Question

What‌ ‌is‌ ‌the‌ ‌area‌ ‌of‌ ‌an‌ ‌equilateral‌ ‌triangle‌ ‌with‌ ‌height‌ ‌of‌ ‌9‌ ‌inches?‌ ‌

Answer 1

Hint: First, we will think of a triangle about how it will look and which would be the base of the triangle. Then from base to the top vertex we will draw a line which would be the height of the triangle which we need to use as it's given in the question.

Complete step-by-step solution:
Now let ABC we the triangle and BC be the base of the triangle and now you need to draw a line from vertex A to the base BC which would be perpendicular to the base since it is an equilateral triangle

We are given that height of triangle, h=9 inches.
Since the given triangle is an equilateral triangle, using its properties will make this question easy.
In this triangle height is also the median of the triangle and the height h divides the base BC of the triangle into two halves of side let $\dfrac{a}{2}$ each if sides of triangle are a.
Now we got two triangles which are congruent to each other and Pythagoras theorem hold for one of the right-angled triangles: ${{a}^{2}}={{h}^{2}}+{{\left( \dfrac{a}{2} \right)}^{2}}$
So, $\dfrac{3}{4}{{a}^{2}}={{h}^{2}}$ that is ${{a}^{2}}=\dfrac{4}{3}{{h}^{2}}$ and now the side is given by $a=\dfrac{2\sqrt{3}}{3}h$
And substituting value of h gives
$\begin{align}
  &=\dfrac{2\sqrt{3}}{3}\times 9 \\
 & =6\sqrt{3} \\
\end{align}$
Now, area of the equilateral triangle ABC is
$\begin{align}
 & \dfrac{\sqrt{3}}{4}\times {{\left( a \right)}^{2}} \\
  &= \dfrac{\sqrt{3}}{4}\times {{\left( 6\sqrt{3} \right)}^{2}} \\
 & = 27\sqrt{3} \\
\end{align}$
Therefore, the area of the given triangle ABC is $27\sqrt{3}$ .

Note: Explore all the properties of equilateral, isosceles and scalene triangle in order to apply them and to be clear with the concepts. Apart from this we must know how to do calculations with irrational numbers.