# A way to find this shaded area without calculus?

This is a popular problem spreading around. Solve for the shaded reddish/orange area. (more precisely: the area in hex color #FF5600) $$ABCD$$ is a square with a side of $$10$$, $$APD$$ and $$CPD$$ are semicircles, and $$ADQB$$ is a quarter circle. The problem is to find the shaded area $$DPQ$$.

I was able to solve it with coordinate geometry and calculus, and I verified the exact answer against a numerical calculation on Desmos.

Ultimately the result is 4 terms and not very complicated. So I was wondering: Is there was a way to solve this using trigonometry? Perhaps there is a way to decompose the shapes I am not seeing.

A couple of years ago there was a similar "Find the shaded area" problem for Chinese students. I was able to solve that without calculus, even though it was quite an involved calculation.

Disclosure: I run the YouTube channel MindYourDecisions. I plan to post a video on this topic. I'm okay posting only the calculus solution, but it would be nice to post one using only trigonometry as many have not taken calculus. I will give proper credit to anyone that helps, thanks!

Update: Thanks for everyone's help! I prepared a video for this and presented 3 methods of solving it (the short way like Achille Hui's answer, a slightly longer way like David K and Seyed's answer, and a third way using calculus). I thanked those people in the video on screen, see around 1:30 in this link: https://youtu.be/cPNdvdYn05c.

• Out of curiosity, is there a particular "audience" that this question is aimed at? Given the relatively simple formulation of the question, I almost feel like that using trig and calculus (while sufficient) might be overkill. ... though obviously I'm getting nowhere using more elementary techniques, at least at the moment, so I could easily be wrong. Nov 28, 2018 at 4:40
• The shaded area in orange? Am I color blind? Nov 28, 2018 at 4:54
• Thanks for the early responses! I will work on them to upvote and accept an answer. To Eevee--you are right, usually these should be simple problems. But there was one weird, cruel meme that required elliptic curves! (See quora.com/…) Nov 28, 2018 at 8:53
• The question is wrong. Those are not semi-circles, otherwise they would intersect at the centre of the square! Nov 28, 2018 at 9:28
• @user21820 You trust a drawn figure more than the formal description of the problem? Nov 28, 2018 at 10:18

The area is equal to difference between the area of two lenses.

It is easy to find the area of lenses like the one I did in this question before: How to find the shaded area • The graphic helps a lot, thanks! I did prepare a video and credited your user name (and others who helped) at around 1:30, see: youtu.be/cPNdvdYn05c?t=90. Dec 3, 2018 at 4:20

The area can be simplified to $$75\tan^{-1}\left(\frac12\right) - 25 \approx 9.773570675060455$$.

It come down to finding the area of lens $$DP$$ and $$DQ$$ and take difference.

What you need is the area of the lens formed by intersecting two circles, one centered at $$(a,0)$$ with radius $$a$$, another centered on $$(0,b)$$ with radius $$b$$. It is given by the expression.

\begin{align}\Delta(a,b) \stackrel{def}{=} & \overbrace{a^2\tan^{-1}\left(\frac{b}{a}\right)}^{I} + \overbrace{b^2\tan^{-1}\left(\frac{a}{b}\right)}^{II} - ab\\ = & (a^2-b^2) \tan^{-1}\left(\frac{b}{a}\right) + \frac{\pi}{2} b^2 - ab \end{align}

In above expression,

• $$I$$ is area of the circular sector span by the lens at $$(a,0)$$ (as a convex hull).
• $$II$$ is area of the circular sector span by the lens at $$(0,b)$$ (as a convex hull).
• $$ab$$ is area of union of these two sectors, a right-angled kite with sides $$a$$ and $$b$$.

Apply this to problem at hand, we get

\begin{align}\verb/Area/(DPQ) &= \verb/Area/({\rm lens}(DQ)) - \verb/Area/({\rm lens}(DP))\\[5pt] &= \Delta(10,5) - \Delta(5,5)\\ &= \left((10^2-5^2)\tan^{-1}\left(\frac12\right) + 5^2\cdot\frac{\pi}{2} - 5\cdot 10\right) - \left( 5^2\cdot\frac{\pi}{2} - 5^2\right)\\ &= 75\tan^{-1}\left(\frac12\right) - 25 \end{align}

• Incredible! I did prepare a video and credited your user name (and others who helped) at around 1:30, see: youtu.be/cPNdvdYn05c?t=90. Dec 3, 2018 at 4:20
• But, how did you obtained that formula for the lenses ? If you used calculus to obtain it, than I think this still wouldn't be an answer to this question. If not, it would be really good if you could give us some reference for it.
– Our
Dec 5, 2018 at 4:31
• @onurcanbektas the area of a lens is the sum of area of two circular sector minus that of a right angled kite. The argument is essentially the one presented in Presh's youtube video. Dec 5, 2018 at 4:44

For fun, I did the old chemist trick of printing out the diagram, cutting it apart, then weighing the pieces on a milligram scale. No calculus!

The total diagram weighed 720 mg, and the sliver weighed 77 mg. Then, $$\frac{77\,\mathrm{mg}}{720\,\mathrm{mg}}\cdot 10^2\,\mathrm{cm}^2\approx 10.7\,\mathrm{cm}^2$$ is the estimated area. This is about $$9.5\%$$ greater than the analytical solution. Not that good, but still not bad for something quick.

One source of error was the extra weight of the toner on the sliver, which printed out rather dark gray. If I knew where my compasses were, I could make a more accurate construction.

• Why couldn't we do fun stuff like that in math class? Thanks, this method would be fun to present in a video sometime for a challenging integral. Dec 4, 2018 at 9:54
• Also probably doesn't help that the circles in the diagram aren't exactly placed. You can see that the semicircle on the left is slightly too small and doesn't quite make it to the top on the left, and doesn't quite reach the halfway point in the middle. And the big quarter circle kinda sticks out on the right side, and starts coming back before it reaches the bottom of the square. May 3, 2020 at 9:44
• And the square isn't really a square. Opening it up in paint and cropping the image down to just the "square" shows me that it's a rectangle about 700 pixels wide and about 738 pixels tall. May 3, 2020 at 9:46

Let $$E$$ be the midpoint of the edge $$CD.$$ Then $$\triangle ADE$$ and $$\triangle AQE$$ are congruent right triangles, and we find that $$\angle DAQ = 2\arctan\left(\frac12\right).$$ Moreover, $$\angle CEQ = \angle DAQ$$ and therefore $$\angle DEQ = \pi - 2\arctan\left(\frac12\right).$$ And of course each of the arcs from $$D$$ to $$P$$ has angle $$\frac\pi2.$$

Knowing the radius and angle of an arc you can find the area of the circular segment bounded by the arc and the chord between the arc's endpoints without calculus. The area of the red region is the sum of the areas of the segments bounded by the arcs between $$D$$ and $$Q,$$ minus the sum of the areas of the segments bounded by the arcs between $$D$$ and $$P.$$ Note that one of the arcs from $$D$$ to $$Q$$ has radius $$10$$, but the other three arcs all have radius $$5.$$

• Yours was the first answer I read that helped. I did prepare a video and credited your user name (and others who helped) at around 1:30, see: youtu.be/cPNdvdYn05c?t=90. Dec 3, 2018 at 4:21

The complete solution can be watched here : https://youtu.be/4Yrk-UNfAis  • I watched this only after I prepared my video. I like the idea of using similar triangles to figure out the lengths of the kite's diagonals. That didn't make it into my video but it is a useful principle. Dec 3, 2018 at 4:25

The answer is $$5+\frac34 (10^2\arctan(\frac12)-40) = 75\arctan(\frac12)-25 =\pm9.774$$.

First I divided the large square in five equal areas of twenty (see image). A part of our shape occupies one fourth of the smaller central square with area 20. The remaining part of our shape consists of half a lense with radius 10 (A), minus a scaled copy of that lense with radius 5 (B). Hence the remaining part of our shape is $$\frac34$$ of area A. Area A is obviously a circle segment with area $$\frac{2\arctan(\frac12)}{2\pi}\times \pi 10^2$$ minus a triangle of area $$40$$. $$\blacksquare$$

• I will have to work through the construction myself, but at first glance this is an incredible method--thanks for the write-up! I will definitely keep in mind for similar problems. Dec 30, 2020 at 5:43

Rotate the shape around P 90, 180 and 270 degrees. The area of the shape can be expressed as 1/4*(the total area (100)- 4*shape DQC)= 25-Shape DQC (yes, I forgot to write down the letters in the image, I'm sorry).

The area of the shape DQC is Area of triangle DQC (20)- The arc DQ + arc QC (which is 1/4 the size of arc DQ). Arc DQ= Angle DAQ/2*r^2-triangle DAQ (40), where angle DAQ= sin^-1 (.8). Therfore the Shape DQC= 20-(.75*(.5*sin^-1(.8)*100-40))= 50-37.5sin^-1(.8)

That brings Shape DPQ to be 25-(50-37.5sin^-1(.8))= 37.5sin^-1(.8)-25

I think this is a good step to find the result without any coordinates, while it is actually not the full solution.

You have six non intersecting subareas, say:

• S1 is DPD
• S2 is DQPD
• S3 is DCQD
• S4 is CBQC
• S5 is BAPQD

Also say that L is the length of the square.

You can at least state these equations :

• S1+S2+S3+S4+S5+S6 = $$L^2$$
• S1+S6 = $$\frac{1}{2}\pi\left(\frac{L}{2}\right)^2$$
• S1+S2+S5+S6 = $$\frac{\pi L^2}{4}$$
• S1+S2+S3 = $$\frac{1}{2}\pi\left(\frac{L}{2}\right)^2$$
• S3+S4 = $$\frac{(2L)^2-\pi L^2}{4}$$
• S2+S5 = $$\frac{\pi L^2-\pi\left(\frac{L}{2}\right)^2}{4}$$

Alas these are not independent, but I'm pretty sure that you can find six independent ones like this.

• This approach cannot possibly work unless you find an equation involving $\tan^{-1}(1/2)$.
– user856
Dec 3, 2018 at 4:17

If you add area of blue circle slices and subtract area of green shapes you'll get the area of the shape you want. You (trivially) know the area of green shapes.

And you know the size of blue areas because by looking at triangles you can find out that the angles are $$2 \ atan \frac{1}{2}$$ (larger one) and $$\frac { \pi } { 2 } - 2\ atan \frac{1}{2}$$ (smaller one).

You just need to divide the angles by $$2\pi$$ and multiply by area of respective full circles.

The response is:

$$Area = 100\ (\frac {2\ atan \frac 1 2} { 2 \pi }\pi + \frac { \pi - 2\ atan \frac 1 2} {2\pi} \frac \pi 4 - \pi/16 - 1/4)$$ $$Area = 100\ (atan \frac 1 2 + \frac { \pi - 2\ atan \frac 1 2} {8} - \pi/16 - 1/4) \approx 9.7735707$$

PS. I found this solution by trying to make geometric sense of the analytical result found using integrals.

• Proof that we have right angle at G is left as exercise for the reader. ;-) Aug 3, 2020 at 21:38 The area can be expressed through rectangles, rectangular triangles and sectors:

$$S(BIJG)=S(BIKA)-S(GJKA)$$

$$S(BEI)=S(BIKA)-S(BEA)-S(AEK)$$

$$S(HEF)=S(HCF)-S(HCE)=S(BFG)-S(HCE)$$

$$S(EJF)=S(HIJF)-S(HEF)-S(HIE)=S(HIJF)-S(BFG)+S(HCE)-S(HIE)$$

$$S(BEF)=S(BIJG)-S(BEI)-S(BFG)-S(EJF)=$$ $$S(BIKA)-S(GJKA)-S(BIKA)+S(BEA)+S(AEK)-S(BFG)-S(HIJF)+S(BFG)-S(HCE)+S(HIE)=$$ $$-S(GJKA)+S(BEA)+S(AEK)-S(HIJF)-S(HCE)+S(HIE)$$

$$HI=x$$, $$AK=5+x$$, $$IE=\sqrt{25-x^2}$$, $$EK=\sqrt{100-(5+x)^2}$$, $$IK=10=\sqrt{25-x^2}+\sqrt{100-(5+x)^2} \Rightarrow$$ $$10-\sqrt{25-x^2}=\sqrt{75-10x-x^2} \Rightarrow$$ $$100-20\sqrt{25-x^2}+25-x^2=75-10x-x^2 \Rightarrow$$ $$50+10x=20\sqrt{25-x^2} \Rightarrow$$ $$5+x=2\sqrt{25-x^2} \Rightarrow$$ $$\sqrt{5+x}=2\sqrt{5-x} \Rightarrow$$ $$5+x=20-4x \Rightarrow x=3$$

$$S(GJKA)=5\cdot 8=40$$, $$S(BEA)=50\cdot \sin^{-1}(AK/AB)=50\cdot\sin^{-1}(0.8)$$, $$S(AEK)=(8\cdot 6)/2=24$$, $$S(HIJF)=3\cdot 5=15$$, $$S(HCE)=25/2\cdot \sin^{-1}(IE/HE)=12.5\cdot\sin^{-1}(0.8)$$, $$S(HIE)=(3\cdot 4)/2=6$$

$$S(BEF)=-40+50\cdot\sin^{-1}(0.8)+24-15-12.5\cdot\sin^{-1}(0.8)+6=$$ $$37.5\cdot\sin^{-1}(0.8)-25$$

Tried BFI numerical sums, didn't get back to error again for the needed extra independent condition .. Let's assume we have 4 semi circles inside ABCD. There will be 4 intersections and all will be equal to DP.So we can calculate DP by 4 times area of DP + the area of square = 2 times the area of circle (r=5) so DP= 50/4 (pi=3) İF we assume one more semi circle in ABCD (APB) and one more quarter circle in ABCD (CDHB) so area of 2 quarter circles - the area of the square will give us the area of intersection DB 10*10*3/2= 150 10*10=100 150-100= 50 In half the area of intersection there are 2 red areas and DP so 50/2= 25 and (25-(25/2))/2=7.5 will be our answer

• 7.5 is the answer?? We're dealing with circles here, the result is most likely an irrational number involving a trig function, or a number involving pi directly. Not sure how you got to 7.5...
– Abel
Apr 2, 2021 at 22:35