A right triangle has one vertex on the graph of

y = 9 - x^2, x > 0, at (x, y), another at the origin, and the thiurd on the positive x-axis at (x, 0). Express the area A of the triangle as a function of x.

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- Oct 29th 2008, 08:00 AMmagentaritaRight Triangle
A right triangle has one vertex on the graph of

y = 9 - x^2, x > 0, at (x, y), another at the origin, and the thiurd on the positive x-axis at (x, 0). Express the area A of the triangle as a function of x.

- Oct 29th 2008, 08:17 AMHenderson
The base of your triangle is your x-coordinate, and the height of the triangle is your y-coordinate. For a triangle, the area is given by

$\displaystyle A = \frac{1}{2}(base)(height)$, so

$\displaystyle A = \frac{1}{2}(x)(y)$

$\displaystyle A = \frac{1}{2}(x)(9-x^2)$

$\displaystyle

A = \frac{9x - x^3}{2}

$ - Oct 29th 2008, 08:54 AMSoroban
Hello, magentarita!

Henderson is absolutely correct!

Did you make a sketch? . . . It's very simple.

Quote:

A right triangle has one vertex on the graph of $\displaystyle y \:=\: 9 - x^2,\;\;x > 0,\text{ at }(x, y),$

another at the origin, and the third on the positive x-axis at (x, 0).

Express the area $\displaystyle A$ of the triangle as a function of $\displaystyle x.$

Code:`|`

9 **

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| *(x,y)

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The area of a triangle is: .$\displaystyle A \;=\;\tfrac{1}{2}\text{(base)(height)} $

The base is $\displaystyle x.$

The height is $\displaystyle y$, where $\displaystyle y \:=\:9-x^2$

Therefore: .$\displaystyle A \;=\;\tfrac{1}{2}x(9-x^2)$

- Oct 29th 2008, 06:13 PMmagentaritagreat work....
- Oct 29th 2008, 06:14 PMmagentaritaSoroban....