# Puzzles/Geometric Puzzles/Rectangle and Circle/Solution

30 inches

## Solution

Given:

$x=12in$ $y=6in$ Find:

$r=?$ We realize that every point on a circle is equidistant from the origin. This implies that the rectangle's corner touching the circle must be r inches away from the origin, where r is the radius. We can then draw a right triangle with sides $r-y$ and $r-x.$ Now apply the Pythagorean Theorem to this triangle and solve for r.

$(r-x)^{2}+(r-y)^{2}=r^{2}$ $r^{2}-2r(x+y)+x^{2}+y^{2}=0$ The above equation is quadratic and can be solved by applying the Quadratic Formula.

$r={\frac {2(x+y)\pm {\sqrt {(4(x+y)^{2}-4(x^{2}+y^{2})\ }}}{2}}$ Which simplifies to,

$r=(x+y)\pm {\sqrt {2xy}}$ Now we can plug in our numbers and solve,

$r=(6+12)\pm {\sqrt {2(6)(12)}}$ $r=18\pm 12$ $r=30$ or $r=6$ Thus the radius of our circle is 30 inches. Notice that 6 inches is not a valid answer. Why?

Comment: Simply making a statement that "we can then draw..." is confusing. Need more explanation on why r-x and r-y are valid descriptors for the triangle sides. This is easier to see with the r-x side than with the r-y.