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 45 feet?

Answer 1

To solve this problem, let's break it down step by step.

Let ( r ) be the radius of the semicircle, which is also the width of the rectangle.

The perimeter of this figure (Norman window) is given as 45 feet:

The perimeter, ( P ), is the sum of the parts: the semicircle's circumference and the perimeter of the rectangle.

The semicircle's circumference: ( \pi r )

The perimeter of the rectangle: ( 2r + 2r = 4r )

So, the total perimeter equation is:

[ \pi r + 4r = 45 ]

This simplifies to:

[ \pi r + 4r = 45 ]

[ (\pi + 4) r = 45 ]

[ r = \frac{45}{\pi + 4} ]

Now, we need to find the total area of the figure.

The area of the semicircle is:

[ \frac{\pi r^2}{2} ]

The area of the rectangle is:

[ r \times 2r = 2r^2 ]

The total area, ( A ), is the sum of these two parts:

[ A = \frac{\pi r^2}{2} + 2r^2 ]

Substitute the value of ( r ) derived earlier:

[ A = \frac{\pi (\frac{45}{\pi + 4})^2}{2} + 2(\frac{45}{\pi + 4})^2 ]

Calculating this would give the area of the largest possible Norman window with a perimeter of 45 feet.