Finding Area of B for Sector Area Homework

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The shaded area is thenArea (shaded) = Area(A+) + Area (C+) + Area (B-)= (1/12)*pi*r^2 + (1/12)*pi*r^2 + (1/4)*sqrt(3)*r^2= (1/6)*pi*r^2 + (1/4)*sqrt(3)*r^2= (pi/6)*r^2 + (1/4)*sqrt(3)*r^2In summary, the shaded area in the given figure can
  • #1
l46kok
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Homework Statement


http://img16.imageshack.us/img16/9616/area.jpg [Broken]


The Attempt at a Solution


Area of square = 25cm^2 therefore
Area of the shaded area = 25 - (Area of A + Area of B + Area of C)
Area of A = Area of C

Therefore, Area of the shaded area = 25 - (2 * Area of A + Area of B)
Area of A = [(5 * 5 * 3.14) / 4] - Area of B
Area of A = 19.625 - Area of B

Plug back into the original equation
Area of the shaded area = 25 - (39.25 - Area of B)


How on Earth would you find the area of B? I was thinking about applying some formulas regarding finding an area of a sector ([tex]r^2 * theta * 0.5[/tex]), then subtracting it by the triangular area, but it involves trigonometry and this problem does not require the knowledge of trigonometry. Any ideas?
 
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  • #2
Your figure is distorted with complicates your problem. The "square" shown is not square as it should be. If it were truly square, it would be evident that each of the circular arcs marks off exactly pi/6 radians on the other arc, thus defining a 60 deg angle at the bottom of the figure to the point of intersection at the top of area B.

Then consider the figure bounded by a line from the lower left corner to the point of intersection at the top of area B, straight up to the top of the square, over the upper left corner of the square, and back to the bottom left of the square. The area A and a small part of B are included, and their combined area is calculated as
Aab = (pi/6)*r^2/2
where r is the 5 cm radius.
The total area in the figure previously described is
Atot = (1/2)*r*cos(pi/6)*r*sin(pi/6)+r*(1-cos(pi/6))*(r/2)

The difference between these two is half the area of the shaded sector:
Ahs = Atot = Aab
= (1/2)*r^2*cos(pi/6)*sin(pi/6)+(1/2)*r^2*(1-cos(pi/6))-(pi/6)*r^2/2
= (1/2)*r^2*(cos(pi/6)*sin(pi/6)+1-cos(pi/6)-pi/6)

For the complete sector, this has to be doubled to get
Asect = r^2*(cos(pi/6)*sin(pi/6)+1-cos(pi/6)-pi/6)

I won't promise that there are no algebraic errors in the foregoing, but this is the approach to working this problem.
 
  • #3
I thank you for your reply, but that is exactly what I was trying to avoid. This problem was shown on 6th grade exam paper, and should not require any trigonometry to solve this. Are you absolutely certain that is the only way to go about this problem?
 
  • #4
I tried constructing two different equations and substituting but I keep ending up with Shaded AR = Shaded AR, since I can't seem to find two different ways of stating either Shaded AR or B.

If you can find a way, then that should solve it.

k
 
  • #5
No, I am not absolutely certain that is the only way to go about that problem. You did not say that this was a 6th grade problem when you asked. I think it is pretty absurd to ask this of a sixth grader, quite frankly. I could not have solved this problem in 6th grade, nor in high school, and I was a top math student 50 years ago. I realize that today's students are soooo... much smarter than I am, but I still think it is absurd.

Please post the 6th grade version of the solution when you have it in hand. I'm really interested in seeing it.
 
  • #6
How about this?

Again, draw the figure without distortion, and draw straight lines from the lower corners to the intersection of the two arcs. Let the two pie shaped areas thus formed be denoted as A+ and C+ because they represent the orginal areas A with a bit more and C with a bit more.

Area (A+) = (1/3)*(1/4)*circle = (1/12)*pi*r^2 where r is the side, 5 cm
Area (C+) = Area (A+)

The area inside the triangle call B-, this is the original area B reduced on both edges. It is an equilateral triangle. The area of an equilateral triangle with side r is
Area (B-) =(1/4)*sqrt(3)*r^2

Asect = the area that was to be determined
= Atotal - Area(A+) - Area(C+) - Area(B-)
= r^2 - 2*(1/12)*pi*r^2-(1/4)*sqrt(3)*r^2
 

1. How do I find the area of a sector?

To find the area of a sector, you need to know the measure of the central angle and the radius of the circle. Then, you can use the formula A = (θ/360)πr^2, where A is the area, θ is the measure of the central angle in degrees, and r is the radius of the circle.

2. Can I find the area of a sector without knowing the central angle?

No, the central angle is necessary to find the area of a sector. You can use other information, such as the arc length and circumference, to find the central angle and then use the formula A = (θ/360)πr^2 to find the area.

3. What units should the radius be in when finding the area of a sector?

The radius should be in the same units as the central angle. For example, if the central angle is given in degrees, the radius should be in degrees as well. If the central angle is given in radians, the radius should be in radians.

4. How is the formula for finding the area of a sector derived?

The formula for finding the area of a sector is derived from the formula for finding the area of a circle, A = πr^2. This formula is multiplied by the ratio of the central angle to a full circle, which is θ/360. This gives us the formula A = (θ/360)πr^2 for finding the area of a sector.

5. Can the area of a sector be larger than the area of the entire circle?

No, the area of a sector can never be larger than the area of the entire circle. The area of a sector is always a fraction of the area of the circle, with the fraction being determined by the measure of the central angle. Therefore, the area of a sector can only be equal to or less than the area of the circle.

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