a) 64cm
b) 64 square cm
c) 4√3 cm
d) 16√3 square cm
A regular quadrilateral pyramid with equilateral triangles is as shown below;
1) The pyramid has 8 side lengths, hence the sum of the length of all the sides is given as:
[tex]\begin{gathered} Sum\text{ of side lengths}=8\times8cm \\ Sum\text{ of side lengths}=64cm \end{gathered}[/tex]2) Since the triangular sides are equilateral, the base of the pyramid will be a square with side length of 8cm. The area of the base is expressed as:
[tex]\begin{gathered} A=length\times length \\ A=8cm\times8cm \\ A=64cm^2 \end{gathered}[/tex]3) Since one side wall is an equilateral triangle, the height will be perpendicular to the base as shown:
In order to determine the height, we will use the Pythagorean theorem as shown:
[tex]\begin{gathered} 8^2=h^2+4^2 \\ h^2=8^2-4^2 \\ h^2=64-16 \\ h^2=48 \\ h=\sqrt{48}=4\sqrt{3}cm \\ \end{gathered}[/tex]4) The area of the side wall is equivalent to the area of the triangle expressed as:
[tex]\begin{gathered} Area\text{ of side wall}=\frac{1}{2}\times base\times height \\ Area\text{ of side wall}=\frac{1}{2}\times8cm\times4\sqrt{3} \\ Area\text{ of side wall}=16\sqrt{3}cm^2 \end{gathered}[/tex]
