Question 1207439
<pre>
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 thepool.If the border is to have a depth of 3 inches,how wide will the border be? (1 cubic yard 27 cubic feet).

I need the set up here. 

Thanks. 


*[illustration ADC_1207439.png].
Since radius of smaller circle (pool) is 5' ({{{10/2}}}), the radius of the larger circle (pool + border) = 5 + w
Area of smaller circle (pool) = πr<sup>2</sup> = 25π
Area of larger circle (pool + border) = πr<sup>2</sup> = π(5 + w)<sup>2</sup> = π(25 + 10w + w<sup>2</sup>)
Area of border = π(25 + 10w + w<sup>2</sup>) - 25π = 25π + 10πw + πw<sup>2</sup> - 25π = 10πw + πw<sup>2</sup>
Since height of border = 3”, or {{{matrix(1,4, 3/12, "=", 1/4, foot)}}}, volume of border = (π10w + πw<sup>2</sup>){{{1/4}}} = {{{(10*pi*w + pi*w^2)/4}}}
As volume of border = 1 cubic yard, or 27 cubic feet, we then get: {{{matrix(1,3, (10*pi*w + pi*w^2)/4, "=", 27)}}}
                              10πw + πw<sup>2</sup> = 108 ---- Cross-multiplying
                        πw<sup>2</sup> + 10πw - 108 = 0
Using the quadratic equation formula: {{{matrix(1,3, w, "=", (- b +- sqrt(b^2 -  4ac))/(2a))}}}, with: a = π
                                                                    b = 10π
                                                                    c = - 108
 {{{matrix(1,3, w, "=", (- b +- sqrt(b^2 -  4ac))/(2a))}}} then becomes: {{{matrix(4,3, w, "=", (- 10pi +- sqrt((- 10pi)^2 - 4(pi)(- 108)))/(2(pi)),
w, "=", (- 10pi +- sqrt(100pi^2 + 432pi))/(2pi),
w, "=", (- 10pi +- sqrt("2,344.128"))/(2pi),
w, "=", (- 10pi +- 48.416)/(2pi))}}}
     <font color = red><font size = 4><b>Width of border</font></font></b>, or {{{highlight_green(matrix(4,3, w, "=", (- 10pi + 48.416)/(2pi), w, "=", (- 31.416 + 48.416)/6.283, w, "=", 17/6.283, highlight(w=2.71), feet, "(Approximately)"))}}}    <font color = red><font size = 4><b>OR</font></font></b>    {{{highlight_green(matrix(4,3, w, "=", (- 10pi - 48.416)/(2pi),
w, "=", (- 31.416 - 48.416)/(2pi), w, "=", (- 79.832)/2, w<0, so, IGNORE)))}}}</pre>