Solving Tangential Circles Inside a Square

Facebook for me is just another platform to learn new things in Mathematics. If you are in the right group, you’re not just sharpen your math skills, it will make you more addicted to solving hard math problems.

Today, we will solve one of the questions I found very interesting to work on. This problem is about two circles tangent externally and inscribed in a square.

Problem: Courtesy of Vincent Earl Ryan Ty of EMC

ABCD  is a square with side 10 units. Circle E is tangent to four sides of the square. Circle G is another circle tangent to two sides of the square and circle E. Find the area of circle G.

circle 2

Solution:

Take note that points E, G, and D are collinear. Also observe that \triangle FED  is an isosceles right triangle with legs of 5 units. Using Pythagorean theorem in FED we have,

DE^2=DF^2+EF^2

DE^2=5^2+5^2

DE^2=50

\sqrt{DE^2}=\sqrt{50}

DE=5\sqrt{2}

Also observe that \triangle DGM is an isosceles right triangle. Thus  DG=MG\sqrt{2}. We know that  MG=r or equal to the radius of the small circle. The one we need to find its area. Now, we just have to find DG.

Solving for DG:

DG=DE-EG

DG=5\sqrt{2}-(5+r)

DG=5\sqrt{2}-5-r

Now, going back to  DG=MG\sqrt{2}  we have,

DG=MG\sqrt{2}

5\sqrt{2}-5-r=r\sqrt{2}

Regrouping the terms we have,

5\sqrt{2}-5=r\sqrt{2}+r

Factor out common terms,

5(\sqrt{2}-1)=r(\sqrt{2}+1)

r=\displaystyle\frac{5(\sqrt{2}-1)}{\sqrt{2}+1}

Rationalizing the denominator we have,

r=\displaystyle\frac{5(\sqrt{2}-1)}{\sqrt{2}+1}\cdot \displaystyle\frac{\sqrt{2}-1}{\sqrt{2}-1}

r=\displaystyle\frac{5(\sqrt{2}-1)^2}{(\sqrt{2})^2-1^2}

r=\displaystyle\frac{5(\sqrt{2}-1)^2}{2-1}

r=5(\sqrt{2}-1)^2

r=5(2-2\sqrt{2}+1)

r=5(3-2\sqrt{2}) units

Solving for the Area of the circle we have,

A_\circ=\pi r^2

A_\circ=\pi (5(3-2\sqrt{2}))^2

A_\circ=\pi 25(9-12\sqrt{2}+8)

A_\circ=25(17-12\sqrt{2})\pi sq. units

Dan

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

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