A 1 m. long piece of wire is cut into 2 pieces one is bent into a circle, one is bent into a triangle.

Let A1 represent the area of the circle and let x represent the length of the
piece of wire used to make the circle. Find an equation giving the area A1 of the
circle as a function of the length x. What is the domain of this function?

Let A represent the sum of the area of the circle and the area of the triangle.
Write A as a function of x. What is the domain of A?

length of wire = one meter
length of circle = x
length of triangle = (1-x)
P=2π R = (x)
radius of circle R =(x)/ 2π
area of circle
A1 = π R² = π ( (x)²/ (2π)²
A1= (x)² / 4π
A1= ( 1/4π)( x)²
since it is quadratic, domain is all real numbers.

A= A1+A2

A2 area of triangle = 1/2 base height
base = (1-x) /3 = perimeter /3 sides
height use pythagorian for the right triangle which is 1/2 of the wired triangle.
hypotenues =(1-x)/3
base = (1-x) /3/2 = (1-x)/6
h = √[((1-x)/3)² - ((1-x)/6)²]
plug h and base into area formula for the triangle. simplify
A =( 1/4π)( x)² +(1/2)[ (1-x)/6]*√[((1-x)/3)² - ((1-x)/6)²]