A way to find this shaded area without calculus?

Question

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.

Answer 1

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

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

Answer 2

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 $5.$

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

The complete solution can be watched here : https://youtu.be/4Yrk-UNfAis

Answer 7

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

Answer 8

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

Answer 9

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.

Answer 10

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 $40$. $\blacksquare$

Answer 11

Tried BFI numerical sums, didn't get back to error again for the needed extra independent condition ..

Answer 12

The area can be expressed through rectangles, rectangular triangles and sectors:

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

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

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

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

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

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

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

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