Question

Identify the lateral area and surface area of a regular pyramid with base edge length 6 cm and slant height 13 cm.

Answer 1

A regular triangular pyramid has equilateral triangles.

The lateral area (Al) of a triangular pyramid is:

`Al=(1)/(2)(perimeter\text{ + }slant\text{ }height)_`

The surface area (As) of a triangular pyramid is:

`\begin{gathered} As=Ab+Al \\ and: \\ Ab=\frac{base\text{ }edge*h}{2} \end{gathered}`

Step 01: Find Al.

Perimeter = sum of the sides = 6 + 6 + 6 = 18 cm

Slant height = 13 cm

`\begin{gathered} Al=(1)/(2)(18*13) \\ Al=(1)/(2)(234) \\ Al=117cm^2 \end{gathered}`

Step 02: Find As.

To find As, first find "h". h can be found according to the figure below:

So, h can be found using the Pythagorean theorem:

`\begin{gathered} 6^2=h^2+3^2 \\ 36=h^2+9 \\ 36-9=h^2+9-9 \\ 27=h^2 \\ Taking\text{ }the\text{ }root\text{ }of\text{ }both\text{ }sides: \\ √(27)=√(h^2) \\ 5.2=h \\ \end{gathered}`

And, find Ab and then find As:

`\begin{gathered} Ab=(6*5.2)/(2) \\ Ab=15.6cm^2 \end{gathered}`
`\begin{gathered} As=Ab+Al \\ As=15.6+117 \\ As=132.6cm^2 \end{gathered}`

Answer:

Lateral area: 117 cm².

Surface area: 132.6 cm².