A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 30 feet?
47.7
To find the area of the largest possible Norman window, we need to maximize the area of the semicircle and the rectangle.
Let's denote the width of the rectangle as 'w' and the radius of the semicircle as 'r'.
Since the diameter of the semicircle is equal to the width of the rectangle, we have:
2r = w
To find the perimeter, we need to consider the circumference of the semicircle and the four sides of the rectangle:
Perimeter (P) = 1/2 * circumference of semicircle + 2 * length + 2 * width
Since the circumference of a circle is given by 2πr, the perimeter equation becomes:
P = πr + 2(2r) + 2r
Given that the perimeter is 30 feet, we can write the equation as:
30 = πr + 6r
Simplifying the equation:
30 = (π + 6) r
r = 30 / (π + 6)
To find the width and length of the rectangle, we use the relation between the radius and the width:
w = 2r
Substituting the value of r, we get:
w = 2 * (30 / (π + 6))
The length of the rectangle is equal to the diameter of the semicircle, which is 2r.
l = 2r = 2 * (30 / (π + 6))
Finally, we can calculate the area of the largest possible Norman window:
Area = area of semicircle + area of rectangle
The area of the semicircle is given by 1/2 * π * r^2:
Semicircle area = 1/2 * π * (30 / (π + 6))^2
The area of the rectangle is given by length multiplied by width:
Rectangle area = 2 * (30 / (π + 6)) * (30 / (π + 6))
Adding the areas of the semicircle and rectangle together:
Area = 1/2 * π * (30 / (π + 6))^2 + 2 * (30 / (π + 6)) * (30 / (π + 6))
Now, you can calculate the exact value of the area using the above equation.