Question

a piece of wire 6x cm long is bent into an equilateral triangle. for which values of x will the area be numerically less than the perimeter?

Answer 1

We need expressions for the area and the perimeter. The perimeter is the sum of the length of the three equal length sides or \[P=3(6x)=18x\]The area of a triangle is (1/2)base*height. The base is easy (=6x). The height is slightly trickier. Use pythagorean theorem \[(6x)^2 = (6x/2)^2 + height^2\]Solving, \\[height=\sqrt{36x^2-9x^2}=\sqrt{9*3x^2}=(3\sqrt{3})x\]So the area of this triangle is:\[A=(1/2)base*height=(1/2)(6x)(3\sqrt{3})x=9\sqrt{3}x^2\]Now we just need to setup an inequality between the expressions for area and the perimeter of the triangle. We want the area to be numerically less than the perimeter. That is:\[A <P\]Inputting our expressions for area and perimeter we have:\[(9\sqrt{3})x^2<18x\]Solving for x: \[x<(18/9\sqrt{3})\]Or,\[x<2/\sqrt{3}\]Usually we don't like roots in the denominator so after rationalizing the denominator we get our final answer:\[x<2\sqrt{3}/3\]The the area be numerically less than the perimeter when the value of x is less than about 1.15cm.

Answer 2

I just realized I screwed up in inputting the numbers of this one. The method is good but the length of each side of the equilateral triangle will be 6x/3=2x NOT 6x as I wrote above.