#a)
Since the net of the regular pyramid is formed from 4 equilateral triangles, then the surface area of the pyramid is 4 x the area of 1 triangle
The rule of the area of the equilateral triangle is
[tex]A=\frac{\sqrt{3}}{4}s^2[/tex]
Where s is the length of the side of the triangle
Since the length of the side of the triangle is 20 cm, then
s = 20
Substitute the value of s in the rule above
[tex]\begin{gathered} A=\frac{\sqrt{3}}{4}\times(20)^2 \\ \\ A=\frac{\sqrt{3}}{4}\times400 \\ \\ A=100\sqrt{3}\text{ cm}^2 \end{gathered}[/tex]
Now to find the surface area of the pyramid multiply the area of 1 triangle by 4
[tex]\begin{gathered} S.A=4\times100\sqrt{3} \\ S.A=400\sqrt{3} \\ S.A=692.820323\text{ cm}^2 \end{gathered}[/tex]
Round it to the nearest square m (whole number)
S.A = 693 cm^2
You need 693 square cm to make the pyramid
