramid whose base is a square.

We can find the surface area of a pyramid by adding up the area of the base and all of the faces. A net can help us visualize this.

Net

A two-dimensional representation of a three-dimensional figure that can be folded into a model of the three-dimensional figure.

Exploration

  • Use the sliders to adjust the height and side length of the square-based pyramid
  • Then drag the Slide to unravel slider to create the net
Loading interactive...
  1. What is the area of the base?
  2. What is the area of one of the triangular sides?
  3. What is the total surface area?
  4. Write a formula for surface area.
  5. Write another formula using the variables B for base area and p for perimeter of the base.

The surface area of a square pyramid can be found as:

\text{SA of a square pyramid} = \text{Area of square base}+ \text{Area of all triangles}

Because the base is a square all 4 of the triangular faces are congruent.

A square-based pyramid and its net with labels faces and base.

  • The area of the one triangular face: A_{\text{face}}=\dfrac{1}{2} ls

  • The area of the square base: A_{\text{base}}=l \cdot w

  • Adding these all together we get: SA=\dfrac{1}{2} ls+\dfrac{1}{2} ls+\dfrac{1}{2} ls+\dfrac{1}{2} ls+l w

Rewriting the repeated addition as multiplication shows we have four times the area of one triangle plus the area of the square base: SA=4 \cdot \dfrac{1}{2} l s + ls

Using the commutative property, we can rewrite this as: SA = \dfrac{1}{2} \cdot l \cdot 4s + ls

Notice that 4s is the perimeter of the square base, which we can call p, and ls is the area of the square base, which we can call B.

Putting that all together, we can develop a formula for the surface area of a square-based pyramid:

\displaystyle SA=\dfrac{1}{2}lp+B
\bm{l}
slant height
\bm{s}
side length
\bm{p}
perimeter of base
\bm{B}
area of base
A square-based pyramid with a right triangle with labels slant height and heigh. Ask your teacher for more information.

Sometimes, instead of the slant height of the pyramid we are given the perpendicular height of the pyramid instead. Luckily, we can use the perpendicular height to calculate the slant height using the Pythagorean theorem: a^2+b^2=c^2

Because we are working with right pyramids, the shorter leg of the right triangle created will be half of the side length of the square base.

Examples

Example 1

A square-based pyramid has a base side length of 30 meters and a slant height of 25 meters.

A square-based pyramid with base length of 30 meters and slant height of 25 meters. Ask your teacher for more information.
a

Create the net of the pyramid.

Worked Solution
Create a strategy

To create the net of a pyramid, draw the base square and attach four triangles along each side of the square.

Apply the idea
A net of square-based pyramid with label of s=30 meters and l=25 meters. Ask your teacher for more information.
b

Use the net to find the surface area.

Worked Solution
Create a strategy

We need to find the area of one triangle and then multiply that by 4 because there are 4 congruent, triangular faces. Then add that to the area of the base.

Apply the idea
A triangle with length of 25 meters and base of 30 meters.

Area of one triangle:

\begin{aligned}\text{Area} =&\left( \dfrac{1}{2} l s\right) \\=&\left(\dfrac{1}{2} \cdot 25 \cdot 30\right)\\=&375\text{ m}^2 \end{aligned}

\begin{aligned} \text{Area of all } 4 \text{ triangles}=4 \cdot 375 = 1500 \text{ m}^2 \end{aligned}

A square with side length of 30 meteres.

Square base:

\begin{aligned}\text{Area} =& s^2 \\=& 30^2\\=&900\text{ m}^2 \end{aligned}

A net of square-based pyramid with label of s=30 meters and l=25 meters. Ask your teacher for more information.

Surface area of the square based pyramid:

\begin{aligned}\text{Surface area} =& 1500 + 900 \\=&2400\text{ m}^2 \end{aligned}

Example 2

A square-based pyramid has a base side length of 9 centimeters and a height of 3 centimeters. Calculate the surface area of the pyramid.

A square pyramid with square side length of 9 centimetres and a height of 3 centimetres.
Worked Solution
Create a strategy

Calculate the area of the square base using the formula A_{\text{base}} = s^2. For the triangular sides, use the formula A_{\text{triangle}} = \dfrac{1}{2} \cdot \text{base} \cdot \text{slant height}. The slant height can be found with the Pythagorean theorem: a^2 + b^2 = c^2.

Apply the idea

The area of the square base:

\displaystyle A_{\text{base}}\displaystyle =\displaystyle 9^2Substitute s = 9
\displaystyle =\displaystyle 81 \, \text{cm}^2Evaluate

The slant height can be found with the Pythagorean theorem a^2 + b^2 =c^2. First we need to find the length of the base of the triangle. We can do this by dividing the length of the square side by 2. The base of the triangle is \dfrac{9}{2}=4.5

\displaystyle 3^2 + 4.5^2\displaystyle =\displaystyle c^2Substitute a=3 ,\, b=4.5
\displaystyle 9+20.25\displaystyle =\displaystyle c^2Evaluate the exponents
\displaystyle 29.25\displaystyle =\displaystyle c^2Evaluate the addition
\displaystyle \sqrt{29.25}\displaystyle =\displaystyle \sqrt{c^2}Take the square root of each side
\displaystyle \approx\displaystyle 5.41 \text{ cm}^2Evaluate the radical

The area of the four triangles:

\displaystyle A_{\text{triangle}}\displaystyle =\displaystyle 4 \cdot \dfrac{1}{2} \cdot 9 \cdot 5.41Substitute \text{base}=9,\,\text{slant height}=5.41
\displaystyle =\displaystyle 97.38 \, \text{cm}^2Evaluate

To find the surface area of the square based pyramid, we need to add the area of the square base and the area of the triangles.

\displaystyle \text{Surface Area}\displaystyle =\displaystyle 81 + 97.38Add the A_{\text{base}} and A_{\text{triangle}}
\displaystyle =\displaystyle 178.38 \, \text{cm}^2Evaluate
Reflect and check

We could have also used the formula for the surface area of a pyramid: SA= lp+B

First we will find the perimeter of the square base. In this pyramid, each side length is 9, so the perimeter of the square is 36.

Now we can substitute the values into the formula.

\displaystyle SA\displaystyle =\displaystyle \dfrac{1}{2}(5.41)(36)+81Substitute the values
\displaystyle =\displaystyle 91.38+81Evaluate the multiplication
\displaystyle =\displaystyle 178.38Evaluate the addition

The surface area of the pyramid is 178.38 \text{cm}^2

Example 3

A square pyramid has a surface area of 96\text{ ft}^{2}. Each triangular face has an area of 15\text{ ft}^2.

Find the side length of the base of the pyramid, correct to the nearest foot.

Worked Solution
Create a strategy

The area of the base can be found by subtracting the surface area by the area of the triangles. There are four congruent triangular faces in a square pyramid.

Once we have the area of the square base, we can solve for the side length of the square.

Apply the idea

There are four triangular faces in a square pyramid. The total area of the triangualar faces is given by:

4\cdot 15 \text{ft}^2= 60\text{ ft}^2

\displaystyle \text{Area of the base}\displaystyle =\displaystyle \text{Surface area} - \text{Area of the triangles}
\displaystyle =\displaystyle 96 - \, 60Substitute the values
\displaystyle =\displaystyle 36Evaluate

\text{Area of the square base}=s^2, where s is the side length of the square base.

\displaystyle {s^2}\displaystyle =\displaystyle s \cdot sRewrite in expanded form
\displaystyle s \cdot s\displaystyle =\displaystyle 36Consider what number times itself is equal to 36
\displaystyle s\displaystyle =\displaystyle 6Evaluate

The side length is 6 \text{ ft}.

Example 4

Joan visited Pyramid of Giza, which is a square pyramid. At the gift shop, she bought a miniature replica of the pyramid that has a slant height of 6 inches and a base length of 4 inches. What is the surface area of the replica?

A pyramid.
Worked Solution
Create a strategy

We can use the formula \dfrac{1}{2}lp+B to find the surface area

Apply the idea

First we will find the area of the base, we have:

\displaystyle B\displaystyle =\displaystyle \left(\text{side length}\right)^2Area of a square
\displaystyle =\displaystyle 4^2Substitute the side length
\displaystyle =\displaystyle 16Simplify

We can now will calculate the perimeter of the base:

\displaystyle p\displaystyle =\displaystyle \left(\text{side length}\right) \cdot 4Perimeter of a square
\displaystyle =\displaystyle 4 \cdot 4Substitute known values
\displaystyle =\displaystyle 16Evaluate the multiplication

Now we can substitute those values into the fomula for surface area of a square pyramid: SA=\dfrac{1}{2}lp+B.

\displaystyle SA\displaystyle =\displaystyle \dfrac{1}{2}lp+BSurface area of a square pyramid
\displaystyle =\displaystyle \dfrac{1}{2}(6)(16)+(16)Substitute known values
\displaystyle =\displaystyle 48+16Evaluate the multiplication
\displaystyle =\displaystyle 64Evaluate the addition

The surface area of the replica is 64 square inches.

Idea summary

A square pyramid is a polyhedron with a square base and four faces that are congruent triangles with a common vertex.

To find the surface area of a square pyramid, we can use the equation:

\text{Surface area of a square pyramid} = \dfrac{1}{2}lp + B

If we are given the perpendicular height, we can find the slant height by using the Pythagorean theorem: a^2+b^2=c^2.

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.2a

Determine the surface area of square-based pyramids by using concrete objects, nets, diagrams, and formulas.

8.MG.2d

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

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