Question

Find the total surface area of a tetrahedron whose base is an equilateral triangle of edge a and whose lateral edges are each equal to b. show your solutions. illustrate if possible.

Answer 1

Answer:

$\frac{a^2\sqrt[]{3}+3a\sqrt[]{4b^2-a^2}}{4}$

Explanation:

The diagram of the tetrahedron showing the given dimensions is attached below.

The tetrahedron has 4 faces:

• One equilateral triangle with side lengths a.

,

• Three Isosceles triangles with side lengths b-b-a.

First, we find the area of the base.

$\begin{gathered} \text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2\text{ where s=side length} \\ \implies\text{Area of the base}=\frac{\sqrt[]{3}}{4}a^2\text{ units squared.} \end{gathered}$

Next, we begin finding the area of one isosceles triangle.

First, find the perpendicular height, h using the Pythagorean Theorem.

$\begin{gathered} b^2=(\frac{a}{2})^2+h^2 \\ \implies h^2=b^2-\frac{a^2}{4}\implies h^{}=\sqrt{\frac{4b^2-a^2}{4}} \\ \implies h^{}=\frac{\sqrt[]{4b^2-a^2}}{2} \end{gathered}$

Thus, the area of one isosceles triangle is:

$A=\frac{1}{2}\times a\times\frac{\sqrt{4b^2-a^2}}{2}=\frac{a\sqrt[]{4b^2-a^2}}{4}$

Finally, the area of the tetrahedron will be:

$\begin{gathered} \text{TSA}=\frac{a^2\sqrt[]{3}}{4}+\frac{3a\sqrt[]{4b^2-a^2}}{4} \\ =\frac{a^2\sqrt[]{3}+3a\sqrt[]{4b^2-a^2}}{4} \end{gathered}$