Question

One side of an equilateral triangle is 30 cm. The area of the triangle is
(a) 225 sq. cm
(b) \[225\sqrt{3}\] sq. cm
(c) \[225\sqrt{2}\] sq. cm
(d) 112.5 sq. cm

Answer 1

Hint: We solve this problem by using the general formula of the area of the triangle.
The formula of area of the triangle is given as
\[A=\dfrac{1}{2}\left( base \right)\left( height \right)\]
We find the height of the triangle by using the condition that the height of the equilateral triangle cuts the base into two equal parts and Pythagoras theorem.
The Pythagoras Theorem states that the square of the hypotenuse is equal to the sum of squares of other two sides that is for the triangle shown below

The Pythagoras theorem is given as\[{{b}^{2}}={{a}^{2}}+{{c}^{2}}\].

Complete step by step answer:
We are given that the side of equilateral triangle is 30 cm.
Let us take a rough figure of an equilateral triangle along with height as follows

We know that the condition that the height of the equilateral triangle cuts the base into two equal parts
By using the above condition to the figure then we get
\[\begin{align}
  & \Rightarrow DC=\dfrac{BC}{2} \\
 & \Rightarrow DC=\dfrac{30}{2} \\
 & \Rightarrow DC=15 \\
\end{align}\]
Now, let us consider the triangle \[\Delta ADC\]
We know that the Pythagoras Theorem states that the square of the hypotenuse is equal to the sum of squares of the other two sides that is for the triangle shown below

The Pythagoras theorem is given as\[{{b}^{2}}={{a}^{2}}+{{c}^{2}}\].
By using the Pythagoras theorem to triangle \[\Delta ADC\] we get
\[\begin{align}
  & \Rightarrow A{{C}^{2}}=A{{D}^{2}}+D{{C}^{2}} \\
 & \Rightarrow A{{D}^{2}}=A{{C}^{2}}-D{{C}^{2}} \\
\end{align}\]
Now, by substituting the required values in above equation then we get
\[\begin{align}
  & \Rightarrow A{{D}^{2}}={{\left( 30 \right)}^{2}}-{{\left( 15 \right)}^{2}} \\
 & \Rightarrow A{{D}^{2}}={{\left( 15 \right)}^{2}}\left( {{2}^{2}}-1 \right) \\
 & \Rightarrow A{{D}^{2}}={{\left( 15 \right)}^{2}}\left( 3 \right) \\
\end{align}\]
By applying the square root on both sides then we get
\[\Rightarrow AD=15\sqrt{3}\]
Now, let us find the area of the triangle \[\Delta ABC\]
We know that the formula of area of the triangle is given as
\[A=\dfrac{1}{2}\left( base \right)\left( height \right)\]
By using the above formula to triangle \[\Delta ABC\] then we get
\[\begin{align}
  & \Rightarrow A=\dfrac{1}{2}\left( BC \right)\left( AD \right) \\
 & \Rightarrow A=\dfrac{1}{2}\left( 30 \right)\left( 15\sqrt{3} \right) \\
 & \Rightarrow A=225\sqrt{3} \\
\end{align}\]
Therefore we can conclude that the area of the given equilateral triangle as \[225\sqrt{3}\] sq. cm
So, option (b) is correct answer.

Note:
 We can solve this problem in a shortcut method.
We have the formula that the area of equilateral triangle of side \[a\] is given as
\[A=\dfrac{\sqrt{3}}{4}{{a}^{2}}\]
By using the above formula to given triangle then we get
\[\begin{align}
  & \Rightarrow A=\dfrac{\sqrt{3}}{4}{{\left( 30 \right)}^{2}} \\
 & \Rightarrow A=\dfrac{\sqrt{3}}{4}\left( 900 \right) \\
 & \Rightarrow A=225\sqrt{3} \\
\end{align}\]
Therefore we can conclude that the area of the given equilateral triangle as \[225\sqrt{3}\] sq. cm
So, option (b) is correct answer.