# An isosceles triangle has its base along the x-axis with one base vertex at the origin and its vertex in the first quadrant on the graph of y = 6-x^2. How do you write the area of the triangle as a function of length of the base?

Jan 16, 2017

$\text{Area, expressed as the function of base-length "l," is } \frac{1}{8} l \left(24 - {l}^{2}\right)$.

#### Explanation:

We consider an Isosceles Triangle $\Delta A B C$ with base vertices

$B \left(0 , 0\right) \mathmr{and} C \left(l , 0\right) , \left(l > 0\right) \in \text{ the X-axis=} \left\{\left(x , 0\right) | x \in \mathbb{R}\right\}$

and the third vertex

A in G={(x,y)|y=6-x^2; x,y in RR} sub Q_I......(star).

Obviously, the length $B C$ of the base is $l$.

Let $M$ be the mid-point of the base $B C . \therefore M \left(\frac{l}{2} , 0\right)$.

$\Delta A B C \text{ is isosceles, } \therefore A M \bot B C .$ So, if $h$ is the height of

$\Delta A B C , \text{ then,} \because , B C$ is the X-Axis, $A M = h .$

Clearly, $A = A \left(\frac{l}{2} , h\right) .$

Now, the Area of $\Delta A B C = \frac{1}{2} l h \ldots \ldots \ldots \ldots \ldots \ldots . \left(\ast\right)$

But, $A \left(\frac{l}{2} , h\right) \in G \therefore \left(\star\right) \Rightarrow h = 6 - {l}^{2} / 4$

Therefore, the Area of $\Delta A B C = \frac{1}{2} l \left(6 - {l}^{2} / 4\right) \ldots \left[\because , \left(\ast\right)\right] ,$ or,

$\text{Area, expressed as the function of base-length "l," is } \frac{1}{8} l \left(24 - {l}^{2}\right)$.