MHB Problem of the week #34 - November, 19th 2012

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The problem involves calculating the area of a shaded region in a circle with a radius of 6 and a chord AB of length 6. The solution identifies the area of sector AOB as 6π and the area of the equilateral triangle AOB as 9√3. The area of the segment is derived by subtracting the triangle's area from the sector's area, resulting in approximately 3.261. Several forum members provided correct solutions, highlighting the collaborative nature of the discussion. The problem showcases geometric principles and calculations related to circles and triangles.
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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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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]
 
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