# Find the Area of overlapping Rectangle and Circle

I'm trying to find the total area of this rectangle and circle.

I can get an approximation of it by subtracting the overlapping bit from the area of the circle and rectangle.

Area of Circle:

$$r=C/2\pi$$ $$r=60/2\pi$$ $$r=30/\pi$$ $$r\approx9.5493$$ $$A=\pi r^2$$ $$A=286.48cm^2$$

Area of rectangle:

$$A=l*w$$ $$A=15*40$$ $$A=600cm^2$$

Area of overlapping part (approximate):

$$D^2-15^2=x$$ x = length of the rectangle part that overlaps

$$19.0986^2-15^2=x^2$$ $$x=11.8219$$

$$A=x*l$$ $$A=177.3278cm^2$$

$$\therefore totalA=286.48+600-177.3278$$ $$totalA=709.152cm^2$$

Not sure if the working is correct or if there is a much easier and more accurate way of finding the area.

• why are you approximating the overlapping area you can calculate the exact value just find the angle theta by using height and radius Commented Aug 10, 2018 at 0:45

If we divide up the figure into pieces like this:

Then the total area is $$\text{Total Area} = \text{Area of Circle} + \text{Area of Rectangle} - 2A - 2B.$$

You already found the area of the circle and the area of the rectangle. So now you just need the area of the overlap $(2A + 2B)$. Instead of approximating it, what are the exact areas of the regions $A$ and $B$?

By the way, your calculation for $x$ is good, but I don't follow $A = x \cdot l$, so maybe that is where you are doing an approximation.

For area of $A$: Try using the formula $\frac{\theta}{360^\circ} \cdot \pi r^2$, where $\theta$ is the angle cut out by the sector. To find $\theta$, use law of cosines: $c^2 = a^2 + b^2 - 2ab \cos \theta$.

For area of $B$: Try using the formula $\frac12 \text{base} \cdot \text{height}$. The height of each is $\frac{15}{2}$ from the figure. I think you already figured out the base $x$.

let's calculate $$\theta$$ $$sin\theta=\frac{h/2}{r}=\frac{h}{2r}$$

also $$h^2/4+x^2=r^2$$ h given r given calculate x

you have x thus you can calculate the area of the rectangle that is formed on overlapping the circle and big rectangle area of that rectangle is $$A1 = 2x*h$$

now the remaining area is the only blue part I will calculate it for only one part and double it.

area is $$A2= \frac{\theta r^2} {2} - \frac{hx}{2}$$

now you already have h,r,x thus calculate it double it add it in the area $$A1$$

the total overlapping area becomes $$A1+2*A2$$ hope it helps

• Thanks for your help Commented Aug 10, 2018 at 1:08