Question

The area of an equilateral triangle is $16\sqrt{3}c{{m}^{2}}$ then finding the length of each side of that triangle.

Answer 1

Hint: Derive the formula of the area of an equilateral triangle from the formula of area of any triangle with length of side length $a$ and the perpendicular drawn from the opposite vertex $h$ $A=\dfrac{1}{2}\times a\times h$. Put the given area in the formula of the area of an equilateral triangle and obtain a quadratic equation. Solve the equation to find out the length of the side.

Complete step-by-step solution:
We know that the area of a triangle with any side of length $a$ and the perpendicular drawn from the vertex opposite to the side with length $h$ is given by the formula.
\[A=\dfrac{1}{2}\times a\times h\]
 

We have drawn in the given triangle ABC a perpendicular AD which meets BC at point D. We know that in an equilateral triangle all sides are of equal length. So $AB=BC=CA=a$. We also know that the perpendicular drawn from a vertex will bisect the opposite side. So $BD=CD=\dfrac{a}{2}$. \[\]
Now let us observe the triangle ADC which is a right angled triangle with hypotenuse $AC=a$ and side $CD=\dfrac{a}{2}$. Let $\text{AD}=\text{h}.$ We know from Pythagoras theorem the square of hypotenuse is sum of squares of other two side in a right angled triangle. Then we have,
\[ \begin{align}
  & A{{C}^{2}}=C{{D}^{2}}+A{{D}^{2}} \\
 & \Rightarrow {{a}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}+{{h}^{2}} \\
 & \Rightarrow h=\sqrt{{{a}^{2}}-{{\dfrac{a}{4}}^{2}}}=\dfrac{\sqrt{3}}{2}a \\
\end{align} \]
So the area of an equilateral triangle is
\[A=\dfrac{1}{2}\times a\times \left( \dfrac{\sqrt{3}}{2}a \right)=\dfrac{\sqrt{3}}{4}{{a}^{2}}\]
It is given question that are of the equilateral triangle is $16\sqrt{3}c{{m}^{2}}=A$. Putting it in above equation we have,
\[\begin{align}
  & 16\sqrt{3}=\dfrac{\sqrt{3}}{4}{{a}^{2}} \\
 & \Rightarrow 16\times 4={{a}^{2}} \\
 & \Rightarrow {{a}^{2}}=64 \\
\end{align}\]
We solve the above quadratic equation by linear factorisation and get ,
\[\begin{align}
  & \Rightarrow {{a}^{2}}=64 \\
 & \Rightarrow {{a}^{2}}-8\times 8=0 \\
 & \Rightarrow \left( a+8 \right)\left( a-8 \right)=0 \\
\end{align}\]
So the possible values of $a$ is 8 and $-8$. As the length is always positive we reject the solution $a=-8$. So length of the side of the equilateral triangle is 8cm.

Note: It is important to distinguish between equilateral from isosceles triangles which have the same formula for the area as any triangle. It is also to noted that length and area are always positive and we need to reject negative values arising in quadratic equations.