Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises: 59

Answer

$A=\frac{\sqrt{3}}{4}x^2$, $x>0$

Work Step by Step

We have an equilateral triangle with sides $x$ and we need to find the area. We know that: $A=1/2*base*height=\frac{1}{2}xh$ We need to eliminate $h$. We construct a right triangle in the middle of the equilateral triangle with sides $h$, $x$, and $1/2x$. We use the Pythagorean Theorem with these three sides: $(\frac{1}{2}x)^2+h^2=x^2$ $h^2=x^2-(\frac{1}{2}x)^2$ $h=\pm\sqrt{x^2-(\frac{1}{2}x)^2}$ $h=+\sqrt{\frac{3}{4}x^2}=\frac{\sqrt{3}}{2}x$ (We eliminate the negative because lengths must be positive.) We plug in $h$ in the area formula: $A=\frac{1}{2}x\frac{\sqrt{3}}{2}x=\frac{\sqrt{3}}{4}x^2$ The domain is $x>0$ because lengths must be positive.
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.