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Re: Inscribed circle in an Equilateral triangle [#permalink]
31 Mar 2012, 04:01

2

This post received KUDOS

Expert's post

enigma123 wrote:

In the fi gure below, if the radius of circle O is r and triangle ABC is equilateral, what is the length of AB, in terms of r?

(A) r \(\sqrt{2}\) (B) r \(\sqrt{3}\) (C) 2r \(\sqrt{3}\) (D) \(\frac{3}{2}\) r (E) 2r

Any idea how to solve?

There are many ways to solve this question including using the direct formula saying that the radius of the inscribed circle in an equilateral triangle is \(r=a*\frac{\sqrt{3}}{6}\).

One can also do the following, consider the diagram below:

Attachment:

Circle.gif [ 6.02 KiB | Viewed 2865 times ]

Angle DCO is 60/2=30 degrees, hence triangle DOC is 30°-60°-90° right triangle (angle DOC is 60°). Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\) and the leg opposite 30° (OD=r) corresponds with \(1\) and the leg opposite 60° (DC) corresponds with \(\sqrt{3}\), so \(\frac{r}{DC}=\frac{1}{\sqrt{3}}\) --> \(DC=r\sqrt{3}\). Now, since DC=AC/2 then \(AC=2DC=2r\sqrt{3}\).

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