Circles Within Square

The two circles in the figure are congruent and have their centers on the line \overleftrightarrow{AC}. They are tangent to each other and to the square ABCD. The radius of the circles is 1. Calculate the area of the square ABCD.
image
Source: NCTM Mathematics Teacher, December 2005

SOLUTION
image
\overline{CE} is tangent to the left circle at point E and the secant segment \overline{CO} intersects the circle at points F and O
CE^2=CF\times CO
1^2=CF\times CO
1=CF(CF+FO)
1=CF(CF+2)
1=CF^2+2CF
0=CF^2+2CF-1
Using the quadratic formula
CF=\textrm{-}1+\sqrt 2
CO=CF+FO
=\textrm{-}1+\sqrt 2+2
=1+\sqrt 2
Triangle COB is a 45^\circ\!\textrm{-}45^\circ\!\textrm{-}90^\circ triangle with side CO=1+\sqrt 2
CB=CO\sqrt 2
=(1+\sqrt 2)\sqrt 2
=\sqrt 2+2
Area of square ABCD
CB^2=(\sqrt 2+2)^2
=2+4\sqrt 2+4
=6+4\sqrt 2

Answer: 6+4\sqrt 2 square units

Alternative solution
image
Triangle GHI is a 45^\circ\!\textrm{-}45^\circ\!\textrm{-}90^\circ triangle with hypotenuse GI=2
GH\sqrt 2=GI
GH\sqrt 2=2
GH=\sqrt 2
CB=CJ+JK+KB
=1+\sqrt 2+1
=2+\sqrt 2
Area of square ABCD
CB^2=(2+\sqrt 2)^2
=4+4\sqrt 2+2
=6+4\sqrt 2

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , , , . Bookmark the permalink.

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