Question

A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. Find the maximum and minimum possible area that can be enclosed by the wire.

Answer 1

Given :

A piece of wire 10 m long is cut into two pieces.

To Find :

One piece is bent into a square and the other is bent into an equilateral triangle.

Solution :

So, perimeter of triangle is 5 m :

$6a = 5 \\\\a = \dfrac{5}{6}\ m$

Area of triangle of side a :

$Area = a^2 \\\\Area = (\dfrac{5}{6})^2\\\\Area = 0.694\ m^2$

Now. we know maximum area of a given perimeter is for equilateral triangle.

3s = 5

$s=\dfrac{5}{3}$

$Area = \dfrac{\sqrt3}{4}s^2\\\\Area = \dfrac{\sqrt3}{4}\times (\dfrac{5}{3})^2\\\\Area =1.20 \ m^2$

Hence, this is the required solution.