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31 Dec 2021, 18:00
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Difficulty:
45%
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74% (02:53) correct
26%
(02:42)
wrong
based on 86
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A square with area 16 is perfectly inscribed inside an equilateral triangle. What is the perimeter of the triangle?
A. \(\frac{(83√3)}{3}+4\)
B. \(4√3+4\)
C. \(8√3+12\)
D. \(36\)
E. \(\frac{(32√3)}{3}+12\)
A. \(\frac{(83√3)}{3}+4\)
B. \(4√3+4\)
C. \(8√3+12\)
D. \(36\)
E. \(\frac{(32√3)}{3}+12\)
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31 Dec 2021, 18:55
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C is the solution.
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Screenshot 2022-01-01 at 7.28.21 AM.png [ 138.41 KiB | Viewed 2898 times ]
31 Dec 2021, 23:33
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With reference to the figure given below -
\( \triangle PQR\) has angles \( \angle PQR = \angle QRP = \angle RPQ \) = 60
In \( \triangle QDA \)
\( \angle DAQ\) = 90
\( \angle QDA \) = 30 ( 180 - (90 + 60))
Hence the triangle is a 30 - 60 - 90 triangle.
We know the side opposite to 60 is 4, hence the side opposite to 30 is \( \frac{4 }{ \sqrt 3}\)
Hence x = \( \frac{4 }{ \sqrt 3}\)
RQ = \( \frac{4 }{ \sqrt 3}\) + \( \frac{4 }{ \sqrt 3}\) + 4
RQ = \( \frac{8 }{ \sqrt 3}\) + 4
Perimeter = 3 * RQ
Perimeter = \( \frac{24 }{ \sqrt 3}\) + 12
Perimeter = \( \frac{24 * \sqrt 3 }{ \sqrt 3 * \sqrt 3 }\) + 12
Perimeter = \( 8 * \sqrt 3 \) + 12
\( \triangle PQR\) has angles \( \angle PQR = \angle QRP = \angle RPQ \) = 60
In \( \triangle QDA \)
\( \angle DAQ\) = 90
\( \angle QDA \) = 30 ( 180 - (90 + 60))
Hence the triangle is a 30 - 60 - 90 triangle.
We know the side opposite to 60 is 4, hence the side opposite to 30 is \( \frac{4 }{ \sqrt 3}\)
Hence x = \( \frac{4 }{ \sqrt 3}\)
RQ = \( \frac{4 }{ \sqrt 3}\) + \( \frac{4 }{ \sqrt 3}\) + 4
RQ = \( \frac{8 }{ \sqrt 3}\) + 4
Perimeter = 3 * RQ
Perimeter = \( \frac{24 }{ \sqrt 3}\) + 12
Perimeter = \( \frac{24 * \sqrt 3 }{ \sqrt 3 * \sqrt 3 }\) + 12
Perimeter = \( 8 * \sqrt 3 \) + 12
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GMAT Club - Around the World.JPG [ 20.08 KiB | Viewed 2842 times ]
