Question

The figure below shows part of a stained-glass window depicting the rising sun. Which function can be used to find the area of the region outside the semicircle but inside the rectangle?

Answer 1

That dude is right

Explanation:

Answer 2

`A(w) = w^2 + 5w - (1)/(8)\pi w^2`

Explanation:

A = the area of the region outside the semicircle but inside the rectangle

w = the width of the rectangle or diameter of the semicircle

Since "A" is determined by "w", therefore, "A" is a function of "w" = A(w).

A(w) = (area of rectangle) - (area of semicircle)

`A(w) = (l*w) - ((1)/(2) \pi r^2)`

Where,

lenght of rectangle (l) = w + 5

width of rectangle (w) = w

r = ½*w =
`(w)/(2)`

Plug in the values:

`A(w) = ((w + 5)*w) - ((1)/(2) \pi ((w)/(2))^2)`

`A(w) = ((w + 5)*w) - ((1)/(2) \pi ((w)/(2))^2)`

Simplify

`A(w) = (w^2 + 5w) - ((1)/(2) \pi ((w^2)/(4))`

`A(w) = w^2 + 5w - (1)/(2)*\pi*(w^2)/(4)* \pi`

`A(w) = w^2 + 5w - (1*\pi*w^2)/(2*4)`

`A(w) = w^2 + 5w - (1*\pi w^2)/(8)`

`A(w) = w^2 + 5w - (1)/(8)\pi w^2`