Question

A piece of wire is bent so as to form the boundary of a square with area A. If the wire is then bent into the shape of an equilateral triangle, what will be the area of the triangle thus bounded in terms of A?

Answer 1

Answer:

The area of the triangule is 0.77A

Step-by-step explanation:

Given the area, we can calculate one side of the square, the area of a square is height*width, but in this case both are the same so we can say that the area is $height^{2}$, so that means one side of the square is: $\sqrt{A}$.

If then we have to bend the wire to form a equilateral triangule, all the sides must be equal, so one side of the triangule is $[tex]\sqrt{A} *\frac{4}{3}$[/tex]. We have to multiply $\sqrt{A} *4$ in this case to  obtain the total lenght of the wire.

According to the formula of the equilateral triangule:

$A=\frac{\sqrt{3} }{4} *s^{2}$

where S is one side, The area will be:

$Area=\frac{\sqrt{3} }{4} *(\sqrt{A}*\frac{4}{3} )^{2}=0.77A$