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

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Sun Aug 26, 2018 5:21 pm

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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

Source: Veritas Prep

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Source: Beat The GMAT — Data Sufficiency | Sun Aug 26, 2018 5:21 pm

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

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

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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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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

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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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