I drew a square of side 5 together with its circumcircle. At the middle of the arc between a side and the circle, I drew a smaller square with two vertices on the side, and the other two on the circle.
What was the area of the smaller square?
As the diagonal of the larger square is congruent to the diameter
of the circle, it can be
calculated as 5√2,
thus the radius of the circle is (5/2)√2.
Let the length of a side of the smaller square be a. Then, as half the length
of the bigger
square is 5/2, one vertex of the smaller square will will touch the circle at
point (a/2, (5/2+a)).
Substituting this point in the equation of a circle:
(a/2)
^{2}
+ ((5+2a)/2)
^{2} = ((5/2)√2)
^{2}.
Simplifying and arranging
the quadratic equation, it becomes:
5a
^{2} + 20a – 25 = 0
Solving for the roots of the quadratic:
a
_{1} 1 = 1, a
_{2}
= 5
As the length is positive, the first root, a
_{1},
is the length of the side of the square, i.e., 1.

Posted by Dej Mar
on 20141206 06:46:01 