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.
Best Answer
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 $x$.