In order to solve problems which require application of the

area and

perimeter (circumference) for circles, it is necessary to

A typical problem involving the

area and

circumference of a

circle gives us the area,

circumference and/or lengths of the

radius or diameter. We may also be given a relationship between the

area and

perimeter of other figures inscribed in the circle. We need to calculate some of these quantities given information about the others. Two examples of this type of problem follow:

- Suppose the circumference of a circle is 5p. What are the radius and area of this circle?

The

circumference is represented by the formula C = 2

pr and given as C = 5

p

2pr = 5 p

r = 5/2

Knowing that r = 5/2, we can find

area using

A = pr^{2} = p(5/2)^{2} = 25p/4

We rely on the formulas for

area and circumference. Sometimes we don’t know the

radius or

diameter when starting the problem. The

radius and

diameter are the key measurements in any circle.

- Suppose a circle has a circumference of 28p. A right triangle with an altitude of 4
*x* and a base of 7*x* is inscribed in this circle as shown in the diagram below. What are the diameter of this circle and the area of the shaded region?

Always make sure that you have a diagram in situations like this where another figure is inscribed in a circle. In our diagram, observe that the

base *b* of the

triangle is also a

diameter of the circle. We will use the fact that the

diameter d = 7x to help solve the problem.

pd = 28 p

d = 28

Earlier we learned that the

diameter of this

circle could be represented as 7x since it is the

hypotenuse of an inscribed right triangle.

7x = 28

x = 4

A_{triangle} = ½bh = ½(7x)(4x) = 14x^{2} = 14(4)^{2} = 224

A_{circle} = pr^{2} = p(14)^{2} = 196p

Therefore the

area of the shaded region

A_{circle} – A_{triangle} = 196p – 224 ≈ 391.75