Bunuel wrote:
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?
A. \(\frac{\sqrt{3}}{64}A^2\)
B. \(\frac{\sqrt{3}}{4}A\)
C. \(\frac{9}{16}A\)
D. \(\frac{3}{4}A\)
E. \(\frac{4\sqrt{3}}{9}A\)
Kudos for a correct solution.
In order to solve this question we should know the area of square (we can plug in a value) and the area of the triangle
Assuming a 3 by 3 square the area of the square would be 9 and the perimeter would be 12- an equilateral triangle has 3 congruent sides so we can divide the perimeter of the square by the number of sides of our triangle which equals 4- a triangle with 3 sides each with a length of 4
Now I learned an equation from Bunuel that helps with relating the area of a square to an equilateral triangle- the area of a equilateral triangle can also calculated using the formula s^2 \sqrt{3}/4- in this formula s represents the number of sides
If we plug in the length of any of the sides of our equilateral triangle then the formula is
4^2 x \sqrt{3}/4
16 x \sqrt{3} /4
16\sqrt{3}/4
4\sqrt{3}
If we plug the area of our square, 9 , into equation D then the result equals the area of our triangle
Thus, the correct answer is (D)