VeritasPrepKarishma wrote:

Rocky1304 wrote:

what is the radius (in cm) of the biggest possible circle that can be inscribed in sector of radius 10 cm and a central angle of 120\(^{\circ}\)

A)10(\(\sqrt{3}\) -1)

B)20(\(\sqrt{3}\)-1)/\(\sqrt{3}\)

C) 10(2\(\sqrt{3}\)-3)

D) 5

E) 10\(\sqrt{3}\)

Since I would like to see how various sides are related, I need to first draw the diagram.

The moment I look at the options and see \(\sqrt{3}\) everywhere and a play of 2 on 10 cm (the radius), I start thinking about 30-60-90 triangle.

So this is what I have. Since the circle is inscribed in the sector, it will touch the radii at one point. So the radius of the inscribed (smaller) circle will be tangential to radius of the large circle (the one to which the sector belongs). SO I find the 30-60-90 triangle I was looking for.

\(\sqrt{3}x + 2x = 10\)

\(x = 10*(2 - \sqrt{3})\)

The radius of the small circle is \(\sqrt{3}x = 10*\sqrt{3}*(2 - \sqrt{3})\)

Answer (C)

**Quote:**

Hi Karishma I understood that 10 is the radius of the sector so that is why it is 3‾√x+2x=10

But I did not understand how did you form this equation. x=10∗(2−3‾√)

We have rationalised the denominator to get this:

\(\sqrt{3}x + 2x = 10\)

\(x = \frac{10}{(\sqrt{3} + 2)}\)

\(x = \frac{10*(\sqrt{3} - 2)}{(\sqrt{3} + 2)*(\sqrt{3} - 2)}\)

\(x = \frac{10*(\sqrt{3} - 2)}{\sqrt{3}^2 - 2^2}\) (because (a+b)(a-b) = a^2 - b^2)

\(x = \frac{10*(\sqrt{3} - 2)}{3 - 4}\)

\(x = 10*(2 - \sqrt{3})\)

_________________

Karishma

Veritas Prep GMAT Instructor

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