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. – Eevee Trainer Nov 28 '18 at 4:40
• You can try proving no solution exists from just the square, half-square, semicircle and quartercircle areas (use linear algebra). So if there's an elementary method then you'll need to draw extra lines. – YiFan Nov 28 '18 at 4:41
• The shaded area in orange? Am I color blind? – Mason Nov 28 '18 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/…) – Presh Nov 28 '18 at 8:53
• The question is wrong. Those are not semi-circles, otherwise they would intersect at the centre of the square! – user21820 Nov 28 '18 at 9:28

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. – Presh Dec 3 '18 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. – onurcanbektas Dec 5 '18 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. – achille hui Dec 5 '18 at 4:44

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. – Presh Dec 3 '18 at 4:20

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. – Presh Dec 3 '18 at 4:21

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. – Presh Dec 4 '18 at 9:54

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. – Presh Dec 3 '18 at 4:25

Homemade (hopefully helpful) diagram of the combination of 4 shapes

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
• S6 is ADPA

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 $\tan^{-1}(1/2)$. – Rahul Dec 3 '18 at 4:17

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

protected by Community♦Apr 21 at 1:17

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