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
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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show 11 more comments
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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
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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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.
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– Théophile
Nov 15 '18 at 19:09
8
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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???
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– alephzero
Nov 15 '18 at 20:16
7
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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!
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– 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.
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– 1123581321
Nov 15 '18 at 22:53
9
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@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
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– Teepeemm
Nov 16 '18 at 2:10
|
show 11 more comments
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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
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
$endgroup$
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
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
geometry area
edited Nov 15 '18 at 18:28
user17838
asked Nov 15 '18 at 17:49
user17838user17838
34126
34126
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
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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???
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– alephzero
Nov 15 '18 at 20:16
7
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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.
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– 1123581321
Nov 15 '18 at 22:53
9
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@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
$endgroup$
– Teepeemm
Nov 16 '18 at 2:10
|
show 11 more comments
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
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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
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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
$begingroup$
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
|
show 11 more comments
12 Answers
12
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oldest
votes
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The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
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add a comment |
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Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
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+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.
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– mckenzm
Nov 15 '18 at 23:26
1
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Here is a more dynamic, animated version of the same picture.
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– Xander Henderson
Nov 16 '18 at 4:44
add a comment |
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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.
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Great explanation, but that “it's” is jarring...
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– 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
add a comment |
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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.
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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 $b$.
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– 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):
$hspace{5cm}$
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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.)
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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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Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):
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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.
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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.
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A wall is a 2-dimensional object.
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– Jossie Calderon
Feb 14 at 18:46
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12 Answers
12
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12 Answers
12
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The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
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add a comment |
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The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
$endgroup$
add a comment |
$begingroup$
The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
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The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
answered Nov 15 '18 at 18:04
ThéophileThéophile
20.2k13047
20.2k13047
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add a comment |
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Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
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+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.
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– mckenzm
Nov 15 '18 at 23:26
1
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Here is a more dynamic, animated version of the same picture.
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– Xander Henderson
Nov 16 '18 at 4:44
add a comment |
$begingroup$
Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
$endgroup$
$begingroup$
+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
add a comment |
$begingroup$
Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
$endgroup$
Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
answered Nov 15 '18 at 19:01
community wiki
Xander Henderson
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+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
add a comment |
$begingroup$
+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
add a comment |
$begingroup$
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.
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Great explanation, but that “it's” is jarring...
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– 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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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!
$endgroup$
– AlexanderJ93
Nov 17 '18 at 9:35
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+1 Thank you for the one-line proof!
$endgroup$
– DaG
Nov 17 '18 at 10:01
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Nov 17 '18 at 9:34
answered Nov 15 '18 at 20:21
AlexanderJ93AlexanderJ93
6,193823
6,193823
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Great explanation, but that “it's” is jarring...
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– 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!
$endgroup$
– DaG
Nov 17 '18 at 10:01
add a comment |
$begingroup$
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
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+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
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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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add a comment |
$begingroup$
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 $(a-b)^2ge 0$.
$endgroup$
add a comment |
$begingroup$
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 $(a-b)^2ge 0$.
$endgroup$
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 $(a-b)^2ge 0$.
answered Nov 15 '18 at 17:55
J.G.J.G.
30.8k23149
30.8k23149
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$begingroup$
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.
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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 $b$.
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– 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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Nov 15 '18 at 19:15
Mark BennetMark Bennet
81.6k984181
81.6k984181
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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 $b$.
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– 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
add a comment |
$begingroup$
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.
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– Mark Bennet
Nov 16 '18 at 8:40
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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
add a comment |
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Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
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add a comment |
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Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
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add a comment |
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Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
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Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
answered Nov 17 '18 at 10:14
farruhotafarruhota
21.3k2841
21.3k2841
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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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add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
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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
answered Nov 16 '18 at 19:02
VanVan
1413
1413
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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.)
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add a comment |
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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.)
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add a comment |
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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.)
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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.)
edited Nov 15 '18 at 19:32
answered Nov 15 '18 at 18:10
KM101KM101
6,0901525
6,0901525
add a comment |
add a comment |
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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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add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
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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
answered Nov 15 '18 at 17:55
b00n heTb00n heT
10.5k12235
10.5k12235
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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):
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add a comment |
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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):
$endgroup$
add a comment |
$begingroup$
Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):
$endgroup$
Adjusted the PWW given by farruhota (and ripping off their very image) to improve by a factor of $2$ (cf. also Théophile's answer):
answered Nov 17 '18 at 21:18
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
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$begingroup$
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
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.
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add a comment |
$begingroup$
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
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.
$endgroup$
add a comment |
$begingroup$
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
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.
$endgroup$
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
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.
answered Nov 15 '18 at 20:05
TurlocTheRedTurlocTheRed
916311
916311
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add a comment |
$begingroup$
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.
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A wall is a 2-dimensional object.
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– Jossie Calderon
Feb 14 at 18:46
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
A wall is a 2-dimensional object.
$endgroup$
– Jossie Calderon
Feb 14 at 18:46
add a comment |
$begingroup$
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.
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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.
answered Nov 16 '18 at 23:28
gnasher729gnasher729
6,1001028
6,1001028
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A wall is a 2-dimensional object.
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– Jossie Calderon
Feb 14 at 18:46
add a comment |
$begingroup$
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
add a comment |
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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.
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– Théophile
Nov 15 '18 at 19:09
8
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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???
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– alephzero
Nov 15 '18 at 20:16
7
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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!
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– 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.
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– 1123581321
Nov 15 '18 at 22:53
9
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@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
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– Teepeemm
Nov 16 '18 at 2:10