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: 9

Answer

$a(b)=\frac{\sqrt 3}{4}b^2$ 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: $a=\frac{1}{2}bh=\frac{1}{2}b\frac{\sqrt 3}{2}b=\frac{\sqrt 3}{4}b^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.