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?

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Answer 1 · 2015-12-24T00:14:42

Area Square $= s^{2}$ $=s^2$
Area Equilateral Triangle $= \frac{\sqrt{3}}{4} \times s^{2}$ $=sqrt(3)/4xxs^2$

For the square, $s = \frac{6}{4} = 1.5$ $s=6/4 = 1.5$ inches

Area Square $= {1.5}^{2} = 2.25$ $=1.5^2=2.25$ sq. in.

For the triangle, $s = \frac{6}{3} = 2$ $s=6/3=2$ inches

Area triangle $= \frac{\sqrt{3}}{4} \times {(2)}^{2} = \sqrt{3} \approx 1.7$ $=sqrt(3)/4xx(2)^2=sqrt3~~1.7$ sq. in.

So, the square has the greater area .

Hope that helped