Answer:
Option B is correct.i.e.,
Step-by-step explanation:
Given: Pyramid with equilateral triangle as base
Length of side of equilateral triangle = s unit
to find: height of equilateral triangle
Here we use a property of equilateral triangle.
Perpendicular from a vertex on a side and median of that side of a triangle is same in equilateral triangle.
All heights are of equal length. So, we just need to find one height or length of 1 altitude.
Figure of base triangle is attached
In Δ ABC
AB = BC = AC = s unit
AD is height
BD = \(\frac{BC}{2}\:=\:\frac{s}{2}\:units\)
Now, In Δ ABD
using pythagoras theorem
BD² + AD² = AB²
\((\frac{s}{2})^2+AD^2=s^2\)
\(\frac{s^2}{4}+AD^2=s^2\)
\(AD^2=s^2-\frac{s^2}{4}\)
\(AD^2=\frac{4s^2-s^2}{4}\)
\(AD^2=\frac{3s^2}{4}\)
\(AD=\sqrt{\frac{3s^2}{4}}\)
\(AD=\frac{\sqrt{3}s}{2}\)
\(AD=\frac{s}{2}\sqrt{3}\)
Therefore, Option B is correct.i.e., \(\frac{s}{2}\sqrt{3}\)
