Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter - Section 11.4 - Applications with Acute Triangles - Exercises - Page 529: 46

Answer

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

Work Step by Step

Theorem 11.4.1: The area of an acute triangle equals one-half the product of the lengths of two sides and the sine of the included angle. In an equilateral triangle, each side has the same length of $s$ and each angle has the same measure of $60^{\circ}$. We can find the area: $A = \frac{1}{2}~s \times s \times sin(\theta)$ $A = \frac{1}{2}~s^2 sin(\theta)$ $A = \frac{1}{2}~s^2 sin(60^{\circ})$ $A = \frac{1}{2}~s^2 ~(\frac{\sqrt{3}}{2})$ $A = \frac{s^2}{4}~\sqrt{3}$
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