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 …

Question

asked Jun 25, 2021 103k views

1 Answer

Nathaniel Saxe · Answer 1 · 2021-07-01T04:41:16+0000

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 :

$6a = 5 \\\\a = (5)/(6)\ m$

Area of triangle of side a :

$Area = a^2 \\\\Area = ((5)/(6))^2\\\\Area = 0.694\ m^2$

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

3s = 5

s=(5)/(3)

$Area = (\sqrt3)/(4)s^2\\\\Area = (\sqrt3)/(4)* ((5)/(3))^2\\\\Area =1.20 \ m^2$

Hence, this is the required solution.