A rectangle is placed around a semicircle. The length of the rectangle is 8 yds. Using Pi = 3.14 find the area of the region outside the semicircle.

Question

Need help finding area if rectangle excluding semicircle. 

Answer 1

I am assuming the semi-circle and the rectangle share a side, such that the diameter of the circle is equal to the length of the rectangle.

 

Also assuming that the curved (semi-circle) part touches the opposite side of the rectangle, so that the width of the rectangle is equal to the radius of the circle. (There really should be a way to submit diagrams with the questions and answers - that would be a nice feature for WyzAnt to implement).

 

So, you know that a diameter is twice the radius. You also know that area of the circle is PI * r², where r is the radius. You know that area of semi-circle is half the area of a full circle. You know that the area of the rectangle is length * width. And now all you have to do is subtract area of the semi-circle from the area of the rectangle. 

Answer 2

Formulas:
Area_semicircle = Πr2/2
Area_rectangle = lw

Given:
Semicircle is engraved in the rectangle
length_rectangle = 8

Assumption:
Diameter of semicircle is equivalent to length of rectangle
diameter_semicircle = length_rectangle
diameter_semicircle = 8

Based on assumption, radius_semicircle is equal to the width of the rectangle.
diameter_semicircle/2 = radius_semicircle
radius_semicircle = 4 = width of rectangle

Solve:
Area_rectangle = l*w = 4*8 = 32
Area_semicircle = Πr2/2 = Π(4)2/2 = 8Π = 25.12

Area_rectangle - Area_semicircle = 32 - 25.12 = 6.88

Answer 3

The rectangle has length 8 yds and width 4 yds because the radius of the semicircle(half circle) is the same as the width of the rectangle.

The area of the rectangle is A=lw or A=8*4=32 sq yds.

The area of the whole circle is A=pi*r^2

A=3.14159*4^2

A=3.14159*16

A=50.2654 sq yds

The area of the semicircle is 50.2654/2=25.1327 sq yds

The area of the region outside the semicircle is the area of the rectangle minus the area of the semicircle.

A=32-25.1327=6.8673 sq yds