An annulus is a ring bounded by two concentric circles. In annulus A
has an inner circle of radius 2 cm and an outer circle of radius 6 cm. Annulus B has an outer circle of radius 9 cm. If annulus B has an area equal to that of
annulus A, what is the radius of the inner circle of annulus B?
Area A = pi(36-4)
= pi (81 -r^2)
so
32 = 81 - r^2
seven
To find the radius of the inner circle of annulus B, we need to calculate the area of annulus A and then set it equal to the area of annulus B.
The area of annulus A can be found by subtracting the area of the inner circle from the area of the outer circle.
Area of annulus A = Area of outer circle of A - Area of inner circle of A
Let's calculate this:
Area of outer circle of A = π * (radius of outer circle of A)^2
= π * (6 cm)^2
Area of inner circle of A = π * (radius of inner circle of A)^2
= π * (2 cm)^2
Area of annulus A = π * (6 cm)^2 - π * (2 cm)^2
= π * (36 cm^2 - 4 cm^2)
= π * 32 cm^2
Now, we know that the area of annulus B is equal to the area of annulus A. So:
Area of annulus B = π * (radius of outer circle of B)^2 - π * (radius of inner circle of B)^2
We are given that the radius of the outer circle of B is 9 cm. Substituting this into the equation:
π * (9 cm)^2 - π * (radius of inner circle of B)^2 = π * 32 cm^2
Simplifying the equation:
81π - π * (radius of inner circle of B)^2 = 32π
Subtracting 81π from both sides:
- π * (radius of inner circle of B)^2 = 32π - 81π
= -49π
Dividing both sides by -π:
(radius of inner circle of B)^2 = 49
Taking the square root of both sides:
radius of inner circle of B = √49
= 7 cm
Therefore, the radius of the inner circle of annulus B is 7 cm.
To find the radius of the inner circle of annulus B, we need to compare the areas of the two annuli.
The area of an annulus can be calculated by subtracting the area of the inner circle from the area of the outer circle.
Let's start with annulus A:
The area of annulus A is given by:
Area(A) = Area(outer circle A) - Area(inner circle A)
The area of a circle can be calculated using the formula:
Area = π * r^2
For annulus A:
Area(A) = π * (6^2) - π * (2^2)
Area(A) = π * 36 - π * 4
Area(A) = π * 32
Now, let's move on to annulus B:
Since annulus B has an outer circle of radius 9 cm, we can calculate its area:
Area(outer circle B) = π * (9^2) = π * 81
We know that the area of annulus B is equal to that of annulus A, so:
Area(A) = Area(B)
Thus,
π * 32 = π * 81 - Area(inner circle B)
To find the radius of the inner circle of annulus B, we need to find its area.
π * 32 = π * 81 - π * (radius of inner circle B)^2
Now, rearrange the equation and solve for the radius of the inner circle of annulus B:
π * (radius of inner circle B)^2 = π * 81 - π * 32
π * (radius of inner circle B)^2 = π * (81 - 32)
(radius of inner circle B)^2 = 81 - 32
(radius of inner circle B)^2 = 49
Taking the square root of both sides:
radius of inner circle B = √49 = 7 cm
Therefore, the radius of the inner circle of annulus B is 7 cm.