Question

a normal window has the shape of a rectangle surmounted by a semicircle. if the perimeter of the window is 34 ft, express the area A of the window as a function of the width x of the window

Answer 1

The 34 feet perimeter and width, x, of the window which is in the shape of a rectangle surmounted by a semicircle is, A = 17•x - x²•(1/2 - π/8)

How can the area of the window be expressed as a function of x?

The shape of the window = A rectangle surmounted by a semicircle

Perimeter of the window, P = 34 feet

Width of the window = x

Required; The area, A, of the window as a function of x

Solution:

Diameter of the semicircle = x

Length of the semicircular arc = π•x/2

Let y represent the height of the window, we have;

P = 2•y + x + π•x/2 = 34

Therefore;

y = (34 - (x + π•x/2))/2 = 17 - x•(1 + π/2)/2

Area of the window, A = x × y + π•x²/8

Which gives;

A = x × (17 - x•(1 + π/2)/2) + π•x²/8 = 17•x - x²/2 - x²•π/8

A = 17•x - x²/2 - x²•π/8 = 17•x - x²•(1/2 - π/8)

Therefore;

Window area, A = 17•x - x²•(1/2 - π/8)

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