Question

Find the total area the regular pyramid.

Answer 1

Answer:

The area of the regular pyramid is 16√3 square units.

Step-by-step explanation:

The area of regular pyramid is the sum of base area and the area of all sides.

From the given figure it is clear that it is a regular pyramid with side length 4 units. All sides and base are equilateral triangles of side 4 units.

The area of an equilateral triangle is

$A=\frac{\sqrt{3}}{4}a^2$

Where, a is the side length.

Substitute a=4 to find the area of base is

$Base=\frac{\sqrt{3}}{4}(4)^2=\frac{16\sqrt{3}}{4}=4\sqrt{3}$

The given figure contain 4 equilateral triangles of side 4 units. So the area of  regular pyramid is

$A=4\times Base=4\times 4\sqrt{3}=16\sqrt{3}$

Therefore the area of the regular pyramid is 16√3 square units.

Answer 2

Answer: $14\sqrt{3}$

Step-by-step explanation:

1. You must apply the following formula for calculate  the total surface area the regular pyramid:

$SA=\frac{pl}{2}+B$

Where p is the perimeter of the base, l is the slant height and B is the area of the base.

2. The perimeter is, where s is the side length:

$p=3s=3*4=12$

3. The faces are isosceles traingles. So, to find the slant height you can divide one base into two equal right triangles and calculate the slant height with Pythagorean Theorem:

$l=\sqrt{4^2-2^2}=2\sqrt{3}$

4. The base is equal to any face, then both have equal areas. As previously you divide one base into two equal righ triangle to find the slant height, you can calcualte the area of one right triangle and multply it by 2 to calculate B:

$B=2*\frac{bh}{2}=2*\frac{2\sqrt{3}}{2}=2\sqrt{3}$

Finally substitute values:

$SA=\frac{12*2\sqrt{3}}{2}+2\sqrt{3}=14\sqrt{3}$