# To find: A function that models area of an isoceles triangle.

To find:
A function that models area of an isoceles triangle.
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The model we want is a function that gives the area of an isosceles triangle.
Let x be the length of two equal sides of an isosceles triangle and b is given the length of its base.
Because of the perimeter of the triangle is the sum of all sides and it is 8.
Therefore, $2x+b=8$
Subtract b from both sides,
$2x=8-b$
Divide by 2,
$x=\frac{8-b}{2}$
Let h be the length of the altitude to the base.
$h=\sqrt{{x}^{2}-{\left(\frac{b}{2}\right)}^{2}}$
Substitute $x=\frac{8-b}{2}$ in $h=\sqrt{{x}^{2}-{\left(\frac{b}{2}\right)}^{2}}$
$h=\sqrt{{\left(\frac{8-b}{2}\right)}^{2}-{\left(\frac{b}{2}\right)}^{2}}$
$=\sqrt{\frac{64-16b+{b}^{2}-{b}^{2}}{4}}$
$=\sqrt{\frac{64-16b}{4}}$
$=\sqrt{\frac{16\left(4-b\right)}{4}}$
$=\sqrt{4\left(4-b\right)}$
$=2\sqrt{4-b}$
Therefore, the area A of the isosceles triangle is given by
$A=\frac{1}{2}bh$
$A\left(b\right)=\frac{1}{2}b\left(2\sqrt{4-b}\right)$
$=b\sqrt{4-b},0
Thus, the area A of the isosceles triangle modeled by the function
$A\left(b\right)=b\sqrt{4-b},0.
Final statement:
The area of the isosceles triangle modeled by the function
$A\left(b\right)=b\sqrt{4-b},0.