# Thread: Right Triangle

1. ## Right 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.

2. 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

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

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

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

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

3. Hello, magentarita!

Henderson is absolutely correct!
Did you make a sketch? . . . It's very simple.

A right triangle has one vertex on the graph of $y \:=\: 9 - x^2,\;\;x > 0,\text{ at }(x, y),$
another at the origin, and the third on the positive x-axis at (x, 0).
Express the area $A$ of the triangle as a function of $x.$
Code:
        |
9 **
|    *   P
|       *(x,y)
|     *:| *
|   *:::|  *
| *:::::|
- + - - - + - * - -
O       Q   3

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

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

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

4. ## great work....

Originally Posted by Henderson
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

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

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

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

$
A = \frac{9x - x^3}{2}
$
I thank you very much.

5. ## Soroban....

Originally Posted by Soroban
Hello, magentarita!

Henderson is absolutely correct!
Did you make a sketch? . . . It's very simple.

Code:
        |
9 **
|    *   P
|       *(x,y)
|     *:| *
|   *:::|  *
| *:::::|
- + - - - + - * - -
O       Q   3
The area of a triangle is: . $A \;=\;\tfrac{1}{2}\text{(base)(height)}$

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

Therefore: . $A \;=\;\tfrac{1}{2}x(9-x^2)$
It's always great to receive an answer from you.