Question

A wire when bent in the form of a square encloses an area of $484\,{m^2}$. Find the largest area enclosed by the same wire when bent to form an equilateral triangle.
A) $372.58\,sq.\,m$
B) $337.17\,sq.\,m$
C) $462.13\,sq.\,m$
D) $274.13\,sq.\,m$

Answer 1

Hint: We are given the area of the square from the area, evaluate the side of the square. Since the wire is then bent to an equilateral triangle therefore, the perimeter of the square is equal to the perimeter of an equilateral triangle. Evaluate the side of an equilateral triangle and use the formula of area.

Complete step-by-step answer:
We are given that wire when bent in the form of a square encloses an area of $484\,{m^2}$.
We have to evaluate the largest area enclosed by the same wire when bent to form an equilateral triangle.
First, we evaluate the side of the square.
We know that area of the square of side $a$ is ${a^2}$.
Therefore,
${a^2} = 484$
Take the square root on both sides.
\[
  \sqrt {{a^2}} = \sqrt {484} \\
  a = 22\,m \\
 \]
Therefore, the side of the given square is $22\,m$.
We know that the perimeter of the square is the sum of all the sides.
A square consists of four sides therefore, the perimeter is $4a$.
Let the perimeter of the square be $P$.
Therefore,
$
  P = 4 \times 22 \\
  P = 88\,m \\
 $
Since the wire is now bent to an equilateral triangle therefore, the perimeter of the square is equal to the perimeter of an equilateral triangle.
Let the side of an equilateral triangle is $b\,$.
The perimeter is again the sum of all the sides.
Therefore, the perimeter of an equilateral triangle is $3b$.
According to our question,
$
  3b = 88 \\
  b = \dfrac{{88}}{3}\,m \\
 $
We know that the area of an equilateral triangle of side $x$is $\ dfrac{{\sqrt 3 }}{4}{x^2}$.
Let the area of the required equilateral triangle is $A$.
Therefore,
$
  A = \dfrac{{\sqrt 3 }}{4} \times {\left( {\dfrac{{88}}{3}} \right)^2} \\
  A = \dfrac{{\sqrt 3 }}{4} \times \dfrac{{88}}{3} \times \dfrac{{88}}{3} \\
  A = 372.58\,sq.\,m \\
 $
Hence, the largest area enclosed is $372.58\,sq.\,m$
Therefore, option (A) is correct.
Note: We can evaluate the area of the triangle by another formula which is given below:
$A = \sqrt {s(s - a)(s - b)(s - c)} $
where $s = \dfrac{{a + b + c}}{2}$ and $a,b,c$ are the length of the sides of the triangle.
This formula is called Heron’s formula.