Question

If x is the length of a median of an equilateral triangle, then the area in terms of x is:
[a] ${{x}^{2}}$
[b] $\dfrac{{{x}^{2}}\sqrt{3}}{2}$
[c] $\dfrac{{{x}^{2}}\sqrt{3}}{3}$
[d] $\dfrac{{{x}^{2}}}{2}$

Answer 1

Hint: Use the property that the median of an equilateral triangle is also an altitude of the triangle. Hence use Pythagoras theorem to find the length of each side of the equilateral triangle in terms of x.
Hence find the area of the triangle using the area of a triangle $=\dfrac{1}{2}base\times height$

Complete step-by-step answer:

Given an equilateral triangle ABC. CE is the median of the triangle. CE = x
To determine The area of the triangle in terms of x.
Since the median of an equilateral triangle is also an altitude of the triangle, we have
$\text{CE}\bot \text{AB}$.
We know that in a right-angled triangle, the sum of the squares of legs is equal to the square of the hypotenuse. This is known as Pythagoras Theorem.
Using Pythagoras Theorem in triangle ACE, we have
$\text{A}{{\text{E}}^{2}}+\text{C}{{\text{E}}^{2}}=\text{A}{{\text{C}}^{2}}\text{ (i)}$
Now since E is the midpoint of AB, we have $\text{AE=}\dfrac{1}{2}\text{AB}$
Also, since ABC is an equilateral triangle, we have $\text{AC=AB}$
Hence $\text{AE=}\dfrac{1}{2}\text{AC}$
Substituting the value of AE and AC in equation (i), we get
\[\begin{align}
  & {{\left( \dfrac{\text{AC}}{2} \right)}^{2}}+{{x}^{2}}={{\left( \text{AC} \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+\dfrac{\text{A}{{\text{C}}^{2}}}{4}=\text{A}{{\text{C}}^{2}} \\
 & \Rightarrow {{x}^{2}}=\dfrac{3}{4}\text{A}{{\text{C}}^{2}} \\
 & \Rightarrow \dfrac{4{{x}^{2}}}{3}=\text{A}{{\text{C}}^{2}} \\
\end{align}\]
Taking square roots on both sides, we get
$\dfrac{2x}{\sqrt{3}}=\text{AC}$
Hence, we have AB = AC = BC $=\dfrac{2x}{\sqrt{3}}$
Hence area of triangle ABC $=\dfrac{1}{2}\text{AB}\times \text{CE=}\dfrac{1}{2}\times \dfrac{2x}{\sqrt{3}}\times x=\dfrac{{{x}^{2}}}{\sqrt{3}}$
Rationalising the denominator by multiplying the numerator and the denominator by $\sqrt{3}$, we get
area of triangle ABC =$\dfrac{{{x}^{2}}}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{3}{{x}^{2}}}{3}$
Hence option [c] is correct.

Note: In an equilateral triangle, the median is also an altitude.
Consider an equilateral triangle ABC, with median AE.
Claim : $\text{AE}\bot \text{AB}$
Proof:

In triangle ACE and triangle BCE, we have
AE = BE , since E is the midpoint of AB.
AC = BC since ABC is an equilateral triangle.
CE = CE (common side)
Hence $\Delta \text{AEC}\cong \text{BEC}$ by S.S.S congruencey criterion.
Hence $\angle \text{AEC=}\angle \text{BEC}$ (corresponding parts of congruent triangles).
Now we know that $\angle \text{AEC+}\angle \text{BEC=180}{}^\circ $ by linear pair axioms.
Hence, we have
$\begin{align}
  & 2\angle \text{AEC=180}{}^\circ \\
 & \Rightarrow \angle \text{AEC=90}{}^\circ \\
 & \Rightarrow \text{AE}\bot \text{AB} \\
\end{align}$
Hence proved.