Diagram of the circle & cross

I need to find the area of the shaded (grey) region in the above picture. All sides of the cross are 4. I feel like this question has an easy answer, but I cannot seem to figure it out. I tried to find the area of the circle, but I can't find the radius/diameter. Also, I couldn't find what part of the area each shaded region is. I'm at a total loss with this one. If anyone could help, it'd be much appreciated. (If there is a similar question anywhere on Stack Exchange or elsewhere, that would also be appreciated) Thank you!

So when I again attempted the question, I found the diameter, and so the area of the circle. However, what I fail to see is how to only subtract the shaded areas. Would just finding the area of the cross and then the area of the segments at the end of each end of the cross work? How would one do so?

Update: The ideal answer would be about as simple as what you might expect in a high school geometry class. Thank you all for your answers.

  • 1
    $\begingroup$ The diameter of the circle is the hypotenuse of a right-angled triangle whose other sides are $4$ and $12$. $\endgroup$ – peterwhy Jun 5 '17 at 1:19
  • $\begingroup$ Thank you so much! Would that make the diameter is sqrt 160 ? @peterwhy $\endgroup$ – Johnny Jun 5 '17 at 1:25

The radius of the circle is $r=\sqrt{2^2+6^2}=\sqrt{40}$. The area of the circle is $A_c=\pi r^2=40\pi$.

The area of the squares is $A_sq=5 \cdot 4\cdot 4=80.$

The area of a single segment is $A_{seg}=\frac{1}{2}r^2(\theta-\sin \theta)$, where $\theta$ is the included angle from the origin to both corners. i.e., $\theta=2\tan^{-1}(2/6).$

Can you take it from here?

  • $\begingroup$ On line 3, how did you get the area of a single segment? Also, what is theta? (Or where did you get theta?) Thank you very much for your response. $\endgroup$ – Johnny Jun 5 '17 at 1:48
  • $\begingroup$ It's from the CRC Handbook of Mathematics. Theta is the included angle from the origin to both ends of the segment. Don't forget to tip the help! $\endgroup$ – Cye Waldman Jun 5 '17 at 2:05
  • $\begingroup$ Thank you very much for all of your help. However, I just have one more quick question. That would mean I substitute 2tan-1(2/6) for theta in the Aseg= formula, and multiply that by four for my answer? Thank you again. @Cye Waldman $\endgroup$ – Johnny Jun 5 '17 at 2:11
  • $\begingroup$ Yes. That's all that you have to do $\endgroup$ – Cye Waldman Jun 5 '17 at 2:19

Here is my take on this: img

Lets say the green area is $G$ the blue area is $B$ and red area is $R$. The shaded region we want is $4G$.

The circle area is $5R+4G+4B$. Subtracting $5R+4B$ from the circle area which is found by $\pi r^2$ will give us the shaded region.

We find $r$ to be $2\sqrt{10}$ using the triangle shown in this picture.

We know $R$ (the red area) is $4^2$ because it is a square with side $4$.

To find $B$ we can focus on a certain triangle, seen in the figure below: img2

The area of the sector with angle $\alpha$ is $\pi (2\sqrt{10})^2 \times \left(\frac{\alpha}{360^{\circ}} \right)$. The triangle area is $2\times\left(\frac{2\times 6}{2}\right)=12$. Therefore the blue area will be given by $$ B = \pi (2\sqrt{10})^2 \times \left(\frac{\alpha}{360^{\circ}} \right) - 12 $$ Now we have to know $\alpha$. We can see that at the triangle $\sin(\alpha/2)=2/2\sqrt{10}$. Therefore $\alpha=2\times\arcsin(1/\sqrt{10})$. Formula for $B$ now becomes $$ B = \pi (2\sqrt{10})^2 \times \left(\frac{2\times\arcsin(1/\sqrt{10})}{360^{\circ}} \right) - 12 $$ Writing all these into the equality $5R+4G+4B=\pi (2\sqrt{10})^2 $ we get $$ 5\times16+4G+4\times\left(\pi (2\sqrt{10})^2 \times \left(\frac{2\times\arcsin(1/\sqrt{10})}{360^{\circ}} \right) - 12\right) = \pi (2\sqrt{10})^2 $$ $$ 80-48+4G+40\pi \times \left(\frac{\arcsin(1/\sqrt{10})}{45^{\circ}} \right) = 40\pi $$ Leaving the shaded region which is $4G$ on one side we get $$ 4G = 40\pi\left(1- \frac{\arcsin(1/\sqrt{10})}{45^{\circ}}\right)-32 $$ This is the shaded region we are looking for!


The area of five squares are clear. Now to compute one part of the curved shape. To compute that, first compute an Arc area the minus from the its above triangle. We know that $r=2\sqrt{10}$, so, the area of one of the curved shape are equal:

$$2\sqrt{10}\times (180 - 2\arccos(\frac{\sqrt{10}}{10}))\times \frac{\pi}{180} - 12 = (90 - \arccos(\frac{\sqrt{10}}{10}))\times \frac{\pi\sqrt{10}}{45} - 12$$

I should have mention that the area arc shape of a circle is eqaul to the $\alpha\times r$ which $\alpha$ is the angle of that arc.

So, the area of the shaded is equal to the area of the circle minus 5 squares plus the four curved shapes: $$S = 40\pi - 16\times 5 - 4 \times ((90 - \arccos(\frac{\sqrt{10}}{10}))\times \frac{\pi\sqrt{10}}{45} - 12) = 40\pi - 80- 4(90 - \arccos(\frac{\sqrt{10}}{10}))\times \frac{\pi\sqrt{10}}{45} + 48 = 40\pi - 32 + \frac{4\pi\sqrt{10}}{45}\times \arccos{\frac{\sqrt{10}}{10}} - 8\pi\sqrt{10}$$

  • $\begingroup$ Except that $r\ne $6. $\endgroup$ – Cye Waldman Jun 5 '17 at 1:37
  • $\begingroup$ That was my question, how did you find the radius? I thought that the diameter was sqrt160. If I am wrong, please show my how you got 6. Thank you! $\endgroup$ – Johnny Jun 5 '17 at 1:41
  • $\begingroup$ @Johnny so simple. imagine a rectangle from center of the circle to the bottom corner of the top right shaded shape. so the radius equal $\sqrt{2^2 + (4+2)^2} = \sqrt{40}$. $\endgroup$ – OmG Jun 5 '17 at 1:52
  • $\begingroup$ Thank you @OmG I thought that you had said that the radius is 6. Thanks for your response! Would the last string of numbers after the = be the simplified answer? (40π-32...-4πsqrt10) $\endgroup$ – Johnny Jun 5 '17 at 1:56
  • $\begingroup$ @Johnny If the answer is helpful, please accept it as an answer. $\endgroup$ – OmG Jun 5 '17 at 1:58

I think that the answer might be found by finding the segements above each edge of the cross (non-shaded) and subtracting those (plus the cross) from the total circle area.


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