# CAT Practice : Coordinate Geometry

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One needs to have a clear understanding of circles, squares and equilateral triangles. Lets see how good you are!

## Equilateral triangle and circle

Q.4: Two circles are placed in an equilateral triangle as shown in the figure. What is the ratio of the area of the smaller circle to that of the equilateral triangle?

1. π : 36$\sqrt {3}$
2. π : 18$\sqrt {3}$
3. π : 27$\sqrt {3}$
4. π : 42$\sqrt {3}$

Choice (C). π : 27$\sqrt {3}$

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## Detailed Solution

In-radius of equilateral triangle of side a = ${a \over 2{\sqrt {3}}}$
Diameter of larger circle = ${a \over 2{\sqrt {3}}}$
Let us say common tangent PQ touches the two circle at R, center of smaller circle is I.

Now, PQ is parallel to BC. AR is perpendicular to PQ. Triangle PQR is also an equilateral triangle and AORID is a straight line. (Try to establish each of these observations. Just to maintain the rigour.)
AD = ${{\sqrt {3}} \over 2}$a
RD = ${a \over {\sqrt {3}}}$
AR = ${{{\sqrt {3}} \over 2}{a} - {a \over {\sqrt {3}}}}$
= ${{3a - 2a \over 2{\sqrt {3}}} = {a \over 2{\sqrt {3}}}}$
AR = ${1 \over 3}$ AD.
Radius of smaller circle = ${1 \over 3}$ radius of larger circle
Radius of smaller circle = ${{1 \over 3} * {a \over 2{\sqrt {3}}} = {a \over 6{\sqrt {3}}}}$

Area of smaller circle = πr2
π${\left( {{a \over 6{\sqrt {3}}}} \right)^2}$ = ${{{\pi}a^2} \over 108}$
Area of △ = ${{\sqrt {3}} \over 4}$a2
Ratio = ${{{\pi}a^2} \over 108}$ : ${{\sqrt {3}} \over 4}$a2
π : 27$\sqrt {3}$. Answer choice (c)

Correct Answer: π : 27$\sqrt {3}$

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Geometry is probably the most vital topic as far as CAT preparation is concerned. Geometry sets the stage for Trigonometry, Cogeo and Mensuration as well.