SOLUTION: A wire is bent into a square enclosing an Area of 48cm^2. If it is to be reshaped as an equilateral triangle, how much araea will be lost?

Algebra ->  Polygons -> SOLUTION: A wire is bent into a square enclosing an Area of 48cm^2. If it is to be reshaped as an equilateral triangle, how much araea will be lost?      Log On


   

Question 924758: A wire is bent into a square enclosing an Area of 48cm^2. If it is to be reshaped as an equilateral triangle, how much araea will be lost?
Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
4%2Asqrt%2848%29 is the perimeter, or the length of the wire.

4%2Asqrt%283%2A4%5E2%29
4%2A4%2Asqrt%283%29
highlight_green%2816%2Asqrt%283%29%29, the wire length.

The equilateral triangle formed with that length of wire is of side length and base length %2816%2F3%29sqrt%283%29. This is also then the hypotenuse of a special 30-60-90 triangle. You can consider %281%2F2%29%2816%2F3%29sqrt%283%29 to be the short leg of a 30-60-90 triangle; The height of the equilateral triangle, being the other leg of the special 30-60-90, can be called y.

This y is found, using Pythagorean Theorem, y%5E2%2B%28%288%2F3%29sqrt%283%29%292=%28%2816%2F3%29sqrt%283%29%29. Find the area of the equilateral triangle from base and height y.

Continuing for y,
y%5E2%2B%288%2F3%29%5E2%2A3=%2816%2F3%29%5E2%2A3
y%5E2=3%2816%2F3%29%5E2-3%288%2F3%29%5E2
y%5E2=%2816%29%5E2%2F3-64%2F3
y%5E2=192%2F3
y%5E2=%283%2A64%29%2F3
y%5E2=64
highlight_green%28y=8%29

The area OF the equilateral triangle here is
%281%2F2%29%28%2816%2F3%29sqrt%283%29%29%288%29, which is half of base*times*height.
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4%2816%2F3%29sqrt%283%29
highlight_green%28%2864%2F3%29sqrt%283%29%29
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The question asks for the difference or how much area is lost in forming the triangle, so you want to find highlight%2848-%2864%2F3%29sqrt%283%29%29.