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The figure belkow shoes an inscribed triangle
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08 Feb 2024, 00:56
08 Feb 2024, 00:56
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The figure below shows an inscribed triangle ABC in a circle. If AB=6, what is the combined area of the shaded regions
GRE circle geometry.jpg [ 15.7 KiB | Viewed 1008 times ]
A. \(4 \pi - \sqrt{3 }\)
B. \(6 \pi - 6 \sqrt{3}\)
C. \(9 \pi - 4 \sqrt{3}\)
D. \(12 \pi - 9 \sqrt{3}\)
E. \(18 \pi - 16 \sqrt{3} \)
Attachment:
GRE circle geometry.jpg [ 15.7 KiB | Viewed 1008 times ]
A. \(4 \pi - \sqrt{3 }\)
B. \(6 \pi - 6 \sqrt{3}\)
C. \(9 \pi - 4 \sqrt{3}\)
D. \(12 \pi - 9 \sqrt{3}\)
E. \(18 \pi - 16 \sqrt{3} \)
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
The figure belkow shoes an inscribed triangle
[#permalink]
15 Feb 2024, 15:00
15 Feb 2024, 15:00
1
First of all The question is wrong. Insufficient information.
The question should be stated as Equilateral triangle and thereafter,
The area of triangle inside of circle = (\sqrt{3}/4) * a^2 = (\sqrt{3}/4) * 36 = 9* \sqrt{3}
If we draw lines from centre of the circle to the two corner of given triangle it will form an isoscales triangle with radius both sides and 6 other side. Then we need to draw perpendicular line from base to the centre of the circle to create, 30:60:90 triangle.
With that we can calculate the radius of the circle, r = 6/ \sqrt{3}
total area of circle is = \pi * r^2 = 12 * \pi
area of shaded region = 12 * \pi - 9 * \sqrt{3}
D.
The question should be stated as Equilateral triangle and thereafter,
The area of triangle inside of circle = (\sqrt{3}/4) * a^2 = (\sqrt{3}/4) * 36 = 9* \sqrt{3}
If we draw lines from centre of the circle to the two corner of given triangle it will form an isoscales triangle with radius both sides and 6 other side. Then we need to draw perpendicular line from base to the centre of the circle to create, 30:60:90 triangle.
With that we can calculate the radius of the circle, r = 6/ \sqrt{3}
total area of circle is = \pi * r^2 = 12 * \pi
area of shaded region = 12 * \pi - 9 * \sqrt{3}
D.
gmatclubot
The figure belkow shoes an inscribed triangle [#permalink]
15 Feb 2024, 15:00
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