A way to find this shaded area without calculus?











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This is a popular problem spreading around. Solve for the shaded area in red.



find area dpq



$ABCD$ 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!










share|cite|improve this question
























  • 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 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 at 4:41






  • 13




    The shaded area in orange? Am I color blind?
    – Mason
    Nov 28 at 4:54






  • 1




    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 at 8:53








  • 2




    The question is wrong. Those are not semi-circles, otherwise they would intersect at the centre of the square!
    – user21820
    Nov 28 at 9:28















up vote
34
down vote

favorite
13












This is a popular problem spreading around. Solve for the shaded area in red.



find area dpq



$ABCD$ 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!










share|cite|improve this question
























  • 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 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 at 4:41






  • 13




    The shaded area in orange? Am I color blind?
    – Mason
    Nov 28 at 4:54






  • 1




    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 at 8:53








  • 2




    The question is wrong. Those are not semi-circles, otherwise they would intersect at the centre of the square!
    – user21820
    Nov 28 at 9:28













up vote
34
down vote

favorite
13









up vote
34
down vote

favorite
13






13





This is a popular problem spreading around. Solve for the shaded area in red.



find area dpq



$ABCD$ 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!










share|cite|improve this question















This is a popular problem spreading around. Solve for the shaded area in red.



find area dpq



$ABCD$ 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!







geometry trigonometry area






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edited Nov 28 at 11:53









amWhy

191k27223438




191k27223438










asked Nov 28 at 4:30









Presh

527515




527515












  • 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 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 at 4:41






  • 13




    The shaded area in orange? Am I color blind?
    – Mason
    Nov 28 at 4:54






  • 1




    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 at 8:53








  • 2




    The question is wrong. Those are not semi-circles, otherwise they would intersect at the centre of the square!
    – user21820
    Nov 28 at 9:28


















  • 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 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 at 4:41






  • 13




    The shaded area in orange? Am I color blind?
    – Mason
    Nov 28 at 4:54






  • 1




    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 at 8:53








  • 2




    The question is wrong. Those are not semi-circles, otherwise they would intersect at the centre of the square!
    – user21820
    Nov 28 at 9:28
















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 at 4:40




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 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 at 4:41




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 at 4:41




13




13




The shaded area in orange? Am I color blind?
– Mason
Nov 28 at 4:54




The shaded area in orange? Am I color blind?
– Mason
Nov 28 at 4:54




1




1




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 at 8:53






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 at 8:53






2




2




The question is wrong. Those are not semi-circles, otherwise they would intersect at the centre of the square!
– user21820
Nov 28 at 9:28




The question is wrong. Those are not semi-circles, otherwise they would intersect at the centre of the square!
– user21820
Nov 28 at 9:28










6 Answers
6






active

oldest

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up vote
8
down vote



accepted










The area can be simplified to $75tan^{-1}left(frac12right) - 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.



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



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(frac12right) + 5^2cdotfrac{pi}{2} - 5cdot 10right)
- left( 5^2cdotfrac{pi}{2} - 5^2right)\
&= 75tan^{-1}left(frac12right) - 25
end{align}
$$






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    up vote
    29
    down vote













    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
    enter image description here






    share|cite|improve this answer






























      up vote
      8
      down vote













      Let $E$ be the midpoint of the edge $CD.$
      Then $triangle ADE$ and $triangle AQE$ are congruent right triangles,
      and we find that $angle DAQ = 2arctanleft(frac12right).$
      Moreover, $angle CEQ = angle DAQ$ and therefore
      $angle DEQ = pi - 2arctanleft(frac12right).$
      And of course each of the arcs from $D$ to $P$ has angle $fracpi2.$



      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 $5.$






      share|cite|improve this answer




























        up vote
        2
        down vote













        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, $frac{77,mathrm{mg}}{720,mathrm{mg}}cdot 10^2,mathrm{cm}^2approx 10.7,mathrm{cm}^2$ 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.






        share|cite|improve this answer




























          up vote
          1
          down vote













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



          enter image description here



          enter image description here






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            up vote
            0
            down vote













            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 = $frac{1}{2}pileft(frac{L}{2}right)^2$

            • S1+S2+S5+S6 = $frac{pi L^2}{4}$

            • S1+S2+S3 = $frac{1}{2}pileft(frac{L}{2}right)^2$

            • S3+S4 = $frac{(2L)^2-pi L^2}{4}$

            • S2+S5 = $frac{pi L^2-pileft(frac{L}{2}right)^2}{4}$


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






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              6 Answers
              6






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              6 Answers
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              up vote
              8
              down vote



              accepted










              The area can be simplified to $75tan^{-1}left(frac12right) - 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.



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



              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(frac12right) + 5^2cdotfrac{pi}{2} - 5cdot 10right)
              - left( 5^2cdotfrac{pi}{2} - 5^2right)\
              &= 75tan^{-1}left(frac12right) - 25
              end{align}
              $$






              share|cite|improve this answer



























                up vote
                8
                down vote



                accepted










                The area can be simplified to $75tan^{-1}left(frac12right) - 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.



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



                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(frac12right) + 5^2cdotfrac{pi}{2} - 5cdot 10right)
                - left( 5^2cdotfrac{pi}{2} - 5^2right)\
                &= 75tan^{-1}left(frac12right) - 25
                end{align}
                $$






                share|cite|improve this answer

























                  up vote
                  8
                  down vote



                  accepted







                  up vote
                  8
                  down vote



                  accepted






                  The area can be simplified to $75tan^{-1}left(frac12right) - 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.



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



                  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(frac12right) + 5^2cdotfrac{pi}{2} - 5cdot 10right)
                  - left( 5^2cdotfrac{pi}{2} - 5^2right)\
                  &= 75tan^{-1}left(frac12right) - 25
                  end{align}
                  $$






                  share|cite|improve this answer














                  The area can be simplified to $75tan^{-1}left(frac12right) - 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.



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



                  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(frac12right) + 5^2cdotfrac{pi}{2} - 5cdot 10right)
                  - left( 5^2cdotfrac{pi}{2} - 5^2right)\
                  &= 75tan^{-1}left(frac12right) - 25
                  end{align}
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 28 at 6:26

























                  answered Nov 28 at 5:28









                  achille hui

                  94.3k5129252




                  94.3k5129252






















                      up vote
                      29
                      down vote













                      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
                      enter image description here






                      share|cite|improve this answer



























                        up vote
                        29
                        down vote













                        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
                        enter image description here






                        share|cite|improve this answer

























                          up vote
                          29
                          down vote










                          up vote
                          29
                          down vote









                          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
                          enter image description here






                          share|cite|improve this answer














                          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
                          enter image description here







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 28 at 14:12

























                          answered Nov 28 at 12:53









                          Seyed

                          6,56331423




                          6,56331423






















                              up vote
                              8
                              down vote













                              Let $E$ be the midpoint of the edge $CD.$
                              Then $triangle ADE$ and $triangle AQE$ are congruent right triangles,
                              and we find that $angle DAQ = 2arctanleft(frac12right).$
                              Moreover, $angle CEQ = angle DAQ$ and therefore
                              $angle DEQ = pi - 2arctanleft(frac12right).$
                              And of course each of the arcs from $D$ to $P$ has angle $fracpi2.$



                              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 $5.$






                              share|cite|improve this answer

























                                up vote
                                8
                                down vote













                                Let $E$ be the midpoint of the edge $CD.$
                                Then $triangle ADE$ and $triangle AQE$ are congruent right triangles,
                                and we find that $angle DAQ = 2arctanleft(frac12right).$
                                Moreover, $angle CEQ = angle DAQ$ and therefore
                                $angle DEQ = pi - 2arctanleft(frac12right).$
                                And of course each of the arcs from $D$ to $P$ has angle $fracpi2.$



                                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 $5.$






                                share|cite|improve this answer























                                  up vote
                                  8
                                  down vote










                                  up vote
                                  8
                                  down vote









                                  Let $E$ be the midpoint of the edge $CD.$
                                  Then $triangle ADE$ and $triangle AQE$ are congruent right triangles,
                                  and we find that $angle DAQ = 2arctanleft(frac12right).$
                                  Moreover, $angle CEQ = angle DAQ$ and therefore
                                  $angle DEQ = pi - 2arctanleft(frac12right).$
                                  And of course each of the arcs from $D$ to $P$ has angle $fracpi2.$



                                  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 $5.$






                                  share|cite|improve this answer












                                  Let $E$ be the midpoint of the edge $CD.$
                                  Then $triangle ADE$ and $triangle AQE$ are congruent right triangles,
                                  and we find that $angle DAQ = 2arctanleft(frac12right).$
                                  Moreover, $angle CEQ = angle DAQ$ and therefore
                                  $angle DEQ = pi - 2arctanleft(frac12right).$
                                  And of course each of the arcs from $D$ to $P$ has angle $fracpi2.$



                                  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 $5.$







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Nov 28 at 5:03









                                  David K

                                  51.6k340114




                                  51.6k340114






















                                      up vote
                                      2
                                      down vote













                                      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, $frac{77,mathrm{mg}}{720,mathrm{mg}}cdot 10^2,mathrm{cm}^2approx 10.7,mathrm{cm}^2$ 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.






                                      share|cite|improve this answer

























                                        up vote
                                        2
                                        down vote













                                        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, $frac{77,mathrm{mg}}{720,mathrm{mg}}cdot 10^2,mathrm{cm}^2approx 10.7,mathrm{cm}^2$ 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.






                                        share|cite|improve this answer























                                          up vote
                                          2
                                          down vote










                                          up vote
                                          2
                                          down vote









                                          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, $frac{77,mathrm{mg}}{720,mathrm{mg}}cdot 10^2,mathrm{cm}^2approx 10.7,mathrm{cm}^2$ 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.






                                          share|cite|improve this answer












                                          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, $frac{77,mathrm{mg}}{720,mathrm{mg}}cdot 10^2,mathrm{cm}^2approx 10.7,mathrm{cm}^2$ 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.







                                          share|cite|improve this answer












                                          share|cite|improve this answer



                                          share|cite|improve this answer










                                          answered 2 days ago









                                          Kyle Miller

                                          7,807827




                                          7,807827






















                                              up vote
                                              1
                                              down vote













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



                                              enter image description here



                                              enter image description here






                                              share|cite|improve this answer

























                                                up vote
                                                1
                                                down vote













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



                                                enter image description here



                                                enter image description here






                                                share|cite|improve this answer























                                                  up vote
                                                  1
                                                  down vote










                                                  up vote
                                                  1
                                                  down vote









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



                                                  enter image description here



                                                  enter image description here






                                                  share|cite|improve this answer












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



                                                  enter image description here



                                                  enter image description here







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered 2 days ago









                                                  user9077

                                                  794312




                                                  794312






















                                                      up vote
                                                      0
                                                      down vote













                                                      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 = $frac{1}{2}pileft(frac{L}{2}right)^2$

                                                      • S1+S2+S5+S6 = $frac{pi L^2}{4}$

                                                      • S1+S2+S3 = $frac{1}{2}pileft(frac{L}{2}right)^2$

                                                      • S3+S4 = $frac{(2L)^2-pi L^2}{4}$

                                                      • S2+S5 = $frac{pi L^2-pileft(frac{L}{2}right)^2}{4}$


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






                                                      share|cite|improve this answer

























                                                        up vote
                                                        0
                                                        down vote













                                                        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 = $frac{1}{2}pileft(frac{L}{2}right)^2$

                                                        • S1+S2+S5+S6 = $frac{pi L^2}{4}$

                                                        • S1+S2+S3 = $frac{1}{2}pileft(frac{L}{2}right)^2$

                                                        • S3+S4 = $frac{(2L)^2-pi L^2}{4}$

                                                        • S2+S5 = $frac{pi L^2-pileft(frac{L}{2}right)^2}{4}$


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






                                                        share|cite|improve this answer























                                                          up vote
                                                          0
                                                          down vote










                                                          up vote
                                                          0
                                                          down vote









                                                          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 = $frac{1}{2}pileft(frac{L}{2}right)^2$

                                                          • S1+S2+S5+S6 = $frac{pi L^2}{4}$

                                                          • S1+S2+S3 = $frac{1}{2}pileft(frac{L}{2}right)^2$

                                                          • S3+S4 = $frac{(2L)^2-pi L^2}{4}$

                                                          • S2+S5 = $frac{pi L^2-pileft(frac{L}{2}right)^2}{4}$


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






                                                          share|cite|improve this answer












                                                          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 = $frac{1}{2}pileft(frac{L}{2}right)^2$

                                                          • S1+S2+S5+S6 = $frac{pi L^2}{4}$

                                                          • S1+S2+S3 = $frac{1}{2}pileft(frac{L}{2}right)^2$

                                                          • S3+S4 = $frac{(2L)^2-pi L^2}{4}$

                                                          • S2+S5 = $frac{pi L^2-pileft(frac{L}{2}right)^2}{4}$


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







                                                          share|cite|improve this answer












                                                          share|cite|improve this answer



                                                          share|cite|improve this answer










                                                          answered Nov 28 at 10:00









                                                          Jean-Baptiste Yunès

                                                          1673




                                                          1673






























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