Geometry Problem

[dr|ft]

Ok i've attatched the shapes. Now what that is, is a square with 4 1/4 circles going through it. The aim is to find the shaded area and we know that the sides of the square are 10cm. Good Luck.

[Dapthar]

Ok i've attatched the shapes. Now what that is, is a square with 4 1/4 circles going through it. The aim is to find the shaded area and we know that the sides of the square are 10cm. Good Luck.

We could use Calculus techniques to compute the area, but I don't suppose that you want us to use them, right?

[dr|ft]

if its a solution, i guess post it? dave was using that idea i think he is solving it that way and posting the solution shortly

[Dave]

My solution uses integration, and I'm sure it's not the easiest way of doing it. Look at the image below for what I'm trying to accomplish. Basically, we get the following integral for the shaded area:

 

[math]A = \int_0^5 \sqrt{100-x^2} \, dx = \tfrac{25}{6} (3\sqrt{3} + 2\pi)[/math]

 

Then the area of one unshaded section is [math]50 - A = \tfrac{25}{6}(12 - 3\sqrt{3} - 2\pi)[/math].

 

Multiplying this by 4 and then subtracting everything from 100 will give you the answer, which I get to be:

 

[math]\tfrac{50}{3}(3\sqrt{3} +2\pi -6)[/math]

[mezarashi]

Interesting question. Reminds me of my geometry days. Let me have a go at it using the classical ways >_>

 

Updated:

 

Hopefully that will be a bit helpful. It took me sometime after all. This problem was more confusing than I thought. Anyway, the whole problem here is trying to the find the area of that small unshaded white area in the square box. The picture i made above represents it as yellow. Well obviously if we can find the area for this two circle case, we can apply it to the four circle case easily. So let's start from here.

 

To find the area YELLOW, it will be most convenient by first finding area BLUE.

I couldnt find a way to explain this properly so I drew another picture

 

 

So basically the strategy here as you can see, is to find the dark green region which is basically the BLUE area in the first picture. You will subtract the triangle formed by the red lines from the green circle segment. Now to find the dimensions.

 

We know that since it's symmetrical, the x length of the triangle is 5. The y can be found using the circle equation:

 

y^2 + x^2 = R^2 (in this case, R = 10cm)

 

This will give you y = 8.66cm

 

Now we know that the height total of the triangle = 2y = 17.32cm

Using the 1/2 x base x height formula, the area = 43.3cm^2

 

Now use the tan rule to find the angle. arctan ( 8.66 / 5 ) = 60 degrees, or 120 degrees (out of 360), meaning this is 1/3 the circle's total area. Using the fraction 1/3 x pi x r^2, we get this area to be: 104.72cm^2

 

Subtracting the two from each other: 104.72-43.3=61.42cm^2 = Area BLUE

 

So now that we know area BLUE, we return to the first picture and see that BLUE + RED = 1/4 circle area, we substitute the values in to get:

 

Area RED = 1/4 x pi x r^2 - Area BLUE

Area RED = 17.12 cm^2

 

Lastly, we know that BLUE + RED + GREEN + YELLOW = 100cm^2 (also RED = GREEN)

 

Area YELLOW = 100 - BLUE - 2*RED

Area YELLOW = 4.393 cm^2

 

Hurray! So we now know that tiny area that is not included in the red shaded part of your diagram. All we do now is take the area of the square and minus away 4 of these:

 

Shaded region = 100 cm^2 - 4*(Area YELLOW)

Shaded region = 100 - 4*(4.393) = 82.4 cm^2

 

Pheew. I hope that's correct. I haven't checked, but it makes logical sense to me. /me sleeps

[Dave]

Tsk, all these decimal things

 

The answer I got, however, is 91.3223cm², so I don't know who's right now

[dr|ft]

i THINK i remember the answer being in the 82's actually, that looks like a good solution mezarashi i gotta analyze it, thanks for the help tho