SOLUTION: A circular pool measures 10 feet across. One cubic yard of concrete is to be used to create a circular border of uniform width around the pool. If the border is to have a depth of
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Question 1208738: A circular pool measures 10 feet across. One cubic yard of concrete is to be used to create a circular border of uniform width around the pool. If the border is to have a depth of 3 inches. How wide should the border be?
Note: 1 cubic yard = 27 cubic feet
Let me see.
I need A = πr^2.
I need the area of the pool.
A = π(5)^2
A = 25π ft^2
I need to find the total area including the border.
A = π(x + 5)^2
A = π(x^2 + 10x + 25)
A πx^2 + 10πx + 25π ft^2
Area of border = volume ÷ depth.
The depth is given in inches. I must concert to feet.
Let me see
3 inches of a foot = 3/12 = 1/4 feet.
Area of border = 27 feet ÷ (1/4) feet
Area of border = 108 feet^2.
Stuck here....
You say? Answer by math_tutor2020(3816) (Show Source):
The circular pool has diameter 10 feet.
That cuts in half to 5 feet which is the radius.
A = area of the pool only
A = pi*r^2 = pi*5^2 = 25pi square feet
B = combined area of the pool and walkway
B = pi*r^2 = pi*(x+5)^2 = pi*(x^2+10x+25) square feet
C = area of just the walkway only
C = B - A
C = pi*(x^2+10x+25) - pi*25
C = pi*(x^2+10x+25 - 25)
C = pi*(x^2+10x)
Multiplying the walkway's area with its height gives the volume of this concrete ring.
volume = area*height
27 cubic feet = pi*(x^2+10x)*(3/12 of a foot)
27 = pi*(x^2+10x)*(0.25)
0.25pi*x^2 + 2.5pi*x - 27 = 0
Use the quadratic formula to solve that equation.
Plug in a = 0.25pi, b = 2.5pi, c = -27
I'll let the student do the scratch work.
An alternative approach is to use a graphing calculator like a TI83, GeoGebra or Desmos (to name a few options).
When solving 0.25pi*x^2+2.5pi*x - 27 = 0, you should get these approximate solutions
{x = -12.705678, x = 2.705678}
I used the calculator's stored version of pi to get the most accuracy possible.
Ignore the negative result since a negative width makes no sense.
Recall that x > 0.
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