Question

If ABCD is a regular tetrahedron with the length of edge as $l$,then
(1) Volume of the tetrahedron is $${\displaystyle \frac{l^3}{6\sqrt{2}}}$$
(2) Volume of the tetrahedron is $${\displaystyle \frac{l^3}{6\sqrt{3}}}$$
(3) The minimum distance of any vertex from the opposite face is $${\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}l$$
(4) The minimum distance of any vertex from the opposite face is $${\displaystyle \frac{\sqrt{3}}{\sqrt{2}}}l$$

Answer 1

Hint: This question is based on the basic concepts of triangles. The formula required for solving this question is that the area of the triangle is equal to, ${\displaystyle \frac{1}{2}}ab\sin (\theta )$, where $\theta$ is the angle enclosed within the two sides of the triangle. The volume of the tetrahedron is ${\displaystyle \frac{1}{3}}\times B\times H$, where ${\displaystyle \frac{\sqrt{3}}{4}}{a}^2$ is the base area and $H$ is the height.
Also, knowing the basics of the equilateral triangle would help, like the area of the triangle is ${\displaystyle \frac{\sqrt{3}}{4}}{a}^2$ and the length of the median is ${\displaystyle \frac{\sqrt{3}}{2}}a$.

Complete step-by-step answer:
Now, we have a regular tetrahedron, that means we have all the sides equal.
Thus, the base of the tetrahedron is nothing but an equilateral triangle. Now, for calculating the volume of the figure we need to calculate ${\displaystyle \frac{1}{3}}\times B\times H$.
Now, let’s calculate the base area,
Base area = ${\displaystyle \frac{1}{2}}ab\sin (\theta )$
We know that ABC is an equilateral triangle.
$={\displaystyle \frac{1}{2}}{l}^2\sin \left({\displaystyle \frac{\pi }{3}}\right)$
Substituting the value of $\sin \left({\displaystyle \frac{\pi }{3}}\right)$ as $\left({\displaystyle \frac{1}{2}}\right)$.
$={\displaystyle \frac{\sqrt{3}}{4}}{l}^{{}^2}$
Now, we need to calculate the height and for that let’s draw a perpendicular from the other vertex to the base triangle.

As the figure is symmetric, so the foot of the perpendicular coincides with the centroid of the base triangle as shown in the figure.
The perpendicular DE is drawn to the base of triangle ABC. Let’s join B to E and observe the triangle DBE.
DB =$l$
$\mathrm{\angle }DEB={\displaystyle \frac{\pi }{2}}$,
$$BE={\displaystyle \frac{2}{3}}$$ of the median drawn from B to the side AC.
The length of the Median is ${\displaystyle \frac{\sqrt{3}}{2}}l$. (From the concepts of equilateral triangle)
Therefore, the length of BE = ${\displaystyle \frac{\sqrt{3}}{2}}l\times {\displaystyle \frac{2}{3}}$
$={\displaystyle \frac{l}{\sqrt{3}}}$.
Now, let’s calculate the value of the perpendicular DE, by using the Pythagoras theorem,
Therefore, $(DE{)}^2+(EB{)}^2=(DB{)}^2$
$(DE{)}^2=(DB{)}^2-(EB{)}^2$
Substituting the values of DB and EB, we get,
$$(DE{)}^2=l^2-{\displaystyle \frac{l^2}{3}}$$
Simplifying the expression, we get,
$$(DE{)}^2={\displaystyle \frac{2l^2}{3}}$$
Taking square root on both sides of the equation, we get,
$$DE={\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}l$$
DE or the length of the perpendicular is ${\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}l$.
Now, the volume $V$ of the tetrahedron is
$V={\displaystyle \frac{1}{3}}\times B\times H$
$\Rightarrow V={\displaystyle \frac{1}{3}}\times \left({\displaystyle \frac{\sqrt{3}}{4}}{l}^2\right)\times \left({\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}l\right)$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow V={\displaystyle \frac{l^3}{6\sqrt{2}}}$

Note: The question can get tedious if the calculations and geometrical mistakes are not taken in account. Please be careful while drawing the figure, keep it neat and mark the angles and length of the sides correctly. Use simplification rules to ease the arithmetic operations. One should remember the formulae for volume of the tetrahedron.