A square and a equilateral triangle are to be formed out of the same piece of wire. The wire is 6 inches long. What is the maximum of the total area the square and the triangle?

Question

Ella Armstrong · Accepted Answer

Area Square #=s^2#
Area Equilateral Triangle #=sqrt(3)/4xxs^2# For the square, #s=6/4 = 1.5# inches Area Square #=1.5^2=2.25# sq. in. For the triangle, #s=6/3=2# inches Area triangle #=sqrt(3)/4xx(2)^2=sqrt3~~1.7# sq. in. So, the square has the greater area . Hope that helped

Theodore Abad · Answer

undefined To maximize the total area of the square and the equilateral triangle formed from the same piece of wire, we need to allocate the wire in such a way that it optimizes the area of both shapes.

For the square, since all four sides are equal, we can divide the wire into 4 equal parts, each forming one side of the square. Therefore, each side of the square would be $ \frac{6}{4} = 1.5 $ inches long.

For the equilateral triangle, all three sides are equal in length. Given that the wire's total length is 6 inches, we can allocate the entire length of the wire to form the three sides of the equilateral triangle.

Now, to find the maximum total area, we calculate the areas of the square and the equilateral triangle:

1. Area of the square: $ (\text{side})^2 = (1.5)^2 = 2.25 $ square inches.
2. Area of the equilateral triangle: We know that the perimeter of the equilateral triangle (the sum of its three sides) is equal to the total length of the wire, which is 6 inches. Therefore, each side of the equilateral triangle would be $ \frac{6}{3} = 2 $ inches long (since all sides are equal).

To find the area of the equilateral triangle, we use the formula $ \frac{\sqrt{3}}{4} \times (\text{side})^2 $, where the side represents the length of one side of the equilateral triangle. So, the area of the equilateral triangle would be:

$$ \frac{\sqrt{3}}{4} \times (2)^2 = \frac{\sqrt{3}}{4} \times 4 = \sqrt{3} $$ square inches.

To maximize the total area, we sum the areas of the square and the equilateral triangle:

$$ 2.25 + \sqrt{3} $$ square inches.

Therefore, the maximum total area that can be obtained from the square and the equilateral triangle formed from the same piece of wire is $ 2.25 + \sqrt{3} $ square inches.