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 ?











share|cite|improve this question











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  • 22




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    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.
    $endgroup$
    – Théophile
    Nov 15 '18 at 19:09








  • 8




    $begingroup$
    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???
    $endgroup$
    – alephzero
    Nov 15 '18 at 20:16








  • 7




    $begingroup$
    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!
    $endgroup$
    – amalloy
    Nov 15 '18 at 22:37






  • 7




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    The problem with the question is that if you solve for the side lengths you get complex numbers.
    $endgroup$
    – 1123581321
    Nov 15 '18 at 22:53






  • 9




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
















46












$begingroup$



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











$endgroup$








  • 22




    $begingroup$
    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.
    $endgroup$
    – Théophile
    Nov 15 '18 at 19:09








  • 8




    $begingroup$
    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???
    $endgroup$
    – alephzero
    Nov 15 '18 at 20:16








  • 7




    $begingroup$
    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!
    $endgroup$
    – amalloy
    Nov 15 '18 at 22:37






  • 7




    $begingroup$
    The problem with the question is that if you solve for the side lengths you get complex numbers.
    $endgroup$
    – 1123581321
    Nov 15 '18 at 22:53






  • 9




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














46












46








46


7



$begingroup$



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











$endgroup$





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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share|cite|improve this question








edited Nov 15 '18 at 18:28







user17838

















asked Nov 15 '18 at 17:49









user17838user17838

34126




34126








  • 22




    $begingroup$
    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.
    $endgroup$
    – Théophile
    Nov 15 '18 at 19:09








  • 8




    $begingroup$
    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???
    $endgroup$
    – alephzero
    Nov 15 '18 at 20:16








  • 7




    $begingroup$
    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!
    $endgroup$
    – amalloy
    Nov 15 '18 at 22:37






  • 7




    $begingroup$
    The problem with the question is that if you solve for the side lengths you get complex numbers.
    $endgroup$
    – 1123581321
    Nov 15 '18 at 22:53






  • 9




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














  • 22




    $begingroup$
    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.
    $endgroup$
    – Théophile
    Nov 15 '18 at 19:09








  • 8




    $begingroup$
    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???
    $endgroup$
    – alephzero
    Nov 15 '18 at 20:16








  • 7




    $begingroup$
    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!
    $endgroup$
    – amalloy
    Nov 15 '18 at 22:37






  • 7




    $begingroup$
    The problem with the question is that if you solve for the side lengths you get complex numbers.
    $endgroup$
    – 1123581321
    Nov 15 '18 at 22:53






  • 9




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








22




22




$begingroup$
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.
$endgroup$
– Théophile
Nov 15 '18 at 19:09






$begingroup$
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.
$endgroup$
– Théophile
Nov 15 '18 at 19:09






8




8




$begingroup$
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???
$endgroup$
– alephzero
Nov 15 '18 at 20:16






$begingroup$
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???
$endgroup$
– alephzero
Nov 15 '18 at 20:16






7




7




$begingroup$
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!
$endgroup$
– amalloy
Nov 15 '18 at 22:37




$begingroup$
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!
$endgroup$
– amalloy
Nov 15 '18 at 22:37




7




7




$begingroup$
The problem with the question is that if you solve for the side lengths you get complex numbers.
$endgroup$
– 1123581321
Nov 15 '18 at 22:53




$begingroup$
The problem with the question is that if you solve for the side lengths you get complex numbers.
$endgroup$
– 1123581321
Nov 15 '18 at 22:53




9




9




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




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










12 Answers
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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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    74












    $begingroup$

    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











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    • $begingroup$
      +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.
      $endgroup$
      – mckenzm
      Nov 15 '18 at 23:26






    • 1




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      Here is a more dynamic, animated version of the same picture.
      $endgroup$
      – Xander Henderson
      Nov 16 '18 at 4:44



















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    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











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    • $begingroup$
      Great explanation, but that “it's” is jarring...
      $endgroup$
      – DaG
      Nov 17 '18 at 9:34










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      Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
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      – AlexanderJ93
      Nov 17 '18 at 9:35










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      +1 Thank you for the one-line proof!
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      – DaG
      Nov 17 '18 at 10:01



















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    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 $(a-b)^2ge 0$.






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      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









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      • $begingroup$
        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 $b$.
        $endgroup$
        – Ilmari Karonen
        Nov 15 '18 at 21:39






      • 1




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        @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
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        – Mark Bennet
        Nov 16 '18 at 8:40





















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      Another PWW (noted by AlexanderJ93 and others):



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






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        4












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        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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          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











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            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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              2












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              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









              $endgroup$





















                0












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                $A=lw$



                $P=2(l+w)$



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



                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









                $endgroup$





















                  -1












                  $begingroup$

                  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









                  $endgroup$













                  • $begingroup$
                    A wall is a 2-dimensional object.
                    $endgroup$
                    – Jossie Calderon
                    Feb 14 at 18:46











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                  12 Answers
                  12






                  active

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                  12 Answers
                  12






                  active

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                  active

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                  active

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                  88












                  $begingroup$

                  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









                  $endgroup$


















                    88












                    $begingroup$

                    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









                    $endgroup$
















                      88












                      88








                      88





                      $begingroup$

                      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









                      $endgroup$



                      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 '18 at 18:04









                      ThéophileThéophile

                      20.2k13047




                      20.2k13047























                          74












                          $begingroup$

                          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











                          $endgroup$













                          • $begingroup$
                            +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.
                            $endgroup$
                            – mckenzm
                            Nov 15 '18 at 23:26






                          • 1




                            $begingroup$
                            Here is a more dynamic, animated version of the same picture.
                            $endgroup$
                            – Xander Henderson
                            Nov 16 '18 at 4:44
















                          74












                          $begingroup$

                          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











                          $endgroup$













                          • $begingroup$
                            +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.
                            $endgroup$
                            – mckenzm
                            Nov 15 '18 at 23:26






                          • 1




                            $begingroup$
                            Here is a more dynamic, animated version of the same picture.
                            $endgroup$
                            – Xander Henderson
                            Nov 16 '18 at 4:44














                          74












                          74








                          74





                          $begingroup$

                          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











                          $endgroup$



                          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 '18 at 19:01


























                          community wiki





                          Xander Henderson













                          • $begingroup$
                            +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.
                            $endgroup$
                            – mckenzm
                            Nov 15 '18 at 23:26






                          • 1




                            $begingroup$
                            Here is a more dynamic, animated version of the same picture.
                            $endgroup$
                            – Xander Henderson
                            Nov 16 '18 at 4:44


















                          • $begingroup$
                            +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.
                            $endgroup$
                            – mckenzm
                            Nov 15 '18 at 23:26






                          • 1




                            $begingroup$
                            Here is a more dynamic, animated version of the same picture.
                            $endgroup$
                            – Xander Henderson
                            Nov 16 '18 at 4:44
















                          $begingroup$
                          +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.
                          $endgroup$
                          – mckenzm
                          Nov 15 '18 at 23:26




                          $begingroup$
                          +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.
                          $endgroup$
                          – mckenzm
                          Nov 15 '18 at 23:26




                          1




                          1




                          $begingroup$
                          Here is a more dynamic, animated version of the same picture.
                          $endgroup$
                          – Xander Henderson
                          Nov 16 '18 at 4:44




                          $begingroup$
                          Here is a more dynamic, animated version of the same picture.
                          $endgroup$
                          – Xander Henderson
                          Nov 16 '18 at 4:44











                          32












                          $begingroup$

                          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











                          $endgroup$













                          • $begingroup$
                            Great explanation, but that “it's” is jarring...
                            $endgroup$
                            – DaG
                            Nov 17 '18 at 9:34










                          • $begingroup$
                            Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            $endgroup$
                            – AlexanderJ93
                            Nov 17 '18 at 9:35










                          • $begingroup$
                            +1 Thank you for the one-line proof!
                            $endgroup$
                            – DaG
                            Nov 17 '18 at 10:01
















                          32












                          $begingroup$

                          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











                          $endgroup$













                          • $begingroup$
                            Great explanation, but that “it's” is jarring...
                            $endgroup$
                            – DaG
                            Nov 17 '18 at 9:34










                          • $begingroup$
                            Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            $endgroup$
                            – AlexanderJ93
                            Nov 17 '18 at 9:35










                          • $begingroup$
                            +1 Thank you for the one-line proof!
                            $endgroup$
                            – DaG
                            Nov 17 '18 at 10:01














                          32












                          32








                          32





                          $begingroup$

                          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











                          $endgroup$



                          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 '18 at 9:34

























                          answered Nov 15 '18 at 20:21









                          AlexanderJ93AlexanderJ93

                          6,193823




                          6,193823












                          • $begingroup$
                            Great explanation, but that “it's” is jarring...
                            $endgroup$
                            – DaG
                            Nov 17 '18 at 9:34










                          • $begingroup$
                            Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            $endgroup$
                            – AlexanderJ93
                            Nov 17 '18 at 9:35










                          • $begingroup$
                            +1 Thank you for the one-line proof!
                            $endgroup$
                            – DaG
                            Nov 17 '18 at 10:01


















                          • $begingroup$
                            Great explanation, but that “it's” is jarring...
                            $endgroup$
                            – DaG
                            Nov 17 '18 at 9:34










                          • $begingroup$
                            Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                            $endgroup$
                            – AlexanderJ93
                            Nov 17 '18 at 9:35










                          • $begingroup$
                            +1 Thank you for the one-line proof!
                            $endgroup$
                            – DaG
                            Nov 17 '18 at 10:01
















                          $begingroup$
                          Great explanation, but that “it's” is jarring...
                          $endgroup$
                          – DaG
                          Nov 17 '18 at 9:34




                          $begingroup$
                          Great explanation, but that “it's” is jarring...
                          $endgroup$
                          – DaG
                          Nov 17 '18 at 9:34












                          $begingroup$
                          Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                          $endgroup$
                          – AlexanderJ93
                          Nov 17 '18 at 9:35




                          $begingroup$
                          Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
                          $endgroup$
                          – AlexanderJ93
                          Nov 17 '18 at 9:35












                          $begingroup$
                          +1 Thank you for the one-line proof!
                          $endgroup$
                          – DaG
                          Nov 17 '18 at 10:01




                          $begingroup$
                          +1 Thank you for the one-line proof!
                          $endgroup$
                          – DaG
                          Nov 17 '18 at 10:01











                          22












                          $begingroup$

                          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









                          $endgroup$


















                            22












                            $begingroup$

                            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









                            $endgroup$
















                              22












                              22








                              22





                              $begingroup$

                              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









                              $endgroup$



                              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 '18 at 17:55









                              J.G.J.G.

                              30.8k23149




                              30.8k23149























                                  9












                                  $begingroup$

                                  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









                                  $endgroup$













                                  • $begingroup$
                                    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 $b$.
                                    $endgroup$
                                    – Ilmari Karonen
                                    Nov 15 '18 at 21:39






                                  • 1




                                    $begingroup$
                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    $endgroup$
                                    – Mark Bennet
                                    Nov 16 '18 at 8:40


















                                  9












                                  $begingroup$

                                  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









                                  $endgroup$













                                  • $begingroup$
                                    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 $b$.
                                    $endgroup$
                                    – Ilmari Karonen
                                    Nov 15 '18 at 21:39






                                  • 1




                                    $begingroup$
                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    $endgroup$
                                    – Mark Bennet
                                    Nov 16 '18 at 8:40
















                                  9












                                  9








                                  9





                                  $begingroup$

                                  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









                                  $endgroup$



                                  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 '18 at 19:15









                                  Mark BennetMark Bennet

                                  81.6k984181




                                  81.6k984181












                                  • $begingroup$
                                    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 $b$.
                                    $endgroup$
                                    – Ilmari Karonen
                                    Nov 15 '18 at 21:39






                                  • 1




                                    $begingroup$
                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    $endgroup$
                                    – Mark Bennet
                                    Nov 16 '18 at 8:40




















                                  • $begingroup$
                                    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 $b$.
                                    $endgroup$
                                    – Ilmari Karonen
                                    Nov 15 '18 at 21:39






                                  • 1




                                    $begingroup$
                                    @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                    $endgroup$
                                    – Mark Bennet
                                    Nov 16 '18 at 8:40


















                                  $begingroup$
                                  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 $b$.
                                  $endgroup$
                                  – Ilmari Karonen
                                  Nov 15 '18 at 21:39




                                  $begingroup$
                                  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 $b$.
                                  $endgroup$
                                  – Ilmari Karonen
                                  Nov 15 '18 at 21:39




                                  1




                                  1




                                  $begingroup$
                                  @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                  $endgroup$
                                  – Mark Bennet
                                  Nov 16 '18 at 8:40






                                  $begingroup$
                                  @IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
                                  $endgroup$
                                  – Mark Bennet
                                  Nov 16 '18 at 8:40













                                  6












                                  $begingroup$

                                  Another PWW (noted by AlexanderJ93 and others):



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






                                  share|cite|improve this answer









                                  $endgroup$


















                                    6












                                    $begingroup$

                                    Another PWW (noted by AlexanderJ93 and others):



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






                                    share|cite|improve this answer









                                    $endgroup$
















                                      6












                                      6








                                      6





                                      $begingroup$

                                      Another PWW (noted by AlexanderJ93 and others):



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






                                      share|cite|improve this answer









                                      $endgroup$



                                      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 '18 at 10:14









                                      farruhotafarruhota

                                      21.3k2841




                                      21.3k2841























                                          4












                                          $begingroup$

                                          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









                                          $endgroup$


















                                            4












                                            $begingroup$

                                            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









                                            $endgroup$
















                                              4












                                              4








                                              4





                                              $begingroup$

                                              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









                                              $endgroup$



                                              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 '18 at 19:02









                                              VanVan

                                              1413




                                              1413























                                                  3












                                                  $begingroup$

                                                  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











                                                  $endgroup$


















                                                    3












                                                    $begingroup$

                                                    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











                                                    $endgroup$
















                                                      3












                                                      3








                                                      3





                                                      $begingroup$

                                                      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











                                                      $endgroup$



                                                      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 '18 at 19:32

























                                                      answered Nov 15 '18 at 18:10









                                                      KM101KM101

                                                      6,0901525




                                                      6,0901525























                                                          2












                                                          $begingroup$

                                                          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









                                                          $endgroup$


















                                                            2












                                                            $begingroup$

                                                            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









                                                            $endgroup$
















                                                              2












                                                              2








                                                              2





                                                              $begingroup$

                                                              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









                                                              $endgroup$



                                                              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 '18 at 17:55









                                                              b00n heTb00n heT

                                                              10.5k12235




                                                              10.5k12235























                                                                  2












                                                                  $begingroup$

                                                                  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









                                                                  $endgroup$


















                                                                    2












                                                                    $begingroup$

                                                                    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









                                                                    $endgroup$
















                                                                      2












                                                                      2








                                                                      2





                                                                      $begingroup$

                                                                      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









                                                                      $endgroup$



                                                                      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 '18 at 21:18









                                                                      Hagen von EitzenHagen von Eitzen

                                                                      283k23272507




                                                                      283k23272507























                                                                          0












                                                                          $begingroup$

                                                                          $A=lw$



                                                                          $P=2(l+w)$



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



                                                                          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









                                                                          $endgroup$


















                                                                            0












                                                                            $begingroup$

                                                                            $A=lw$



                                                                            $P=2(l+w)$



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



                                                                            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









                                                                            $endgroup$
















                                                                              0












                                                                              0








                                                                              0





                                                                              $begingroup$

                                                                              $A=lw$



                                                                              $P=2(l+w)$



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



                                                                              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









                                                                              $endgroup$



                                                                              $A=lw$



                                                                              $P=2(l+w)$



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



                                                                              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 '18 at 20:05









                                                                              TurlocTheRedTurlocTheRed

                                                                              916311




                                                                              916311























                                                                                  -1












                                                                                  $begingroup$

                                                                                  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









                                                                                  $endgroup$













                                                                                  • $begingroup$
                                                                                    A wall is a 2-dimensional object.
                                                                                    $endgroup$
                                                                                    – Jossie Calderon
                                                                                    Feb 14 at 18:46
















                                                                                  -1












                                                                                  $begingroup$

                                                                                  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









                                                                                  $endgroup$













                                                                                  • $begingroup$
                                                                                    A wall is a 2-dimensional object.
                                                                                    $endgroup$
                                                                                    – Jossie Calderon
                                                                                    Feb 14 at 18:46














                                                                                  -1












                                                                                  -1








                                                                                  -1





                                                                                  $begingroup$

                                                                                  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









                                                                                  $endgroup$



                                                                                  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 '18 at 23:28









                                                                                  gnasher729gnasher729

                                                                                  6,1001028




                                                                                  6,1001028












                                                                                  • $begingroup$
                                                                                    A wall is a 2-dimensional object.
                                                                                    $endgroup$
                                                                                    – Jossie Calderon
                                                                                    Feb 14 at 18:46


















                                                                                  • $begingroup$
                                                                                    A wall is a 2-dimensional object.
                                                                                    $endgroup$
                                                                                    – Jossie Calderon
                                                                                    Feb 14 at 18:46
















                                                                                  $begingroup$
                                                                                  A wall is a 2-dimensional object.
                                                                                  $endgroup$
                                                                                  – Jossie Calderon
                                                                                  Feb 14 at 18:46




                                                                                  $begingroup$
                                                                                  A wall is a 2-dimensional object.
                                                                                  $endgroup$
                                                                                  – Jossie Calderon
                                                                                  Feb 14 at 18:46


















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