Question

In an equilateral triangle of side $3\sqrt{3}$, find the length of the altitude.

Answer 1

Hint:In order to solve this question, we should know that the area of an equilateral triangle can be calculated by the formula, $\dfrac{\sqrt{3}}{4}{{a}^{2}}$, if we have been given side length, as a represents the length of the side. We also know that the area of a triangle can be calculated as $\dfrac{1}{2}\times base\times height$. So, we will equate the area by calculating it using both the formulas to get the length of the altitude.

Complete step-by-step answer:
In this question, we have been given an equilateral triangle of side $3\sqrt{3}$ and we have been asked to find the length of the altitude of the triangle. Let us first draw the figure as per the conditions given in the question. So, we get the figure as below,

Here, we have represented triangle ABC as an equilateral triangle and AD as the altitude of the triangle ABC on side BC. Now, from the conventional formula, we know that area of a triangle can be calculated as,
Area of triangle = $\dfrac{1}{2}\times base\times height$
Now, we know that BC is the base and AD is the height, So, we can write it as,
Area of triangle = $\dfrac{1}{2}\times BC\times AD$
Now, we have been given that the side of the equilateral triangle is $3\sqrt{3}$. So, we can write,
Area of triangle = $\dfrac{1}{2}\times 3\sqrt{3}\times AD=\dfrac{3\sqrt{3}}{2}AD\ldots \ldots \ldots \left( i \right)$
We also know that the area of an equilateral triangle is given as $\dfrac{\sqrt{3}}{4}{{a}^{2}}$, where a is the side. So, we can write the area of the triangle for side BC as,
Area of triangle = $\dfrac{\sqrt{3}}{4}{{\left( BC \right)}^{2}}$
Now, we will put the value of BC, that is $3\sqrt{3}$. So, we get,
Area of triangle = $\dfrac{\sqrt{3}}{4}{{\left( 3\sqrt{3} \right)}^{2}}$
And we can further write it as,
Area of triangle = $\dfrac{\sqrt{3}}{4}\times 27=\dfrac{27\sqrt{3}}{4}\ldots \ldots \ldots \left( ii \right)$
As, we have been talking about the same triangle in both the formulas, we can equate both the values of the areas from equation (i) and (ii). Therefore, we get,
Area of triangle = $\dfrac{3\sqrt{3}}{2}\left( AD \right)=\dfrac{27\sqrt{3}}{4}$
Now, we will simplify it to get the value of AD. So, we will write AD on one side and all the rest terms on the other side. So, we get,
$AD=\dfrac{27\sqrt{3}}{4}\times \dfrac{2}{3\sqrt{3}}$
We know that the common terms of the numerator and denominator gets cancelled out, so we get,
$AD=\dfrac{9}{2}$
Hence, we get the length of the altitude as $\dfrac{9}{2}$.

Note: We can solve this question by applying the Pythagoras theorem in triangle ADC, where AD and DC are the perpendicular and the base respectively and AC is the hypotenuse. SO, we will get the value of AD, because AC is the side of the triangle, that is $3\sqrt{3}$ and DC is the half of side BC, because altitude of right angled triangle bisects the side, so we can say $DC=\dfrac{3\sqrt{3}}{2}$.