Answer to Question #168254 in Geometry for AYRIAH

Question

Find the total area of a regular triangular pyramid if each side of the base is 6 m. and the slant height forms an angle of 30º with the base.

Answer 1

A pyramid whose base is a regular polygon and congruent isosceles triangles as lateral faces. In a regular pyramid the axis is perpendicular to the base.

Thus in a regular pyramid th axis and the altitude are identical.

The slant height of a regular pyramid is the length of the median through the apex of any lateral face. In the Fig DM is the slant height.

Given the regular triangular pyramid. Its base is equilateral triangle.

If $a$ is the side of the polygon base, $h$ is the height, $l$ is the slant height of a regular pyramid and the slant height forms an angle of $\alpha$ with the base, then 

$$AB=AC=BC=a, DO=h, DM=l,$$

$$\angle DMO=\alpha$$

Consider the right triangle DMO

$$OM=\dfrac{a}{2\sqrt{3}}, l=\dfrac{OM}{\cos\alpha}=\dfrac{a}{2\sqrt{3}\cos\alpha}$$

$$S_{CDB}=\dfrac{1}{2}DM\cdot BC=\dfrac{a^2}{4\sqrt{3}\cos\alpha}$$

$$S_{ABC}=\dfrac{a^2\sqrt{3}}{4}$$

The total area of a regular triangular pyramid

$$S=S_{ADB}+S_{ADC}+S_{BDC}+S_{ABC}$$

$$=3\cdot\dfrac{a^2}{4\sqrt{3}\cos\alpha}+\dfrac{a^2\sqrt{3}}{4}$$

$$=\dfrac{a^2\sqrt{3}}{4}(\dfrac{1}{\cos\alpha}+1)$$

$a=6\ m, \alpha=30\degree$

$$S=\dfrac{6^2\sqrt{3}}{4}(\dfrac{1}{\cos30\degree}+1)$$

$$=9\sqrt{3}(\dfrac{2}{\sqrt{3}}+1)=9(2+\sqrt{3})\ (m^2)$$

The total area of the pyramid is $9(2+\sqrt{3})$ m².