Question

Among the following shapes of equal perimeter, which one has the largest area? how?

a A square b An equilateral triangle

c A circle d A regular pentagon

Answer 1

Here we can take any positive integer for proof:
Taking the perimeter of each given figure as 12 cm.
a) Perimeter of square = 12
Then, each side of perimeter = $\frac{12}{4}$ = 3 cm
So, area of square = ${\left(3\right)}^2=9c{m}^2$
b) Perimeter of an equilateral triangle = 12 cm
Then, each side of equilateral triangle = $\frac{12}{3}$ = 4 cm
So, area of an equilateral triangle = $\frac{\sqrt{3}}{4}\times 4^2=\frac{\sqrt{3}}{4}\times 16=4\sqrt{3}{\mathrm{cm}}^2=4\times 1.73=6.92{\mathrm{cm}}^2$
c) Perimeter of circle = 12 cm
Then, radius of circle = $\frac{12}{2\mathrm{\pi }}$ cm
So, area of circle = $\mathrm{\pi }\times {\left(\frac{12}{2\mathrm{\pi }}\right)}^2=\mathrm{\pi }\times \frac{36}{{\mathrm{\pi }}^2}=\frac{36}{\mathrm{\pi }}=36\times \frac{7}{22}=11.45{\mathrm{cm}}^2$
d) Perimeter of regular pentagon = 12 cm
Then, each side of pentagon = $\frac{12}{5}$ cm
Radius of circumscribed circle = $\frac{1}{2}\times lengthofeachside\times \cos ec\frac{180°}{numberofsides}=\frac{1}{2}\times \frac{12}{5}\times \cos ec36°=\frac{1}{2}\times \frac{12}{5}\times 1.71=2.052$cm
So, area of regular pentagon = $\frac{1}{2}\times 5\times {\left(2.052\right)}^2\times \sin \frac{360°}{5}=\frac{1}{2}\times 5\times {\left(2.052\right)}^2\times 0.95=10.00042\approx 10$ sq.cm.

Therefore, from all of the above we can see the area of circle is the largest.
Hence, For equal perimeter circle has the largest area among the given shapes.