Consider a window the shape of which is a rectangle of height?

Area of rectangle is wh
area of triangle is (1/2)wT
total area is wh + (1/2)wT
T = 0.8w
A = wh + (1/2)w(0.8w)
A = wh + 0.4w²

perimeter P = 2h + w + 2x
where x is the angled side of the top
if you divide the top section via a vertical line, you have two right triangles.
x² = T² + (w/2)²
T = 0.8w
x² = 0.64w² + (w/2)² = 1.14w²
x = 1.0677w

P = 2h + 3.135416w
to minimize the P, we need to get P in terms of one variable.
h = (A/w) – (0.4w²/w)
h = (A/w) – (0.4w)
P = 2h + 3.135416w
P = 2((A/w) – (0.4w)) + 3.135416w
P = (2A/w) – (0.8w) + 3.135416w
P = (2A/w) + 2.335416w

now finally we can differentiate P and set equal to 0 to find min
assuming A is fixed
P' = 0 = 2.335416 – 2A/w²
solve for w
2A/w² = 2.335416
w² = 2A/2.335416 = A(0.8563786)
w = 0.9254073√A

h you can get from h = (A/w) – (0.4w)

.

The cross sectional area A will be equal to area of rectangle + area of triangle.

A = w*h*(.5*w*T)
= .5*h*T*w*w
= .5hT(w)^2

Thus,

h = 2A/T(w)^2

and w = sqrt(2A/hT)