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











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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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  • 21




    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
    2 days ago








  • 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
    yesterday








  • 6




    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
    yesterday






  • 7




    The problem with the question is that if you solve for the side lengths you get complex numbers.
    – 1123581321
    yesterday






  • 8




    @1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
    – Teepeemm
    yesterday















up vote
33
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









New contributor




user17838 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 21




    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
    2 days ago








  • 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
    yesterday








  • 6




    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
    yesterday






  • 7




    The problem with the question is that if you solve for the side lengths you get complex numbers.
    – 1123581321
    yesterday






  • 8




    @1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
    – Teepeemm
    yesterday













up vote
33
down vote

favorite
5









up vote
33
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









New contributor




user17838 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












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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edited 2 days ago





















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  • 21




    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
    2 days ago








  • 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
    yesterday








  • 6




    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
    yesterday






  • 7




    The problem with the question is that if you solve for the side lengths you get complex numbers.
    – 1123581321
    yesterday






  • 8




    @1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
    – Teepeemm
    yesterday














  • 21




    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
    2 days ago








  • 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
    yesterday








  • 6




    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
    yesterday






  • 7




    The problem with the question is that if you solve for the side lengths you get complex numbers.
    – 1123581321
    yesterday






  • 8




    @1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
    – Teepeemm
    yesterday








21




21




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
2 days ago






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
2 days ago






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
yesterday






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
yesterday






6




6




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
yesterday




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
yesterday




7




7




The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
yesterday




The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
yesterday




8




8




@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
yesterday




@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
yesterday










11 Answers
11






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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






share|cite|improve this answer




























    up vote
    65
    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
      yesterday






    • 1




      Here is a more dynamic, animated version of the same picture.
      – Xander Henderson
      yesterday


















    up vote
    23
    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
      10 hours ago










    • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
      – AlexanderJ93
      10 hours ago










    • +1 Thank you for the one-line proof!
      – DaG
      9 hours ago


















    up vote
    20
    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
      8
      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
        yesterday






      • 1




        @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
        – Mark Bennet
        yesterday




















      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
        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













          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













            Another PWW (noted by AlexanderJ93 and others):



            $hspace{5cm}$![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





















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                  11 Answers
                  11






                  active

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                  11 Answers
                  11






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                  active

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                  active

                  oldest

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                  up vote
                  69
                  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
                    69
                    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
                      69
                      down vote



                      accepted







                      up vote
                      69
                      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 2 days ago









                      Théophile

                      19.1k12944




                      19.1k12944






















                          up vote
                          65
                          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
                            yesterday






                          • 1




                            Here is a more dynamic, animated version of the same picture.
                            – Xander Henderson
                            yesterday















                          up vote
                          65
                          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
                            yesterday






                          • 1




                            Here is a more dynamic, animated version of the same picture.
                            – Xander Henderson
                            yesterday













                          up vote
                          65
                          down vote










                          up vote
                          65
                          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 2 days ago


























                          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
                            yesterday






                          • 1




                            Here is a more dynamic, animated version of the same picture.
                            – Xander Henderson
                            yesterday


















                          • +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
                            yesterday






                          • 1




                            Here is a more dynamic, animated version of the same picture.
                            – Xander Henderson
                            yesterday
















                          +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
                          yesterday




                          +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
                          yesterday




                          1




                          1




                          Here is a more dynamic, animated version of the same picture.
                          – Xander Henderson
                          yesterday




                          Here is a more dynamic, animated version of the same picture.
                          – Xander Henderson
                          yesterday










                          up vote
                          23
                          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
                            10 hours ago










                          • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            – AlexanderJ93
                            10 hours ago










                          • +1 Thank you for the one-line proof!
                            – DaG
                            9 hours ago















                          up vote
                          23
                          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
                            10 hours ago










                          • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            – AlexanderJ93
                            10 hours ago










                          • +1 Thank you for the one-line proof!
                            – DaG
                            9 hours ago













                          up vote
                          23
                          down vote










                          up vote
                          23
                          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 10 hours ago

























                          answered yesterday









                          AlexanderJ93

                          5,127422




                          5,127422












                          • Great explanation, but that “it's” is jarring...
                            – DaG
                            10 hours ago










                          • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            – AlexanderJ93
                            10 hours ago










                          • +1 Thank you for the one-line proof!
                            – DaG
                            9 hours ago


















                          • Great explanation, but that “it's” is jarring...
                            – DaG
                            10 hours ago










                          • Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            – AlexanderJ93
                            10 hours ago










                          • +1 Thank you for the one-line proof!
                            – DaG
                            9 hours ago
















                          Great explanation, but that “it's” is jarring...
                          – DaG
                          10 hours ago




                          Great explanation, but that “it's” is jarring...
                          – DaG
                          10 hours ago












                          Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                          – AlexanderJ93
                          10 hours ago




                          Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                          – AlexanderJ93
                          10 hours ago












                          +1 Thank you for the one-line proof!
                          – DaG
                          9 hours ago




                          +1 Thank you for the one-line proof!
                          – DaG
                          9 hours ago










                          up vote
                          20
                          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
                            20
                            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
                              20
                              down vote










                              up vote
                              20
                              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 2 days ago









                              J.G.

                              17.9k11830




                              17.9k11830






















                                  up vote
                                  8
                                  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
                                    yesterday






                                  • 1




                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    – Mark Bennet
                                    yesterday

















                                  up vote
                                  8
                                  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
                                    yesterday






                                  • 1




                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    – Mark Bennet
                                    yesterday















                                  up vote
                                  8
                                  down vote










                                  up vote
                                  8
                                  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 2 days ago









                                  Mark Bennet

                                  79.5k978177




                                  79.5k978177












                                  • 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
                                    yesterday






                                  • 1




                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    – Mark Bennet
                                    yesterday




















                                  • 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
                                    yesterday






                                  • 1




                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    – Mark Bennet
                                    yesterday


















                                  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
                                  yesterday




                                  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
                                  yesterday




                                  1




                                  1




                                  @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                  – Mark Bennet
                                  yesterday






                                  @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                  – Mark Bennet
                                  yesterday












                                  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








                                  New contributor




                                  Van is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.






















                                    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








                                    New contributor




                                    Van is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.




















                                      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








                                      New contributor




                                      Van is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      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








                                      New contributor




                                      Van is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      share|cite|improve this answer



                                      share|cite|improve this answer






                                      New contributor




                                      Van is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.









                                      answered yesterday









                                      Van

                                      413




                                      413




                                      New contributor




                                      Van is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.





                                      New contributor





                                      Van is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






                                      Van is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






















                                          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 2 days ago









                                              b00n heT

                                              10.1k12134




                                              10.1k12134






















                                                  up vote
                                                  2
                                                  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, 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










                                                      up vote
                                                      2
                                                      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 2 days ago

























                                                      answered 2 days ago









                                                      KM101

                                                      1,765313




                                                      1,765313






















                                                          up vote
                                                          2
                                                          down vote













                                                          Another PWW (noted by AlexanderJ93 and others):



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






                                                          share|cite|improve this answer

























                                                            up vote
                                                            2
                                                            down vote













                                                            Another PWW (noted by AlexanderJ93 and others):



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






                                                            share|cite|improve this answer























                                                              up vote
                                                              2
                                                              down vote










                                                              up vote
                                                              2
                                                              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 9 hours ago









                                                              farruhota

                                                              17.4k2736




                                                              17.4k2736






















                                                                  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 yesterday









                                                                      TurlocTheRed

                                                                      55819




                                                                      55819






















                                                                          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 20 hours ago









                                                                              gnasher729

                                                                              5,9511028




                                                                              5,9511028






















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