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In the above circle the radius is 6 and chord $AB=6$. What is the area of the shaded region?
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Problem of the week #34 - November, 19th 2012
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SUMMARY
The discussion focuses on solving a geometry problem involving a circle with a radius of 6 and a chord AB measuring 6. The area of the shaded segment is calculated using the formula for the area of a sector and the area of an equilateral triangle. The final area of the segment is determined to be approximately 3.261 square units. Contributors MarkFL, SuperSonic4, soroban, Sudharaka, BAdhi, and caffeinemachine provided correct solutions, with soroban detailing the methodology used to arrive at the answer.
PREREQUISITES- Understanding of circle geometry, specifically sectors and chords.
- Knowledge of calculating areas of triangles, particularly equilateral triangles.
- Familiarity with the mathematical constant π (pi).
- Basic algebra skills for manipulating geometric formulas.
- Study the properties of circles, focusing on sectors and segments.
- Learn how to derive the area of different types of triangles, including equilateral triangles.
- Explore advanced applications of π in various geometric contexts.
- Practice solving similar geometry problems involving circles and chords.
Students studying geometry, educators teaching mathematical concepts, and anyone interested in solving geometric problems involving circles and their properties.
In the above circle the radius is 6 and chord $AB=6$. What is the area of the shaded region?
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- #2
- 4,533
- 13
Congratulations to the following members for their correct solutions:
1) MarkFL
2) SuperSonic4
3) soroban
4) Sudharaka
5) BAdhi
6) caffeinemachine
Solution (from soroban):
[sp]Let $O$ be the center of the circle.
The area of sector $AOB \text{ }$ is: .[/color]$\frac{1}{6}\pi r^2 \:=\:\frac{\pi}{6}(6^2) \:=\:6\pi$
The area of equilateral triangle $AOB \text{ }$ is: .[/color]$\frac{\sqrt{3}}{4}(6^2) \:=\:9\sqrt{3}$
Therefore, area of the segment is: .[/color]$6\pi - 9\sqrt{3} \;\approx\;3.261$
[/size][/sp]
1) MarkFL
2) SuperSonic4
3) soroban
4) Sudharaka
5) BAdhi
6) caffeinemachine
Solution (from soroban):
[sp]Let $O$ be the center of the circle.
The area of sector $AOB \text{ }$ is: .[/color]$\frac{1}{6}\pi r^2 \:=\:\frac{\pi}{6}(6^2) \:=\:6\pi$
The area of equilateral triangle $AOB \text{ }$ is: .[/color]$\frac{\sqrt{3}}{4}(6^2) \:=\:9\sqrt{3}$
Therefore, area of the segment is: .[/color]$6\pi - 9\sqrt{3} \;\approx\;3.261$
[/size][/sp]
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