Can area of rectangle be greater than the square of its diagonal? [closed]











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Q: A wall, rectangular in shape, has a perimeter of 72 m. If the length of its diagonal is 18 m, what is the area of the wall ?




The answer given to me is area of 486 m2. This is the explanation given to me
This is the explanation given to me




Is it possible to have a rectangle of diagonal 18 m and area greater than the area of a square of side 18 m ?











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closed as off-topic by user21820, John B, Holo, TheSimpliFire, Parcly Taxel Nov 18 at 2:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, John B, Holo, TheSimpliFire, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 22




    The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
    – Théophile
    Nov 15 at 19:09








  • 8




    The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
    – alephzero
    Nov 15 at 20:16








  • 7




    In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
    – amalloy
    Nov 15 at 22:37






  • 7




    The problem with the question is that if you solve for the side lengths you get complex numbers.
    – 1123581321
    Nov 15 at 22:53






  • 9




    @1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
    – Teepeemm
    Nov 16 at 2:10















up vote
40
down vote

favorite
5













Q: A wall, rectangular in shape, has a perimeter of 72 m. If the length of its diagonal is 18 m, what is the area of the wall ?




The answer given to me is area of 486 m2. This is the explanation given to me
This is the explanation given to me




Is it possible to have a rectangle of diagonal 18 m and area greater than the area of a square of side 18 m ?











share|cite|improve this question















closed as off-topic by user21820, John B, Holo, TheSimpliFire, Parcly Taxel Nov 18 at 2:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, John B, Holo, TheSimpliFire, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 22




    The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
    – Théophile
    Nov 15 at 19:09








  • 8




    The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
    – alephzero
    Nov 15 at 20:16








  • 7




    In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
    – amalloy
    Nov 15 at 22:37






  • 7




    The problem with the question is that if you solve for the side lengths you get complex numbers.
    – 1123581321
    Nov 15 at 22:53






  • 9




    @1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
    – Teepeemm
    Nov 16 at 2:10













up vote
40
down vote

favorite
5









up vote
40
down vote

favorite
5






5






Q: A wall, rectangular in shape, has a perimeter of 72 m. If the length of its diagonal is 18 m, what is the area of the wall ?




The answer given to me is area of 486 m2. This is the explanation given to me
This is the explanation given to me




Is it possible to have a rectangle of diagonal 18 m and area greater than the area of a square of side 18 m ?











share|cite|improve this question
















Q: A wall, rectangular in shape, has a perimeter of 72 m. If the length of its diagonal is 18 m, what is the area of the wall ?




The answer given to me is area of 486 m2. This is the explanation given to me
This is the explanation given to me




Is it possible to have a rectangle of diagonal 18 m and area greater than the area of a square of side 18 m ?








geometry area






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share|cite|improve this question













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edited Nov 15 at 18:28

























asked Nov 15 at 17:49









user17838

31126




31126




closed as off-topic by user21820, John B, Holo, TheSimpliFire, Parcly Taxel Nov 18 at 2:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, John B, Holo, TheSimpliFire, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by user21820, John B, Holo, TheSimpliFire, Parcly Taxel Nov 18 at 2:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, John B, Holo, TheSimpliFire, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 22




    The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
    – Théophile
    Nov 15 at 19:09








  • 8




    The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
    – alephzero
    Nov 15 at 20:16








  • 7




    In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
    – amalloy
    Nov 15 at 22:37






  • 7




    The problem with the question is that if you solve for the side lengths you get complex numbers.
    – 1123581321
    Nov 15 at 22:53






  • 9




    @1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
    – Teepeemm
    Nov 16 at 2:10














  • 22




    The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
    – Théophile
    Nov 15 at 19:09








  • 8




    The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
    – alephzero
    Nov 15 at 20:16








  • 7




    In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
    – amalloy
    Nov 15 at 22:37






  • 7




    The problem with the question is that if you solve for the side lengths you get complex numbers.
    – 1123581321
    Nov 15 at 22:53






  • 9




    @1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
    – Teepeemm
    Nov 16 at 2:10








22




22




The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
– Théophile
Nov 15 at 19:09






The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
– Théophile
Nov 15 at 19:09






8




8




The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
– alephzero
Nov 15 at 20:16






The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
– alephzero
Nov 15 at 20:16






7




7




In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
– amalloy
Nov 15 at 22:37




In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
– amalloy
Nov 15 at 22:37




7




7




The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
Nov 15 at 22:53




The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
Nov 15 at 22:53




9




9




@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
Nov 16 at 2:10




@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
Nov 16 at 2:10










12 Answers
12






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82
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accepted










The area of the square built on the diagonal must be at least twice the area of the rectangle:



$hskip 4 cm$ enter image description here






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    up vote
    72
    down vote













    Another proof without words, at the suggestion of Semiclassical:



    enter image description here



    The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.






    share|cite|improve this answer























    • +1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
      – mckenzm
      Nov 15 at 23:26






    • 1




      Here is a more dynamic, animated version of the same picture.
      – Xander Henderson
      Nov 16 at 4:44


















    up vote
    31
    down vote













    A simple explanation without proof or pictures:



    The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.






    share|cite|improve this answer























    • Great explanation, but that “it's” is jarring...
      – DaG
      Nov 17 at 9:34










    • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
      – AlexanderJ93
      Nov 17 at 9:35










    • +1 Thank you for the one-line proof!
      – DaG
      Nov 17 at 10:01


















    up vote
    22
    down vote













    In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.






    share|cite|improve this answer




























      up vote
      9
      down vote













      You can prove that no such rectangle exists as follows:



      Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$



      Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.



      The answer given, though arithmetically correct does not represent a real wall.





      I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.






      share|cite|improve this answer





















      • To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
        – Ilmari Karonen
        Nov 15 at 21:39






      • 1




        @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
        – Mark Bennet
        Nov 16 at 8:40




















      up vote
      6
      down vote













      Another PWW (noted by AlexanderJ93 and others):



      $hspace{5cm}$![enter image description here






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        up vote
        4
        down vote













        No. As others have said.



        What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.



        If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.



        If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:



        Total perimeter: 70
        Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)



        This should now give the solution of:




        • $I^2 + B^2 = 25^2 = 625 $

        • $2I + 2B = 70 $

        • $I + B = 35 $

        • $I^2 + 2IB + B^2 = 1,225 $

        • $2IB = 600 $


        • $IB = 300$ ,


        which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)



        Hope that helps!



        -Van






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          up vote
          3
          down vote













          No, use the Pythagorean Theorem.



          $$c^2 = a^2+b^2$$



          $c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).



          Recall for any real number, its square must be non-negative.



          $$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$



          The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.



          Now, to find the area itself.



          For the diagonal:



          $$c^2 = a^2+b^2$$



          $$implies 18^2 = a^2+b^2$$



          $$color{blue}{324 = a^2+b^2} tag{1}$$



          For the perimeter:



          $$2(a+b) = 72$$



          $$a+b = 36$$



          Now, define one variable in terms of the other.



          $$color{purple}{a = 36-b} tag{2}$$



          Combine $(1)$ and $(2)$.



          $$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$



          $$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$



          But $$Delta = b^2-4ac$$



          $$Delta = 72^2-4(2)(972) = -2592$$



          $$implies Delta < 0$$



          Thus, there is no solution. (No such rectangle exists.)






          share|cite|improve this answer






























            up vote
            2
            down vote













            No. Using Pythagoras and a simple inequality we get
            $$d^2=a^2+b^2geq 2abgeq ab$$
            If $a,b$ are the sides and $d$ the diagonal






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              up vote
              2
              down vote













              Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):



              enter image description here






              share|cite|improve this answer




























                up vote
                0
                down vote













                $A=lw$



                $P=2(l+w)$



                $d=sqrt{l^2+w^2}$



                Can $A>d^2$?



                Can $lw>l^2+w^2$?



                $-lw>l^2-2wl+w^2=(l-w)^2$



                Width and length are necessarily positive. The square of their difference also must be positive.



                So we have a negative number that must be greater than a positive number. A contradiction.






                share|cite|improve this answer




























                  up vote
                  0
                  down vote













                  A wall has a thickness.



                  Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.



                  Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.



                  The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.



                  We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.



                  This t can be used to calculate the area of the wall as (72 - 4t) * t.



                  Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.






                  share|cite|improve this answer




























                    12 Answers
                    12






                    active

                    oldest

                    votes








                    12 Answers
                    12






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes








                    up vote
                    82
                    down vote



                    accepted










                    The area of the square built on the diagonal must be at least twice the area of the rectangle:



                    $hskip 4 cm$ enter image description here






                    share|cite|improve this answer

























                      up vote
                      82
                      down vote



                      accepted










                      The area of the square built on the diagonal must be at least twice the area of the rectangle:



                      $hskip 4 cm$ enter image description here






                      share|cite|improve this answer























                        up vote
                        82
                        down vote



                        accepted







                        up vote
                        82
                        down vote



                        accepted






                        The area of the square built on the diagonal must be at least twice the area of the rectangle:



                        $hskip 4 cm$ enter image description here






                        share|cite|improve this answer












                        The area of the square built on the diagonal must be at least twice the area of the rectangle:



                        $hskip 4 cm$ enter image description here







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 15 at 18:04









                        Théophile

                        19.3k12946




                        19.3k12946






















                            up vote
                            72
                            down vote













                            Another proof without words, at the suggestion of Semiclassical:



                            enter image description here



                            The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.






                            share|cite|improve this answer























                            • +1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
                              – mckenzm
                              Nov 15 at 23:26






                            • 1




                              Here is a more dynamic, animated version of the same picture.
                              – Xander Henderson
                              Nov 16 at 4:44















                            up vote
                            72
                            down vote













                            Another proof without words, at the suggestion of Semiclassical:



                            enter image description here



                            The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.






                            share|cite|improve this answer























                            • +1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
                              – mckenzm
                              Nov 15 at 23:26






                            • 1




                              Here is a more dynamic, animated version of the same picture.
                              – Xander Henderson
                              Nov 16 at 4:44













                            up vote
                            72
                            down vote










                            up vote
                            72
                            down vote









                            Another proof without words, at the suggestion of Semiclassical:



                            enter image description here



                            The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.






                            share|cite|improve this answer














                            Another proof without words, at the suggestion of Semiclassical:



                            enter image description here



                            The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            answered Nov 15 at 19:01


























                            community wiki





                            Xander Henderson













                            • +1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
                              – mckenzm
                              Nov 15 at 23:26






                            • 1




                              Here is a more dynamic, animated version of the same picture.
                              – Xander Henderson
                              Nov 16 at 4:44


















                            • +1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
                              – mckenzm
                              Nov 15 at 23:26






                            • 1




                              Here is a more dynamic, animated version of the same picture.
                              – Xander Henderson
                              Nov 16 at 4:44
















                            +1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
                            – mckenzm
                            Nov 15 at 23:26




                            +1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
                            – mckenzm
                            Nov 15 at 23:26




                            1




                            1




                            Here is a more dynamic, animated version of the same picture.
                            – Xander Henderson
                            Nov 16 at 4:44




                            Here is a more dynamic, animated version of the same picture.
                            – Xander Henderson
                            Nov 16 at 4:44










                            up vote
                            31
                            down vote













                            A simple explanation without proof or pictures:



                            The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.






                            share|cite|improve this answer























                            • Great explanation, but that “it's” is jarring...
                              – DaG
                              Nov 17 at 9:34










                            • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                              – AlexanderJ93
                              Nov 17 at 9:35










                            • +1 Thank you for the one-line proof!
                              – DaG
                              Nov 17 at 10:01















                            up vote
                            31
                            down vote













                            A simple explanation without proof or pictures:



                            The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.






                            share|cite|improve this answer























                            • Great explanation, but that “it's” is jarring...
                              – DaG
                              Nov 17 at 9:34










                            • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                              – AlexanderJ93
                              Nov 17 at 9:35










                            • +1 Thank you for the one-line proof!
                              – DaG
                              Nov 17 at 10:01













                            up vote
                            31
                            down vote










                            up vote
                            31
                            down vote









                            A simple explanation without proof or pictures:



                            The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.






                            share|cite|improve this answer














                            A simple explanation without proof or pictures:



                            The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 17 at 9:34

























                            answered Nov 15 at 20:21









                            AlexanderJ93

                            5,314522




                            5,314522












                            • Great explanation, but that “it's” is jarring...
                              – DaG
                              Nov 17 at 9:34










                            • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                              – AlexanderJ93
                              Nov 17 at 9:35










                            • +1 Thank you for the one-line proof!
                              – DaG
                              Nov 17 at 10:01


















                            • Great explanation, but that “it's” is jarring...
                              – DaG
                              Nov 17 at 9:34










                            • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                              – AlexanderJ93
                              Nov 17 at 9:35










                            • +1 Thank you for the one-line proof!
                              – DaG
                              Nov 17 at 10:01
















                            Great explanation, but that “it's” is jarring...
                            – DaG
                            Nov 17 at 9:34




                            Great explanation, but that “it's” is jarring...
                            – DaG
                            Nov 17 at 9:34












                            Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            – AlexanderJ93
                            Nov 17 at 9:35




                            Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            – AlexanderJ93
                            Nov 17 at 9:35












                            +1 Thank you for the one-line proof!
                            – DaG
                            Nov 17 at 10:01




                            +1 Thank you for the one-line proof!
                            – DaG
                            Nov 17 at 10:01










                            up vote
                            22
                            down vote













                            In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.






                            share|cite|improve this answer

























                              up vote
                              22
                              down vote













                              In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.






                              share|cite|improve this answer























                                up vote
                                22
                                down vote










                                up vote
                                22
                                down vote









                                In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.






                                share|cite|improve this answer












                                In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 15 at 17:55









                                J.G.

                                19.4k21932




                                19.4k21932






















                                    up vote
                                    9
                                    down vote













                                    You can prove that no such rectangle exists as follows:



                                    Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$



                                    Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.



                                    The answer given, though arithmetically correct does not represent a real wall.





                                    I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.






                                    share|cite|improve this answer





















                                    • To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
                                      – Ilmari Karonen
                                      Nov 15 at 21:39






                                    • 1




                                      @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                      – Mark Bennet
                                      Nov 16 at 8:40

















                                    up vote
                                    9
                                    down vote













                                    You can prove that no such rectangle exists as follows:



                                    Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$



                                    Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.



                                    The answer given, though arithmetically correct does not represent a real wall.





                                    I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.






                                    share|cite|improve this answer





















                                    • To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
                                      – Ilmari Karonen
                                      Nov 15 at 21:39






                                    • 1




                                      @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                      – Mark Bennet
                                      Nov 16 at 8:40















                                    up vote
                                    9
                                    down vote










                                    up vote
                                    9
                                    down vote









                                    You can prove that no such rectangle exists as follows:



                                    Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$



                                    Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.



                                    The answer given, though arithmetically correct does not represent a real wall.





                                    I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.






                                    share|cite|improve this answer












                                    You can prove that no such rectangle exists as follows:



                                    Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$



                                    Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.



                                    The answer given, though arithmetically correct does not represent a real wall.





                                    I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 15 at 19:15









                                    Mark Bennet

                                    79.9k980178




                                    79.9k980178












                                    • To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
                                      – Ilmari Karonen
                                      Nov 15 at 21:39






                                    • 1




                                      @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                      – Mark Bennet
                                      Nov 16 at 8:40




















                                    • To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
                                      – Ilmari Karonen
                                      Nov 15 at 21:39






                                    • 1




                                      @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                      – Mark Bennet
                                      Nov 16 at 8:40


















                                    To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
                                    – Ilmari Karonen
                                    Nov 15 at 21:39




                                    To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
                                    – Ilmari Karonen
                                    Nov 15 at 21:39




                                    1




                                    1




                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    – Mark Bennet
                                    Nov 16 at 8:40






                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    – Mark Bennet
                                    Nov 16 at 8:40












                                    up vote
                                    6
                                    down vote













                                    Another PWW (noted by AlexanderJ93 and others):



                                    $hspace{5cm}$![enter image description here






                                    share|cite|improve this answer

























                                      up vote
                                      6
                                      down vote













                                      Another PWW (noted by AlexanderJ93 and others):



                                      $hspace{5cm}$![enter image description here






                                      share|cite|improve this answer























                                        up vote
                                        6
                                        down vote










                                        up vote
                                        6
                                        down vote









                                        Another PWW (noted by AlexanderJ93 and others):



                                        $hspace{5cm}$![enter image description here






                                        share|cite|improve this answer












                                        Another PWW (noted by AlexanderJ93 and others):



                                        $hspace{5cm}$![enter image description here







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Nov 17 at 10:14









                                        farruhota

                                        17.8k2736




                                        17.8k2736






















                                            up vote
                                            4
                                            down vote













                                            No. As others have said.



                                            What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.



                                            If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.



                                            If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:



                                            Total perimeter: 70
                                            Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)



                                            This should now give the solution of:




                                            • $I^2 + B^2 = 25^2 = 625 $

                                            • $2I + 2B = 70 $

                                            • $I + B = 35 $

                                            • $I^2 + 2IB + B^2 = 1,225 $

                                            • $2IB = 600 $


                                            • $IB = 300$ ,


                                            which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)



                                            Hope that helps!



                                            -Van






                                            share|cite|improve this answer

























                                              up vote
                                              4
                                              down vote













                                              No. As others have said.



                                              What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.



                                              If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.



                                              If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:



                                              Total perimeter: 70
                                              Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)



                                              This should now give the solution of:




                                              • $I^2 + B^2 = 25^2 = 625 $

                                              • $2I + 2B = 70 $

                                              • $I + B = 35 $

                                              • $I^2 + 2IB + B^2 = 1,225 $

                                              • $2IB = 600 $


                                              • $IB = 300$ ,


                                              which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)



                                              Hope that helps!



                                              -Van






                                              share|cite|improve this answer























                                                up vote
                                                4
                                                down vote










                                                up vote
                                                4
                                                down vote









                                                No. As others have said.



                                                What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.



                                                If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.



                                                If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:



                                                Total perimeter: 70
                                                Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)



                                                This should now give the solution of:




                                                • $I^2 + B^2 = 25^2 = 625 $

                                                • $2I + 2B = 70 $

                                                • $I + B = 35 $

                                                • $I^2 + 2IB + B^2 = 1,225 $

                                                • $2IB = 600 $


                                                • $IB = 300$ ,


                                                which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)



                                                Hope that helps!



                                                -Van






                                                share|cite|improve this answer












                                                No. As others have said.



                                                What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.



                                                If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.



                                                If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:



                                                Total perimeter: 70
                                                Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)



                                                This should now give the solution of:




                                                • $I^2 + B^2 = 25^2 = 625 $

                                                • $2I + 2B = 70 $

                                                • $I + B = 35 $

                                                • $I^2 + 2IB + B^2 = 1,225 $

                                                • $2IB = 600 $


                                                • $IB = 300$ ,


                                                which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)



                                                Hope that helps!



                                                -Van







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Nov 16 at 19:02









                                                Van

                                                413




                                                413






















                                                    up vote
                                                    3
                                                    down vote













                                                    No, use the Pythagorean Theorem.



                                                    $$c^2 = a^2+b^2$$



                                                    $c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).



                                                    Recall for any real number, its square must be non-negative.



                                                    $$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$



                                                    The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.



                                                    Now, to find the area itself.



                                                    For the diagonal:



                                                    $$c^2 = a^2+b^2$$



                                                    $$implies 18^2 = a^2+b^2$$



                                                    $$color{blue}{324 = a^2+b^2} tag{1}$$



                                                    For the perimeter:



                                                    $$2(a+b) = 72$$



                                                    $$a+b = 36$$



                                                    Now, define one variable in terms of the other.



                                                    $$color{purple}{a = 36-b} tag{2}$$



                                                    Combine $(1)$ and $(2)$.



                                                    $$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$



                                                    $$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$



                                                    But $$Delta = b^2-4ac$$



                                                    $$Delta = 72^2-4(2)(972) = -2592$$



                                                    $$implies Delta < 0$$



                                                    Thus, there is no solution. (No such rectangle exists.)






                                                    share|cite|improve this answer



























                                                      up vote
                                                      3
                                                      down vote













                                                      No, use the Pythagorean Theorem.



                                                      $$c^2 = a^2+b^2$$



                                                      $c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).



                                                      Recall for any real number, its square must be non-negative.



                                                      $$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$



                                                      The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.



                                                      Now, to find the area itself.



                                                      For the diagonal:



                                                      $$c^2 = a^2+b^2$$



                                                      $$implies 18^2 = a^2+b^2$$



                                                      $$color{blue}{324 = a^2+b^2} tag{1}$$



                                                      For the perimeter:



                                                      $$2(a+b) = 72$$



                                                      $$a+b = 36$$



                                                      Now, define one variable in terms of the other.



                                                      $$color{purple}{a = 36-b} tag{2}$$



                                                      Combine $(1)$ and $(2)$.



                                                      $$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$



                                                      $$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$



                                                      But $$Delta = b^2-4ac$$



                                                      $$Delta = 72^2-4(2)(972) = -2592$$



                                                      $$implies Delta < 0$$



                                                      Thus, there is no solution. (No such rectangle exists.)






                                                      share|cite|improve this answer

























                                                        up vote
                                                        3
                                                        down vote










                                                        up vote
                                                        3
                                                        down vote









                                                        No, use the Pythagorean Theorem.



                                                        $$c^2 = a^2+b^2$$



                                                        $c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).



                                                        Recall for any real number, its square must be non-negative.



                                                        $$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$



                                                        The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.



                                                        Now, to find the area itself.



                                                        For the diagonal:



                                                        $$c^2 = a^2+b^2$$



                                                        $$implies 18^2 = a^2+b^2$$



                                                        $$color{blue}{324 = a^2+b^2} tag{1}$$



                                                        For the perimeter:



                                                        $$2(a+b) = 72$$



                                                        $$a+b = 36$$



                                                        Now, define one variable in terms of the other.



                                                        $$color{purple}{a = 36-b} tag{2}$$



                                                        Combine $(1)$ and $(2)$.



                                                        $$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$



                                                        $$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$



                                                        But $$Delta = b^2-4ac$$



                                                        $$Delta = 72^2-4(2)(972) = -2592$$



                                                        $$implies Delta < 0$$



                                                        Thus, there is no solution. (No such rectangle exists.)






                                                        share|cite|improve this answer














                                                        No, use the Pythagorean Theorem.



                                                        $$c^2 = a^2+b^2$$



                                                        $c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).



                                                        Recall for any real number, its square must be non-negative.



                                                        $$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$



                                                        The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.



                                                        Now, to find the area itself.



                                                        For the diagonal:



                                                        $$c^2 = a^2+b^2$$



                                                        $$implies 18^2 = a^2+b^2$$



                                                        $$color{blue}{324 = a^2+b^2} tag{1}$$



                                                        For the perimeter:



                                                        $$2(a+b) = 72$$



                                                        $$a+b = 36$$



                                                        Now, define one variable in terms of the other.



                                                        $$color{purple}{a = 36-b} tag{2}$$



                                                        Combine $(1)$ and $(2)$.



                                                        $$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$



                                                        $$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$



                                                        But $$Delta = b^2-4ac$$



                                                        $$Delta = 72^2-4(2)(972) = -2592$$



                                                        $$implies Delta < 0$$



                                                        Thus, there is no solution. (No such rectangle exists.)







                                                        share|cite|improve this answer














                                                        share|cite|improve this answer



                                                        share|cite|improve this answer








                                                        edited Nov 15 at 19:32

























                                                        answered Nov 15 at 18:10









                                                        KM101

                                                        2,742416




                                                        2,742416






















                                                            up vote
                                                            2
                                                            down vote













                                                            No. Using Pythagoras and a simple inequality we get
                                                            $$d^2=a^2+b^2geq 2abgeq ab$$
                                                            If $a,b$ are the sides and $d$ the diagonal






                                                            share|cite|improve this answer

























                                                              up vote
                                                              2
                                                              down vote













                                                              No. Using Pythagoras and a simple inequality we get
                                                              $$d^2=a^2+b^2geq 2abgeq ab$$
                                                              If $a,b$ are the sides and $d$ the diagonal






                                                              share|cite|improve this answer























                                                                up vote
                                                                2
                                                                down vote










                                                                up vote
                                                                2
                                                                down vote









                                                                No. Using Pythagoras and a simple inequality we get
                                                                $$d^2=a^2+b^2geq 2abgeq ab$$
                                                                If $a,b$ are the sides and $d$ the diagonal






                                                                share|cite|improve this answer












                                                                No. Using Pythagoras and a simple inequality we get
                                                                $$d^2=a^2+b^2geq 2abgeq ab$$
                                                                If $a,b$ are the sides and $d$ the diagonal







                                                                share|cite|improve this answer












                                                                share|cite|improve this answer



                                                                share|cite|improve this answer










                                                                answered Nov 15 at 17:55









                                                                b00n heT

                                                                10.2k12134




                                                                10.2k12134






















                                                                    up vote
                                                                    2
                                                                    down vote













                                                                    Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):



                                                                    enter image description here






                                                                    share|cite|improve this answer

























                                                                      up vote
                                                                      2
                                                                      down vote













                                                                      Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):



                                                                      enter image description here






                                                                      share|cite|improve this answer























                                                                        up vote
                                                                        2
                                                                        down vote










                                                                        up vote
                                                                        2
                                                                        down vote









                                                                        Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):



                                                                        enter image description here






                                                                        share|cite|improve this answer












                                                                        Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):



                                                                        enter image description here







                                                                        share|cite|improve this answer












                                                                        share|cite|improve this answer



                                                                        share|cite|improve this answer










                                                                        answered Nov 17 at 21:18









                                                                        Hagen von Eitzen

                                                                        274k21266494




                                                                        274k21266494






















                                                                            up vote
                                                                            0
                                                                            down vote













                                                                            $A=lw$



                                                                            $P=2(l+w)$



                                                                            $d=sqrt{l^2+w^2}$



                                                                            Can $A>d^2$?



                                                                            Can $lw>l^2+w^2$?



                                                                            $-lw>l^2-2wl+w^2=(l-w)^2$



                                                                            Width and length are necessarily positive. The square of their difference also must be positive.



                                                                            So we have a negative number that must be greater than a positive number. A contradiction.






                                                                            share|cite|improve this answer

























                                                                              up vote
                                                                              0
                                                                              down vote













                                                                              $A=lw$



                                                                              $P=2(l+w)$



                                                                              $d=sqrt{l^2+w^2}$



                                                                              Can $A>d^2$?



                                                                              Can $lw>l^2+w^2$?



                                                                              $-lw>l^2-2wl+w^2=(l-w)^2$



                                                                              Width and length are necessarily positive. The square of their difference also must be positive.



                                                                              So we have a negative number that must be greater than a positive number. A contradiction.






                                                                              share|cite|improve this answer























                                                                                up vote
                                                                                0
                                                                                down vote










                                                                                up vote
                                                                                0
                                                                                down vote









                                                                                $A=lw$



                                                                                $P=2(l+w)$



                                                                                $d=sqrt{l^2+w^2}$



                                                                                Can $A>d^2$?



                                                                                Can $lw>l^2+w^2$?



                                                                                $-lw>l^2-2wl+w^2=(l-w)^2$



                                                                                Width and length are necessarily positive. The square of their difference also must be positive.



                                                                                So we have a negative number that must be greater than a positive number. A contradiction.






                                                                                share|cite|improve this answer












                                                                                $A=lw$



                                                                                $P=2(l+w)$



                                                                                $d=sqrt{l^2+w^2}$



                                                                                Can $A>d^2$?



                                                                                Can $lw>l^2+w^2$?



                                                                                $-lw>l^2-2wl+w^2=(l-w)^2$



                                                                                Width and length are necessarily positive. The square of their difference also must be positive.



                                                                                So we have a negative number that must be greater than a positive number. A contradiction.







                                                                                share|cite|improve this answer












                                                                                share|cite|improve this answer



                                                                                share|cite|improve this answer










                                                                                answered Nov 15 at 20:05









                                                                                TurlocTheRed

                                                                                788211




                                                                                788211






















                                                                                    up vote
                                                                                    0
                                                                                    down vote













                                                                                    A wall has a thickness.



                                                                                    Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.



                                                                                    Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.



                                                                                    The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.



                                                                                    We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.



                                                                                    This t can be used to calculate the area of the wall as (72 - 4t) * t.



                                                                                    Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.






                                                                                    share|cite|improve this answer

























                                                                                      up vote
                                                                                      0
                                                                                      down vote













                                                                                      A wall has a thickness.



                                                                                      Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.



                                                                                      Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.



                                                                                      The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.



                                                                                      We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.



                                                                                      This t can be used to calculate the area of the wall as (72 - 4t) * t.



                                                                                      Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.






                                                                                      share|cite|improve this answer























                                                                                        up vote
                                                                                        0
                                                                                        down vote










                                                                                        up vote
                                                                                        0
                                                                                        down vote









                                                                                        A wall has a thickness.



                                                                                        Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.



                                                                                        Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.



                                                                                        The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.



                                                                                        We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.



                                                                                        This t can be used to calculate the area of the wall as (72 - 4t) * t.



                                                                                        Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.






                                                                                        share|cite|improve this answer












                                                                                        A wall has a thickness.



                                                                                        Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.



                                                                                        Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.



                                                                                        The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.



                                                                                        We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.



                                                                                        This t can be used to calculate the area of the wall as (72 - 4t) * t.



                                                                                        Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.







                                                                                        share|cite|improve this answer












                                                                                        share|cite|improve this answer



                                                                                        share|cite|improve this answer










                                                                                        answered Nov 16 at 23:28









                                                                                        gnasher729

                                                                                        5,9411028




                                                                                        5,9411028















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