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Find a formula for the described function and state its domain.

Express the area of an equilateral triangle as a function of the length of a side.

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Let the length of a side of the equilateral triangle be $x$. Then by the Pythagorean Theorem, the height $y$ of the triangle satisfies $y^{2}+\left(\frac{1}{2} x\right)^{2}=x^{2},$ so that $y^{2}=x^{2}-\frac{1}{4} x^{2}=\frac{3}{4} x^{2}$ and $y=\frac{\sqrt{3}}{2} x .$ Using the formula for the area $A$ of a triangle, $A=\frac{1}{2}(\text { base })(\text { height }),$ we obtain $A(x)=\frac{1}{2}(x)\left(\frac{\sqrt{4}}{2} x\right)=\frac{\sqrt{3}}{4} x^{2},$ with domain $x>0$

02:32

Jeffrey Payo

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 1

Four Ways to Represent a Function

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

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All right, let's find a function that represents the area of unequal lateral triangle. So suppose the legs measure X X and X for this equilateral triangle, and we know that to get the area of a triangle, you can multiply 1/2 times the base times height. At this point, we do know the bases X, but we don't know the height. So let's go ahead and draw in a height and unequal lateral triangle has a lot of special properties, one of which is that when you break it up into two triangles like this by dropping a perpendicular and drawing the height, you end up with 30 60 90 triangles on the left and the right, and when you have a 30 60 90 triangle, you have a special side relationship. I'll just draw one of them. We have a 30 60 90 triangle. I noticed that the high pot noose would be X since it was already labeled X, and the short leg would be 1/2 of X because we dropped that perpendicular, which cut that other side in half. And then if you use the Pythagorean theorem, you'll find that the long leg works out to be square root three over two times X It's the short leg, times square root three, and you've probably studied 30 60 90 triangles before. So that should be familiar to you now that we know that we can substitute the square root of 3/2 X and for the height in our area formula on weaken substitute X in for the base so we have area equals 1/2 times x times square root 3/2 X and if we clean that up a little bit, 1/2 times square, 3/2 is square, 3/4 and x times X is X squared. So there's our formula for the area of the equilateral triangle as a function of the side length. Now, how about the domain of that? Well, we know that because X represents the side length of a triangle, it has to be positive. So X is greater than zero

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