Question

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $6\sqrt{2}cm$, then the area of the triangle is
A. $16\sqrt{2}c{{m}^{2}}$
B. $16\sqrt{3}c{{m}^{2}}$
C. $12\sqrt{2}c{{m}^{2}}$
D. $12\sqrt{3}c{{m}^{2}}$

Answer 1

Hint: First consider the square of side $a$ and find its value by using the Pythagoras theorem. Now equate its perimeter to that of the equilateral triangle. Use the formula $\dfrac{\sqrt{3}}{4}{{\left( side\text{ }of\text{ }triangle \right)}^{2}}$ to get the area of the equilateral triangle.

Complete step-by-step answer:

We have been given a square and an equilateral triangle of equal perimeters. The diagonal of the square is given as $6\sqrt{2}cm$. We have to find the area of the triangle. Let us consider the square of side $a$.

We have been given that the length of the diagonal of this square is $6\sqrt{2}cm$. We know that each angle of a square is the right angle. So, we can apply Pythagoras theorem in triangle $ABC$, and we get, ${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}$ .

By substituting the values of $AB=BC=a$ and the value of $AC=6\sqrt{2}$, we get,

$\begin{align}

  & {{a}^{2}}+{{a}^{2}}={{\left( 6\sqrt{2} \right)}^{2}} \\

 & \Rightarrow 2{{a}^{2}}=72\Rightarrow {{a}^{2}}=\dfrac{72}{2}=36 \\

\end{align}$

So, we get $a=\sqrt{36}=6cm$. We know that the perimeter of any polygon is equal to the sum of the length of its sides. So, we get, perimeter of square ${{P}_{S}}=4\left( length\text{ }of\text{ }side \right)\Rightarrow {{P}_{S}}=4a$. By substituting the value of $a=6$ , we get, ${{P}_{S}}=4\times 6=24cm\ldots \ldots \ldots \left( i \right)$.

Now, let us consider an equilateral triangle of sides $b$.

We know that the perimeter of a triangle ${{P}_{T}}=3\left( length\text{ }of\text{ }side \right)\Rightarrow {{P}_{T}}=3b\ldots \ldots \ldots \left( ii \right)$.

We have been given that the perimeter of the square is equal to the perimeter of the equilateral triangle, so we get, ${{P}_{S}}={{P}_{T}}$.

By substituting values of ${{P}_{S}},{{P}_{T}}$ from equation (i) and (ii) in the above equation, we get, $24=3b$. By dividing both the sides by 3, we get, $b=\dfrac{24}{3}\Rightarrow b=8cm$. So, we get the length of the side of the equilateral triangle as $8cm$.

Now, we know that the area of the equilateral triangle is $\dfrac{\sqrt{3}}{4}{{\left( side \right)}^{2}}$. So, we get the area of the given equilateral triangle $=\dfrac{\sqrt{3}}{4}{{\left( b \right)}^{2}}=\dfrac{\sqrt{3}}{4}{{\left( 8 \right)}^{2}}=16\sqrt{3}c{{m}^{2}}$. So, the area of the triangle is $16\sqrt{3}c{{m}^{2}}$. Hence option B is the correct answer.


Note: In this question, many students make mistakes by assuming the length of the sides of the square and triangle are the same, which is wrong because then the perimeter will not be equal, which is already mentioned in the question. It is better to remember the area of equilateral triangle as $\dfrac{\sqrt{3}}{4}{{\left( side\text{ }of\text{ }triangle \right)}^{2}}$ instead of $\dfrac{1}{2}\times base\times height$ as it makes it easy to solve the question.