If a circle is inscribed in an equilateral triangle, what is

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If a circle is inscribed in an equilateral triangle, what is the area of the triangle NOT taken up by the circle?

(1) The area of the circle is 12Ï€
(2) The length of a side of the triangle is 12

OA D

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by Jay@ManhattanReview » Mon Aug 27, 2018 10:57 pm

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BTGmoderatorDC wrote:If a circle is inscribed in an equilateral triangle, what is the area of the triangle NOT taken up by the circle?

(1) The area of the circle is 12Ï€
(2) The length of a side of the triangle is 12

OA D

Source: Veritas Prep
Let's solve this one logically.

Only one circle can be inscribed in an equilateral triangle; thus, if we have the value of the area of the circle, we can get the value of the area of the equilateral triangle, and thereby the area of the triangle NOT taken up by the circle.

Similarly, if we have the value of the area of the equilateral triangle, we can get the value of the area of the circle, and thereby the area of the triangle NOT taken up by the circle.

So, each statement itself is sufficient.

The correct answer: D

Hope this helps!

-Jay
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by fskilnik@GMATH » Mon Sep 03, 2018 1:47 pm

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BTGmoderatorDC wrote:If a circle is inscribed in an equilateral triangle, what is the area of the triangle NOT taken up by the circle?

(1) The area of the circle is 12Ï€
(2) The length of a side of the triangle is 12
\[? = {S_{\,{\text{grey}}}}\,\,\,\,\left( {{\text{figure}}} \right)\]
Using the 30-60-90 shortcut in the triangle shown in the figure, we were able to relate all elements involved, so that:
\[?\,\, = \,\,\frac{{L \cdot {h_{{\text{eq}}}}}}{2} - \pi {r^2}\, = \,\,\,\frac{{{L^2} \cdot \,\sqrt 3 }}{4} - \pi {r^2}\,\,\mathop = \limits^{\left( * \right)} \,3{r^2} \cdot \sqrt 3 - \pi {r^2} = \boxed{{r^2}\left( {3\sqrt 3 - \pi } \right)}\,\,\,\left( {**} \right)\]
\[\left( * \right)\,\,\frac{{L\sqrt 3 }}{2} = {h_{{\text{eq}}}}\mathop = \limits^{figure} \,\,3r\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,2\sqrt 3 \,\,} \,\,3L = 6r\sqrt 3 \,\,\, \Rightarrow \,\,\,\,L = 2r\sqrt 3 \]

\[\left( 1 \right)\,\,\pi {r^2} = 12\pi \,\,\,\, \Rightarrow \,\,\,{r^2} = 12\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,? = \,\,{\text{unique}}\]

\[\left( 2 \right)\,\,L = 12\,\,\, \Rightarrow \,\,\,3r = {h_{{\text{eq}}}}\,\,{\text{unique}}\,\,\, \Rightarrow \,\,\,{r^2} = \,\,{\text{unique}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,? = \,\,{\text{unique}}\]

This solution follows the notations and rationale taught in the GMATH method.

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