SOLUTION: Hi Need some help in solving this beauty. Circle A is shaded and lies inside circle B. The radius of B is 4 inches more than the radius of A. What is the radius of each if the un

Question 1163042: Hi
Need some help in solving this beauty. Circle A is shaded and lies inside circle B. The radius of B is 4 inches more than the radius of A. What is the radius of each if the unshaded area is.
Twice as much as the shaded area
28pi square inches more than the shaded area
16pi square inches more than twice as much as the shaded area.
Thanks
Found 3 solutions by Alan3354, Theo, MathTherapy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Circle A is shaded and lies inside circle B. The radius of B is 4 inches more than the radius of A. What is the radius of each if the unstated [sic] unshaded area is.
Twice as much as the shaded area
---------------
a = radius of smaller circle
Area1 = pi*a^2
b = radius of larger circle = a+4
Area2 = pi*(a+4)^2
Unshaded area = Area2 - Area1 = 2*Area1
---> Area2 = 3*Area1
pi*(a+4)^2 = 3pi*a^2
a^2 + 8a + 4 = 3a^2
2a^2 - 8a - 4 = 0
a^2 - 4a - 2 = 0
Solve for a
etc

28pi square inches more than the shaded area
16pi square inches more than twice as much as the shaded area.
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
this is not as horrible as you might think, at least conceptually.
what is horrible is the arithmetic required.
fortunately, i was able to use some mechanized tools to help with that.
i used a quadratic equation solve that can be found at https://www.mathsisfun.com/quadratic-equation-solver.html
i used an online scientific calculator that can be found at http://www.alcula.com/calculators/scientific-calculator/#gsc.tab=0
the use of these tools, especially the online scientific calculator, does require some knowledge and experience in how to use them, so it's not as straight forward as you might think, but they did provide the results of calculations i required without me having to do those calculations manually.

A = pi * r^2
B = pi * (r + 4)^2

A is the shaded area.
B - A is the unshaded area *****

B - A = pi * (r + 4)^2 - pi * r^2
factor out the pi to get:
B - A = pi * ((r + 4)^2 - r^2
simplify to get:
B - A = pi * (r^2 + 8r + 16 - r^2)
simplify further to get:
B - A = pi * (8r + 16) *****

the unshaded area is therefore equal to pi * (8r + 16)

problem 1:
What is the radius of each if the unshaded area is twice as much as the shaded area?

the unshaded area is equal to pi * (8r + 16)
the shaded area is equal to pi * r^2
if the unshaded area is equal to twice the shaded area, you get:
pi * (8r + 16) = 2 * pi * r^2
divide both sides of this equation by pi to get:
8r + 16 = 2r^2
subtract the left side of the equation from both sides of the equation to get:
0 = 2r^2 - 8r - 16
divide both sides of this equation by 2 to get:
0 = r^2 - 4r - 8
factor this quadratic equation to get:
r = 5.4641016151378 or r = -1.4641016151378
since r can't be negative, you get:
r = 5.4641016151378
when r = 5.4641016151378, .....
A = pi * r^2 = 93.796667199
B = pi * (r+4)^2 = 281.390001597
B - A = 187.593334398
2 * A = 187.593334398
B - A is equal to 2 * A.
your solution is that the radius of A is equal to 5.4641016151378 inches and the radius of B is equal to that plus 4 = 9.4641016151378 inches

problem 2:
What is the radius of each if the unshaded area is 28 * pi square inches more than the shaded area?

the unshaded area is equal to pi * (8r + 16)
the shaded area is equal to pi * r^2
if the unshaded area is equal to 28 * pi square inches more than the shaded area, then you get:
pi * (8r + 16) = pi * r^2 + 28 * pi
factor out the pi on the right side of the equation to get:
pi * (8r + 16) = pi * (r^2 + 28)
divide both sides of the equation by pi to get:
8r + 16 = r^2 + 28
subtract the left side of the equation from both sides of the equation to get:
0 = r^2 + 28 - 8r - 16
combine like terms to get:
0 = r^2 - 8r + 12
solve this quadratic equation to get:
r = 2 or r = 6
when r = 2, ..................
A = pi * r^2 = 12.566370614
B = pi * (r + 4)^2 = 113.097335529
B - A = 100.530964915
A + 28 * pi = 100.530964915
since B - A is the unshaded area and A is the shaded area, then you get:
the unshaded area is equal to the shaded area + 28 * pi.
your solution is that the radius of A is equal to 2 inches and the radius of B is equal to that plus 4 = 6 inches.
when r = 6,........................
A = pi * r^2 = 113.097335529
B = pi * (r + 4)^2 = 314.159265359
B - A = 201.06192983
A + 28 * pi = 201.06192983
since B - A is the unshaded area and A is the shaded area, then you get:
the unshaded area is equal to the shaded area + 28 * pi.
your solution is that the radius of A is equal to 6 inches and the radius of B is equal to that plus 4 = 10 inches.

problem 3:
What is the radius of each if the unshaded area is 16pi square inches more than twice as much as the shaded area?

the unshaded area is equal to pi * (8r + 16)
the shaded area is equal to pi * r^2
if the unshaded area is equal to 16 * pi square inches more than twice the shaded area, then you get:
pi * (8r + 16) = 2 * pi * r^2 + 16 * pi
factor out the pi from the right side of the equation to get:
pi * (8r + 16) = pi * (2 * r^2 + 16)
divide both sides of the equation by pi to get:
8r + 16 = 2 * r^2 + 16
subtract 8r + 16 from both sides of the equation to get:
0 = 2 * r^2 + 16 - 8r - 16
combine like terms to get:
0 = 2 * r^2 - 8r
divide both sides of the equation by 2 to get:
0 = r^2 - 4r
solve this quadratic equation to get:
r = 0 or r = 4
since r can't be equal to 0, then you get:
r = 4
when r = 4, ......................
A = pi * r^2 = 50.265482457
B = pi * (r + 4)^2 = 201.06192983
B - A = 150.796447372
2 * A + 16 * pi = 150.796447372
since B - A is the unshaded area and A is the shaded area, then you get:
the unshaded area is equal to twice the shaded area + 16 * pi.
your solution is that the radius of A is equal to 4 inches and the radius of B is equal to that plus 4 = 8 inches.

to summarize your solutions:

the radius of the larger circle is always 4 inches more than the radius of the smaller circle.
the shaded area is the area of the smaller circle.
the unshaded area is the area of the larger circle minus the area of the smaller circle.

when the radius of the smaller circle is equal to 5.4641016151378 inches, then the unshaded area is equal to twice the shaded area.

when the radius of the smaller circle is equal to 2 inches or 6 inches, then the unshaded area is equal to the shaded area plus 28 * pi square inches.

when the radius of the smaller circle is equal to 4 inches, then the unshaded area is equal to the twice the shaded area + 16 * pi square inches.

i also solved the problem using excel.
the solution from excel are shown below:

you are right.
this was definitely a bear, mainly because the calculations required were horrendous.

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Hi
Need some help in solving this beauty. Circle A is shaded and lies inside circle B. The radius of B is 4 inches more than the radius of A. What is the radius of each if the unshaded area is.
Twice as much as the shaded area
28pi square inches more than the shaded area
16pi square inches more than twice as much as the shaded area.
Thanks

Let radius of A be r Then radius of B is r + 4 a.) What is the radius of each if the unshaded area is twice as much as the shaded area? Unshaded area (see above) = Unshaded area (GIVEN): 2 * shaded area, or Therefore, ------- Factoring out GCF, 2π Solve, using the quadratic equation formula or by completing the square. Select the POSITIVE “r” ROOT to get: Radius of smaller/shaded circle, or Radius of larger circle: b.) What is the radius of each if the unshaded area is 28π sq. inches more than the shaded area? Unshaded area (see above) = Unshaded area (GIVEN): 28π + shaded area, or Therefore, ------- Factoring out GCF, π 0 = (r - 2)(r - 6) r = 2 OR r = 6 c.) What’s the radius of each if unshaded area is 16π sq. inches more than twice the shaded area? Unshaded area (see above) = Unshaded area (GIVEN): 16π + 2 * shaded area, or Therefore, ------- Factoring out GCF, 2π 0 = r(r - 4) r - 4 = 0 OR r = 0 (ignore) r = 4

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