Question

A right triangle has one vertex on the graph of $y=9-x^2,x>0$, at $\left(x,y\right)$, another at the origin, and the third on the positive x-axis at $\left(x,0\right)$. Express the area Aof the triangle as a function of x.

Short answer

The area of the given triangle as a function is$A=\frac{1}{2}\times x\times \left(9-x^2\right)$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

One vertex is at $\left(x,y\right)$.

Second vertex is at $\left(0,0\right)$.

Third vertex is at $\left(x,0\right)$.

We know the area of the triangle is localid="1645800398027" $A=\frac{1}{2}\times b\times h......\left(i\right)$.

Distance between points $\left(x,y\right),\left(0,0\right)$, which is the base of the triangle,

$$\begin{gathered} b=\sqrt{{\left(x-0\right)}^2+{\left(0-0\right)}^2} \\ =\sqrt{x^2} \\ =x \end{gathered}$$

Step 2. Find the height.

Consider the given question,

Distance between points $\left(x,0\right),\left(x,y\right)$, which is the height of the triangle,

localid="1645800355147" role="math" $$\begin{gathered} h=\sqrt{{\left(x-x\right)}^2+{\left(y-0\right)}^2} \\ =\sqrt{0^2+y^2} \\ =y \\ =9-x^2 \end{gathered}$$

Substitute the values in equation (i),

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