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Ratio of Circles | AMC-10A, 2009 | Problem 21

Try this beautiful problem from Geometry: Ratio of area of Circles from AMC-10A, 2009, Problem 21. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on ratio of Circles from AMC 10A, 2009, Problem 21.

Ratio of Circles – AMC-10A, 2009- Problem 21


Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

Figure of the problem
  • $3-2 \sqrt{2}$
  • $2-\sqrt{2}$
  • $4(3-2 \sqrt{2})$
  • $\frac{1}{2}(3-\sqrt{2})$
  • $2 \sqrt{2}-2$

Key Concepts


Geometry

Circle

Pythagoras

Check the Answer


Answer: \(4(3-2 \sqrt{2})\)

AMC-10A (2009) Problem 21

Pre College Mathematics

Try with Hints


Circles in Circle

We have to find out the ratio of the sum of the areas of the four smaller circles to the area of the larger circle. To find out the area any circle,we need radius.so at first we have to find out radius of two types circle.Can you find out the radius of two type circle i.e small circle and big circle………………

Can you now finish the problem ……….

finding ratio of area of circles

Let the radius of the Four small circles be \(r\).Therfore from the above diagram we can say \(CD=DE=EF=CF=2r\). Now the quadrilateral \(CDEF\) in the center must be a square. Therefore from Pythagoras theorm we can say \(DF=\sqrt{(2r)^2 + (2r)^2}=2r\sqrt 2\). So \(AB=AD+DF+BF=r+2r\sqrt 2+r=2r+2r\sqrt 2\)

Therefore radius of the small circle is \(r\) and big circle is\(R=r+r \sqrt{2}=r(1+\sqrt{2})\)

Can you now finish the problem ……….

shaded circles

Therefore the area of the large circle is \(L=\pi R^{2}=\pi r^{2}(1+\sqrt{2})^{2}=\pi r^{2}(3+2 \sqrt{2}) \)and the The area of four small circles is \(S=4 \pi r^{2}\)

The ratio of the area will be \(\frac{S}{L}=\frac{4 \pi r^{2}}{\pi r^{2}(3+2 \sqrt{2})}\)

=\(\frac{4}{3+2 \sqrt{2}}\)

=\(\frac{4}{3+2 \sqrt{2}} \cdot \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}\)

=\(\frac{4(3-2 \sqrt{2})}{3^{2}-(2 \sqrt{2})^{2}}\)

=\(\frac{4(3-2 \sqrt{2})}{1}\)

=\(4(3-2 \sqrt{2})\)

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