The Area of Pyramids

What is the area of the base of the following regular pyramid?

1. Identify the relevant shapes
   The base is a regular hexagon.

2. Apply the appropriate formula
   Since this plane figure is a regular polygon, apply this formula: ||\begin{align} A_b &=A_{\text{regular polygons}}\\\\&= \dfrac{\color{#51B6C2}{s} \times \color{#3A9A38}{a_b} \times n}{2}\\\\&= \dfrac{\color{#51B6C2}{5} \times \color{#3A9A38}{4{.}33} \times 6}{2}\\\\&= 64{.}95 \ \text{cm}^2 \end{align}||

3. Interpret the answer
   Since there is no context involved in this problem, simply state that the pyramid’s base area is |64.95\ \text{cm}^2.|

Since a regular pyramid is, in fact, a pyramid with a regular polygon as its base, be careful not to confuse the base’s apothem with that of the pyramid.

The apothem is usually identified by the variable |a.| To differentiate between the two apothems, add a subscript. Thus, the pyramid’s apothem becomes |a_p| and the base’s apothem becomes |a_b.| The choice of the subscript or the way of identifying the 2 measures may vary in different contexts.

||A_L = \dfrac{P_b \times a_p}{2}|| where ||\begin{align}A_L&=\text{Lateral area}\\P_b&=\text{Base perimeter}\\a_p &= \text{Pyramid apothem}\end{align}||

The facades of some of Egypt's pyramids need to be restored in order to keep them open to the public. The surface area of the square pyramid of Khephren needs to be found before companies can be sent a call for tender.

1. Identify the solid
   It is a square-based pyramid. Therefore, we are looking for the lateral area of a regular pyramid.

2. Apply the lateral area formula of the identified solid
   ||\begin{align} A_L &=\dfrac{P_b \times \color{#FA7921}{a_p}}{2}\\\\&=\dfrac{(215+215+215+215) \times \color{#FA7921}{179{.}30}}{2}\\\\&= 77\ 099 \ \text{m}^2\end{align}||

3. Interpret the answer
   The surface area to be restored is equivalent to |77\ 099\ \text{m}^2.|

Note: Because it is a square-based pyramid, the same answer can be obtained by calculating the area of a single lateral face (a triangle) and multiplying it by 4.

Demonstrating the formula for the lateral area of a regular pyramid

Therefore, the usual formula for the area of a triangle is used: |A = \dfrac{b\times h}{2}.| Since the large triangle’s base corresponds to the base’s perimeter, replace |b| with |P_b|, and since the triangle’s height corresponds to the pyramid’s apothem, replace |h| with |a.| Thus, we obtain the formula for the lateral area of the above regular pyramid. ||\begin{align} \color{#333FB1}{A_L} &= \text{Area of the large triangle} \\ &= \dfrac{\color{#FA7921}b \times \color{#EC0000}{h}}{2} \\ &= \dfrac{\color{#FA7921}{P_b}\times \color{#EC0000}{a}}{2} \end{align}||

Calculate the lateral area of the following rectangular pyramid.

1. Identify the solid
   It is a rectangular-based pyramid, so we cannot use the formula for the lateral area of a regular pyramid. The four lateral faces must be calculated separately. ||\begin{align} A_{\triangle\ \text{front}} &= A_{\triangle\ \text{back}}= \dfrac{\color{#3A9A38}{b_1}\times \color{#EC0000}{h_1}}{2} \\\\ A_{\triangle\ \text{right side}} &= A_{\triangle\ \text{left side}}= \dfrac{\color{#51b6c2}{b_2}\times \color{#FA7921}{h_2}}{2} \\\\ A_L &= 2\times A_{\triangle\ \text{front}} + 2 \times A_{\triangle\ \text{the right}}\end{align}||

2. Apply the lateral area formula of the identified solid
   ||\begin{align} A_{\triangle\ \text{front}} &= \dfrac{\color{#3A9A38}{15}\times \color{#EC0000}{{9{.}60}}}{2} \\ &= 72\ \text{cm}^2 \\\\ A_{\triangle\ \text{right side}} &= \dfrac{\color{#51b6c2}{5}\times \color{#FA7921}{11{.}92}}{2} \\ &=29{.}8\ \text{cm}^2 \\\\ A_L &= 2\times A_{\triangle\ \text{front}} + 2 \times A_{\triangle\ \text{the right}} \\ &= 2 \times 72 + 2 \times 29{.}8 \\ &= 203{.}6\ \text{cm}^2 \end{align}||

3. Interpret the answer
   The lateral area of the pyramid is |203{.}6\ \text{cm}^2.|

||A_T = A_L + ​A_b|| where ||A_T=\text{Total area}||

We want to cover the dice used to play Dungeons & Dragons with a particular material. Each die is in the shape of a regular tetrahedron.

How much material will be needed if we want to cover 150 dice?

1. Identify the faces concerned
   Since the dice are to be completely covered, the area of the 4 faces must be calculated (i.e., the total area).

2. Calculate the base area
   Since each die is a regular tetrahedron, all the faces are congruent equilateral triangles. Thus, ||\begin{align} A_b &= \dfrac{\color{#3B87CD}b \times \color{#EC0000}h}{2} \\ &= \dfrac{\color{#3B87CD}{1{.}5} \times \color{#EC0000}{1{.}3}}{2} \\ &= 0{.}975 \ \text{cm}^2\end{align}||

3. Calculate the lateral area
   A tetrahedron is part of the group of regular pyramids, so: ||\begin{align} A_L &= \dfrac{\color{#3b87cd}{P_b} \times \color{#ec0000}a}{2} \\ &= \dfrac{(\color{#3b87cd}{1{.}5} + \color{#3b87cd}{1{.}5} + \color{#3b87cd}{1{.}5}) \times \color{#ec0000}{1{.}3}}{2}\\ &= 2{.}925 \ \text{cm}^2\end{align}||

4. Calculate the total area ||\begin{align} A_T &= A_L + A_b \\ &= 2{.}925 + 0{.}975\\ &= 3{.}9 \ \text{cm}^2 \end{align}||

5. Interpret the answer
   We want to cover 150 dice, so the total area is |3{.}9\ \text{cm}^2/\text{die} \times 150\ \text{dice}= 585\ \text{cm}^2.|

||A_T=4\times\dfrac{b\times h}{2}|| where ||\begin{align} b &= \text{Triangle’s edge or base} \\ h &= \text{Apothem or triangle’s height} \\ A_T &= \text{Tetrahedron’s total area} \end{align}||

Finding the measurement of the apothem from the height

In the case of a right pyramid, a right triangle can be obtained by tracing the pyramid’s apothem, the pyramid’s height, and the line segment connecting the centre of the base and the midpoint of the base’s side.

Since it is a right pyramid, and the height is located at the base’s centre, the measurement of the leg is half the measurement of the base’s side.

By associating the measurement of one leg of the right triangle with half of one side of the base and associating the other leg with the pyramid’s height, the apothem becomes the hypotenuse. Now there is enough information to use the Pythagorean Theorem: ||\begin{align} \color{#3A9A38}{a}^2 + \color{#EC0000}{b}^2 &= \color{#51B6C2}{c}^2\\\\ \color{#3A9A38}{6}^2 + \color{#EC0000}{8}^2 &= \color{#51B6C2}{a}^2\\ 100 &= \color{#51B6C2}{a}^2\\ 10\ \text{cm} &​= \color{#51B6C2}{a} \end{align}||
Thus, the pyramid’s apothem is |10\ \text{cm}.|

When using several concepts simultaneously, be careful with the variables you are using. On a pyramid, |\boldsymbol{\color{#51B6C2}{a}}| refers to the apothem, while in the Pythagorean theorem, the variable |\boldsymbol{\color{#51B6C2}{c}}| refers to this same segment. To fully understand the two examples, use colours to associate the numbers with the segments they represent.