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

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

Step-by-step 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}=(√(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=((a)/(2))^2+h^2 \\ \implies h^2=b^2-(a^2)/(4)\implies h^{}=\sqrt{(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=(1)/(2)* a*(√(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}`