express the perimeter of triangle P as function of area A
Hello, sluggerbroth!
Who asked this question? . . . It is stupid!
Express the perimeter of triangle $\displaystyle P$ as function of area $\displaystyle A$.All these triangles have an area of 6 square units.Code:- - - - * - - - - - - - - - - - - * - - - - - - - - - - - - * - : * * * * * * : : * * * * * * : 3 * * * * * * 3 : * * * * * * : : * * * * * * : - * * * * * - - - * * * * * - - - * * * * * - - - - : - - 4 - - : : - - 4 - - : : - - 4 - - :
But obviously they have different perimeters.
So $\displaystyle P \:=\:f(6)\,$ has at least three possible values.
. . It is not a function.
How about $\displaystyle P=a+b+c$ and $\displaystyle A=\rho \cdot s$ where $\displaystyle \rho$ is the radius of the inscribed circle, and $\displaystyle s=\frac{a+b+c}{2}=\frac{P}{2}$. Now $\displaystyle P=\frac{2 A}{\rho}$. This can be considered a function in A (not just A, but A nevertheless).
This comes from Schaum’s Outlines Precalculus
It is a two part question and reads as follows:
A) Express the area A of equilateral triangle as function of one side s.
B) Express the perimeter of the triangle P as a function of the area A.
The answers listed are:
A) A(s)=(s squared )(square root 3)/4
and
B) P(A)=(6 square root A)/ (fourth root 3)
How to get to these answers???? HELP!!