Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.1 - Functions and Their Graphs - Exercises 1.1 - Page 11: 9


$a(b)=\frac{\sqrt 3}{4}b^2$ a - area b - base

Work Step by Step

We'll use $a$ for the area, $b$ for the base and $h$ for the height. Because of the properties of an equilateral triangle, we know that $h$ intersects the base in the middle. This means $h^2+(\frac{b}{2})^2=b^2$. From that we get $h=\frac{\sqrt 3}{2}b$. Then, we obtain: $a=\frac{1}{2}bh=\frac{1}{2}b\frac{\sqrt 3}{2}b=\frac{\sqrt 3}{4}b^2$
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