A regular triangular pyramid has equilateral triangles.
The lateral area (Al) of a triangular pyramid is:
[tex]Al=\frac{1}{2}(perimeter\text{ + }slant\text{ }height)_[/tex]
The surface area (As) of a triangular pyramid is:
[tex]\begin{gathered} As=Ab+Al \\ and: \\ Ab=\frac{base\text{ }edge*h}{2} \end{gathered}[/tex]
Step 01: Find Al.
Perimeter = sum of the sides = 6 + 6 + 6 = 18 cm
Slant height = 13 cm
[tex]\begin{gathered} Al=\frac{1}{2}(18*13) \\ Al=\frac{1}{2}(234) \\ Al=117cm^2 \end{gathered}[/tex]
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:
[tex]\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: \\ \sqrt{27}=\sqrt{h^2} \\ 5.2=h \\ \end{gathered}[/tex]
And, find Ab and then find As:
[tex]\begin{gathered} Ab=\frac{6*5.2}{2} \\ Ab=15.6cm^2 \end{gathered}[/tex][tex]\begin{gathered} As=Ab+Al \\ As=15.6+117 \\ As=132.6cm^2 \end{gathered}[/tex]
Answer:
Lateral area: 117 cm².
Surface area: 132.6 cm².
