• know how to use basic formulas for area and circumference of a circle. The area is pr² where r is the radius of the circle. The circumference is found using the formula 2pr or pd, where r is the radius and d is the length of the diameter.
• solve basic linear and quadratic equations.
• use the Pythagorean Theorem in finding the legs or hypotenuse of a right triangle.
• know that the hypotenuse of a right triangle inscribed in a circle is a diameter of the circle, and know that the diagonal of a rectangle inscribed in a circle is a diameter of the circle. These facts are true because the diagonal of an inscribed rectangle is also a hypotenuse of a right triangle (half of the rectangle) and the legs of these triangles intercept half or 180º of the circle.

1. 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 = 2pr and given as C = 5p

 

2pr = 5 p
r = 5/2

 

Knowing that r = 5/2, we can find area using

 

A = pr² = 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.

 

2. Suppose a circle has a circumference of 28p. A right triangle with an altitude of 4x and a base of 7x 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.

 

We were given that the circumference is 28p. We can use the circumference formula C = pd to solve for d.

 

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

 

The strategy to get the area of the shaded region is to subtract the area of the triangle from the area of the circle. The area of the triangle is

 

A_triangle = ½bh = ½(7x)(4x) = 14x² = 14(4)² = 224

 

Since the diameter of the circle is 28, its radius is r = 14. We can now find the area of the circle using A = pr².

 

A_circle = pr² = p(14)² = 196p

 

Therefore the area of the shaded region

 

A_circle – A_triangle = 196p – 224 ≈ 391.75