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Finding sector area

  • Thread starter l46kok
  • Start date
1. Homework Statement
http://img16.imageshack.us/img16/9616/area.jpg [Broken]

3. 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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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 wont promise that there are no algebraic errors in the foregoing, but this is the approach to working this problem.
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?
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.

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.
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

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