## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises - Page 56: 78

#### Answer

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

#### Work Step by Step

We can draw a perpendicular line from the top of the triangle to the base of the triangle that makes two $30^{\circ}-60^{\circ}$ triangles. We know that the lengths of the sides of these triangles have the ratio $1, \sqrt{3},$ and $2$. In this case, the sides are $\frac{s}{2}, \frac{\sqrt{3}~s}{2},$ and $s$. The area of one of these triangles is $\frac{1}{2}bh$. $A = \frac{1}{2}bh$ $A = \frac{1}{2}(\frac{s}{2})(\frac{\sqrt{3}~s}{2})$ $A = \frac{\sqrt{3}s^2}{8}$ Since there are two $30^{\circ}-60^{\circ}$ triangles, the total area is $2\times \frac{\sqrt{3}~s^2}{8}$ which is $\frac{\sqrt{3}~s^2}{4}$ The area of this equilateral triangle is $\frac{\sqrt{3}~s^2}{4}$

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