Question

The figure below is an equilateral triangle with sides of length 6. What is the area of the triangle?

A. 12
B. 18
C. 36
D. ${\text{9}}\sqrt {\text{3}} $

Answer 1

Hint: As the given triangle is equilateral, we can use the concept of trigonometry to find the area of the equilateral triangle, we use the formula of \[{\text{sinC}}\]where \[\angle {\text{C = 60$^\circ$ }}\] to get the length of altitude, then we use the formula of area of triangle i.e. ${\text{Area = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ $\times$ base $\times$ height}}$, and hence we get the required answer.

Complete step by step answer:

We are given an equilateral triangle ABC with side length 6 units. We are also given an altitude of the triangle AX.

Consider the right-angled triangle AXC. We can use trigonometry to get,
${\text{sinC = }}\dfrac{{{\text{AX}}}}{{{\text{AC}}}}$
As ABC is a right triangle, \[\angle {\text{C = 60$^\circ$ }}\] and AC= 6, we get
${\text{sin60 = }}\dfrac{{{\text{AX}}}}{{\text{6}}}$
After rearranging, we get
${\text{AX = 6 $\times$ sin60}}$
We use the value of \[{\text{sin6}}{{\text{0}}^{\text{o}}}{\text{ = }}\dfrac{{\sqrt {\text{3}} }}{{\text{2}}}\], we get,
\[{\text{AX = }}\dfrac{{6\sqrt {\text{3}} }}{{\text{2}}}\]
Using the altitude and base in the equation to find the area, we get,
$
  {\text{Area = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ $\times$ AX $\times$ BC}} \\
  {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ $\times$ }}\dfrac{{{\text{6}}\sqrt {\text{3}} }}{{\text{2}}}{\text{ $\times$ 6}} \\
 $ On simplification, we get,
${\text{Area = 9}}\sqrt {\text{3}} $
Therefore, the area of the triangle is ${\text{9}}\sqrt {\text{3}} $ square units
Hence, the correct answer is option D.

Note: The concept of trigonometry is used in this problem. Trigonometric ratios and their values at common angles must be known. We can also find the area using the direct equation ${\text{A = }}\dfrac{{\sqrt {\text{3}} }}{{\text{4}}}{{\text{a}}^{\text{2}}}$, where a is the side of equilateral triangle.
An alternate method to find the height of the triangle is by using Pythagoras theorem.
Consider the right-angled triangle AXC,
By Pythagoras theorem,
$A{X^2} + X{C^2} = A{C^2}$
We are given $AC = 6$ and AX is the height h.
We know that the altitude of an equilateral triangle bisects the side. So, we can write $XC = \dfrac{a}{2} = \dfrac{6}{2} = 3$
Substituting these values in Pythagoras theorem, we get,
${h^2} + {3^2} = {6^2}$
$ \Rightarrow {h^2} + 9 = 36$
$
   \Rightarrow {h^2} = 36 - 9 \\
   \Rightarrow {h^2} = 27 \\
 $
Taking the positive square root, we get.
$h = \sqrt {27} = \sqrt {3 \times 3 \times 3} $
$ \Rightarrow h = 3\sqrt 3 $
Using the altitude and base in the equation to find the area, we get,
$
  {\text{Area = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ $\times$ h $\times$ a}} \\
  {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ $\times$ 3}}\sqrt {\text{3}} {\text{ $\times$ 6}} \\
 $
On simplification, we get,
${\text{Area = 9}}\sqrt {\text{3}} $