Awire 5 meters long is cut into two pieces. one piece is bent into a square for a frame for a stained glass ornament, while the other piece is bent into a circle for a tv antenna. to reduce storage space, where should the wire be cut to minimize the total area of both figures? give the length of wire used for each:

Let the circumference of the circle be x. Then the perimeter of the triangle is (5-x). 

The diameter of the circle is x/π, and its radius is x/(2π). Thus, its area is: 
Ac = π(x/2π)² = x²/(4π) 

One side of the triangle is (5-x)/3. Its height is √3/2 times as side. Thus, its area is: 
At = (1/2)(5 - x)(√3/2)(7 - x)/3 
At = (√3/12)(25- 10x + x²) 

Now, to minimize total area, take the derivative and set to 0: 
A = Ac + At 
A' = x/(2π) + (√3/12)(2x - 10) 
A' = x/(2π) + x√3/6 - 5√3/6 
0 = x(√3/6 + 1/(2π)) - 5√3/6 
x = (5√3/6) / (√3/6 + 1/(2π)) 
x ≈ 5.16

That makes the length of the triangle piece 5-x = 2.84 

To maximize area, it must be at one of the endpoints, x=0 or x=5.

x=0: A = (1/2)(5)(√3/2)(5)/3 ≈ 3.61
x=5: A = 5²/(4π) ≈ 1.99

  q

step-by-step explanation:

each line from a vertex to the midpoint of the opposite side is called a median. the point where the medians intersect, point q, is the centroid.

the centroid of δabc is point q.