Since the base of the pyramid is a triangle with a side length of 4 cm
That means the base is an equilateral triangle of sides 4 cm
The height of the equilateral triangle is
[tex]h=\frac{s}{2}\sqrt{3}[/tex]
h is the height
s is the side
Since the side is 4 cm, then
s = 4
Substitute it in the rule above to find h
[tex]\begin{gathered} h=\frac{4}{2}\sqrt{3} \\ \\ h=2\sqrt{3} \end{gathered}[/tex]
(a)
[tex]AM=2\sqrt{3}[/tex]
Since the rule of the area of a triangle is
[tex]A=\frac{1}{2}\times b\times h[/tex]
b is the base
h is the height
Since the base is 4, then
b = 4
Substitute the values of b and h in the rule of the area to find it
[tex]\begin{gathered} A=\frac{1}{2}\times4\times2\sqrt{3} \\ \\ A=4\sqrt{3}\text{ cm}^2 \end{gathered}[/tex]
(b)
The area of the base is
[tex]A=4\sqrt{3}\text{ cm}^2[/tex]
Since the altitude of the pyramid is 4 cm
We will use the Pythagoras theorem to find the side SC
From the triangle above we will use the height of the pyramid OS and a part of the height of the base = 2/3 h to find the side SC
[tex]\begin{gathered} SC^2=4^2+(\frac{2}{3}\times2\sqrt{3})^2 \\ \\ SC^2=16+(\frac{4}{3}\sqrt{3})^2 \\ \\ SC^2=16+\frac{16}{3} \\ \\ SC^2=\frac{64}{3} \end{gathered}[/tex]
Take a square root for both sides
[tex]\begin{gathered} SC=\sqrt{\frac{64}{3}} \\ \\ SC=\frac{8}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ SC=\frac{8}{3}\sqrt{3} \end{gathered}[/tex]
(c)
The length of the side edge SC is
[tex]SC=\frac{8}{3}\sqrt{3}\text{ cm}[/tex]
