# Thread: Area allocation with shapes

1. ## Area allocation with shapes

As seen in the attachment I have a 100 x 100 square and a quarter of a circle with radius being 104.032. This radius gives an area of 8500 from 1/4(104.032^2) * Pi

My problem is I need the shaded area to equal 8500. I have to do this while still using a circle radius. So I need to know how, in terms of formulas ect, to find the radius to make the shaded region equal 8500. Any help would be great.

I tried to set up a system of equations with splitting the area into two equal triangles and then a pie section. Areas being B*100 for the area sum of the triangles and (theta * R^2)/2 in radians for the area of the pie section. The sum of these would be set to 8500. Problem I ran into was I ended up with an equation that was unsolvable. I suspect there are easier ways of approaching this like maybe area under a curve or something. I appreciate any help, thank you for your time.

2. so you are trying to find the area of intersection for a circle and square in the 2nd quadrant.

the top of the square is y=100, x starts at -100 and goes to 0.

instead of thinking of the region as three regions (two triangles and a wedge) think of it as two regions (area under curve and area under flat line)

the equation for a circle with radius r is x^2 + Y^2 = r^2
but since your region is in the 2nd quadrant, y is positive and thus:
y= sqrt(r^2-x^2)

to find where the curve stops, set y=100 and get

r^2-x^2=10,000
x^2 = r^2-10,000
x=sqrt(r^2-10,000)

the area of the non curvy region is (100-sqrt(r^2-10,000))*100
the area of the curvey part is: oh boy here we go

the definite integral from x=100 to x=sqrt(r^2-10,000) of sqrt(r^2-x^2) which equals (i believe) 1/r arcsin (sqrt(r^2-10,000)/r) - 1/r arcsin(100/r)

I am really high right now and do not have paper so you might want to do the calculus yourself. Add the area of the curvy and non curvey regions and set equal to 3500, then solve for r. Solving for r will such. Try using inverse funtions on both sides at first and then just resort to newtons method or something. However, rest assured that if a solution to your problem exists, it will be the solution to this equation.

again, i am stoned out of my mind so make sure the integral of sqrt(r^2-x^2) really is what I said it is. Good luck and happy equation solving.

3. Haha I have no clue how to approach that. I see how you are setting up that equation but couldnt even begin to try and solve it.

I do have a side thought to this problem. I know that the quarter circle section area is Pi/4 * R^2. I know that this area minus the two equal areas of the parts that will be outside the square must be 8500. So I was thinking that this equation would apply.

Pi/4 *R^2 - (arccos(100/R)*R^2 - 100sqrt(R^2-100^2)) = 8500

The reasoning for this would be the overall area of the quarter circle minus the area that is outside the box. Arccos(100/R)R^2 is simple the area equation for a pie piece of a circle just not divided by two since there are two sections. This area is reduced by subtracting the 100sqrt(R^2-100^2) which is the area of the two triangles formed between the bottom right corner, top right corner, and the intersection point of the circle.

Only problem is I have no idea how to solve this type equation out.
Any ideas?

4. Originally Posted by smuckers
As seen in the attachment I have a 100 x 100 square and a quarter of a circle ...
My problem is I need the shaded area to equal 8500. I have to do this while still using a circle radius. So I need to know how, in terms of formulas ect, to find the radius to make the shaded region equal 8500. Any help would be great.
...
& the area external to curve and inside the square is 1500.

For some $\displaystyle \theta$
radius = $\displaystyle \dfrac{100}{\cos \left(\dfrac{\pi}{4} -\theta \right)}$

The diagonal of the square: $\displaystyle \sqrt{ 2\cdot 100^2} = 100\sqrt{2}$

The area of $\displaystyle \triangle ACD = \dfrac{r \sin(\theta) 100 \sqrt{2}} {2}$

The area of the sector = $\displaystyle \dfrac {r^2 \theta }{2}$

ShadedArea = $\displaystyle \dfrac{r \sin \theta 100 \sqrt{2}} {2}$ - $\displaystyle \dfrac {r^2 \theta }{2}$ = 750

Area = $\displaystyle r \sin(\theta) 100 \sqrt{2} - r^2 \theta = 1500$

subbing in the definition of r from above
One equation with 1 unknown.
Area = $\displaystyle \left( \dfrac{100}{\cos \left(\dfrac{\pi}{4} -\theta \right)} \right) \sin(\theta) 100 \sqrt{2} - \left( \dfrac{100}{\cos \left(\dfrac{\pi}{4} -\theta \right)}\right)^2 \theta = 1500$

Solve for $\displaystyle \theta$ & then for R.

Spoiler:

$\displaystyle \theta = 0.459289725$ (radians)
R=105.5636