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

Question

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

bahaistag · Accepted Answer

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, $2 x + b = 8$
Subtract b from both sides,
$2 x = 8 - b$
Divide by 2,
$x = \frac{8 - b}{2}$
Let h be the length of the altitude to the base.
$h = \sqrt{x^{2} - {(\frac{b}{2})}^{2}}$
Substitute $x = \frac{8 - b}{2}$ in $h = \sqrt{x^{2} - {(\frac{b}{2})}^{2}}$
$h = \sqrt{{(\frac{8 - b}{2})}^{2} - {(\frac{b}{2})}^{2}}$
$= \sqrt{\frac{64 - 16 b + b^{2} - b^{2}}{4}}$
$= \sqrt{\frac{64 - 16 b}{4}}$
$= \sqrt{\frac{16 (4 - b)}{4}}$
$= \sqrt{4 (4 - b)}$
$= 2 \sqrt{4 - b}$
Therefore, the area A of the isosceles triangle is given by
$A = \frac{1}{2} b h$
$A (b) = \frac{1}{2} b (2 \sqrt{4 - b})$
$= b \sqrt{4 - b}, 0 < b < 4$
Thus, the area A of the isosceles triangle modeled by the function
$A (b) = b \sqrt{4 - b}, 0 < b < 4$ .
Final statement:
The area of the isosceles triangle modeled by the function
$A (b) = b \sqrt{4 - b}, 0 < b < 4$ .

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

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