# Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 360: 25

$\frac{s^{2}}{4}\sqrt 3$

#### Work Step by Step

Using Heron's formula if three sides of a triangle have length a,b,c then area A of a triangle is A = $\sqrt s(s-a)(s-b)(s-c)$ where s = $\frac{a+b+c}{2}$ Lets take a=b=c =s s = $\frac{s+s+s}{2}$ = $\frac{3s}{2}$ A = $\sqrt \frac{3s}{2}(\frac{3s}{2}-s)(\frac{3s}{2}-s)(\frac{3s}{2}-s)$ =$\sqrt \frac{3s}{2} (\frac{3s-2s}{2})^{3}$ =$\sqrt \frac{3s}{2} (\frac{s}{2})^{3}$ =$\sqrt \frac{3}{16}s^{4}$ =$\frac{s^{2}}{4}\sqrt 3$

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