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07 Jul 2022, 12:00
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
80% (02:17) correct
20%
(02:23)
wrong
based on 56
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The figure above represents a square with sides of length 4 surmounted by a circle with center O. What is the outer perimeter of the figure?
(A) \(\frac{49}{9}\) \(π + 12\)
(B) \(\frac{20}{3}\) \(π + 12\)
(C) \(\frac{5}{6}\) \(π + 12\)
(D) \(π + 12\)
(E) \(9π + 12\)
Attachment:
2.jpg [ 4.69 KiB | Viewed 3953 times ]
KabirKanha
GMAT 1: 780 Q51 V46
Posts: 32
Updated on: 07 Jul 2022, 21:20
Originally posted by KabirKanha on 07 Jul 2022, 12:44.
Last edited by KabirKanha on 07 Jul 2022, 21:20, edited 1 time in total.
Last edited by KabirKanha on 07 Jul 2022, 21:20, edited 1 time in total.
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We can see that two sides of the triangle are equal (radii of the same circle) and hence the two corresponding angles will also be equal. Additionally, since the angle so formed at the centre is \(60^{\circ}\), it becomes clear that the triangle is actually equilateral with each side equal to the radius i.e., \(4\) units.
To find the perimeter of the combined figure, let us break it down into two parts:
1. Circular Perimeter:
(Perimeter of \(300^{\circ}\))
= \(\frac{300}{360} (2 \pi r)\)
= \(\frac{5}{6} (8 \pi)\)
= \(\frac{20 \pi}{3}\)
2. Square Perimeter:
(Perimeter of 3 sides)
= \(3(4)\)
= \(12\)
=> Therefore, total perimeter of the figure = \(\frac{20 \pi}{3} + 12\)
To find the perimeter of the combined figure, let us break it down into two parts:
1. Circular Perimeter:
(Perimeter of \(300^{\circ}\))
= \(\frac{300}{360} (2 \pi r)\)
= \(\frac{5}{6} (8 \pi)\)
= \(\frac{20 \pi}{3}\)
2. Square Perimeter:
(Perimeter of 3 sides)
= \(3(4)\)
= \(12\)
=> Therefore, total perimeter of the figure = \(\frac{20 \pi}{3} + 12\)
07 Jul 2022, 12:44
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The outer perimeter of the figure = Perimeter of the major arc AD + Sum of three sides of the square
In \(\triangle AOD\)
\(\angle AOX\) = \(\angle XOD\) = \(30^{\circ}\)
\(\angle XAO\) = \(\angle XDO\) = \(60^{\circ}\)
Hence In \(\triangle AOX\) is a 30 - 60 - 90 triangle
AX = XD = 2 units
Therefore AO = radius of the circle = 4 units
Perimeter of the major arc AD = \(\frac{300 }{ 360 }\) * 2 * \(\pi\) * 4
= \(\frac{5 }{ 6 }\) * 2 * \(\pi\) * 4
= \(\frac{5 }{ 3 }\) * \(\pi\) * 4
= \(\frac{20 }{ 3 }\) * \(\pi\)
Perimeter of the square = sum of three outer sides = 4 * 3 = 12
Total perimeter = \(\frac{20 }{ 3 }\) \(\pi\) + 12
\(Option B\)
In \(\triangle AOD\)
\(\angle AOX\) = \(\angle XOD\) = \(30^{\circ}\)
\(\angle XAO\) = \(\angle XDO\) = \(60^{\circ}\)
Hence In \(\triangle AOX\) is a 30 - 60 - 90 triangle
AX = XD = 2 units
Therefore AO = radius of the circle = 4 units
Perimeter of the major arc AD = \(\frac{300 }{ 360 }\) * 2 * \(\pi\) * 4
= \(\frac{5 }{ 6 }\) * 2 * \(\pi\) * 4
= \(\frac{5 }{ 3 }\) * \(\pi\) * 4
= \(\frac{20 }{ 3 }\) * \(\pi\)
Perimeter of the square = sum of three outer sides = 4 * 3 = 12
Total perimeter = \(\frac{20 }{ 3 }\) \(\pi\) + 12
\(Option B\)
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