Can area of rectangle be greater than the square of its diagonal?
up vote
33
down vote
favorite
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
New contributor
|
show 9 more comments
up vote
33
down vote
favorite
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
New contributor
21
The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
– Théophile
2 days ago
8
The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
– alephzero
yesterday
6
In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
– amalloy
yesterday
7
The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
yesterday
8
@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
yesterday
|
show 9 more comments
up vote
33
down vote
favorite
up vote
33
down vote
favorite
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
New contributor
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
New contributor
New contributor
edited 2 days ago
New contributor
asked 2 days ago
user17838
17326
17326
New contributor
New contributor
21
The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
– Théophile
2 days ago
8
The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
– alephzero
yesterday
6
In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
– amalloy
yesterday
7
The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
yesterday
8
@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
yesterday
|
show 9 more comments
21
The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
– Théophile
2 days ago
8
The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
– alephzero
yesterday
6
In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
– amalloy
yesterday
7
The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
yesterday
8
@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
yesterday
21
21
The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
– Théophile
2 days ago
The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
– Théophile
2 days ago
8
8
The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
– alephzero
yesterday
The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
– alephzero
yesterday
6
6
In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
– amalloy
yesterday
In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
– amalloy
yesterday
7
7
The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
yesterday
The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
yesterday
8
8
@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
yesterday
@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
yesterday
|
show 9 more comments
11 Answers
11
active
oldest
votes
up vote
69
down vote
accepted
The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
add a comment |
up vote
65
down vote
Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
+1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
– mckenzm
yesterday
1
Here is a more dynamic, animated version of the same picture.
– Xander Henderson
yesterday
add a comment |
up vote
23
down vote
A simple explanation without proof or pictures:
The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.
Great explanation, but that “it's” is jarring...
– DaG
10 hours ago
Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
– AlexanderJ93
10 hours ago
+1 Thank you for the one-line proof!
– DaG
9 hours ago
add a comment |
up vote
20
down vote
In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.
add a comment |
up vote
8
down vote
You can prove that no such rectangle exists as follows:
Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$
Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.
The answer given, though arithmetically correct does not represent a real wall.
I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.
To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
– Ilmari Karonen
yesterday
1
@IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
– Mark Bennet
yesterday
add a comment |
up vote
4
down vote
No. As others have said.
What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.
If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.
If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:
Total perimeter: 70
Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)
This should now give the solution of:
- $I^2 + B^2 = 25^2 = 625 $
- $2I + 2B = 70 $
- $I + B = 35 $
- $I^2 + 2IB + B^2 = 1,225 $
- $2IB = 600 $
$IB = 300$ ,
which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)
Hope that helps!
-Van
New contributor
add a comment |
up vote
2
down vote
No. Using Pythagoras and a simple inequality we get
$$d^2=a^2+b^2geq 2abgeq ab$$
If $a,b$ are the sides and $d$ the diagonal
add a comment |
up vote
2
down vote
No, use the Pythagorean Theorem.
$$c^2 = a^2+b^2$$
$c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).
Recall for any real number, its square must be non-negative.
$$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$
The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.
Now, to find the area itself.
For the diagonal:
$$c^2 = a^2+b^2$$
$$implies 18^2 = a^2+b^2$$
$$color{blue}{324 = a^2+b^2} tag{1}$$
For the perimeter:
$$2(a+b) = 72$$
$$a+b = 36$$
Now, define one variable in terms of the other.
$$color{purple}{a = 36-b} tag{2}$$
Combine $(1)$ and $(2)$.
$$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$
$$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$
But $$Delta = b^2-4ac$$
$$Delta = 72^2-4(2)(972) = -2592$$
$$implies Delta < 0$$
Thus, there is no solution. (No such rectangle exists.)
add a comment |
up vote
2
down vote
Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
add a comment |
up vote
0
down vote
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
Can $A>d^2$?
Can $lw>l^2+w^2$?
$-lw>l^2-2wl+w^2=(l-w)^2$
Width and length are necessarily positive. The square of their difference also must be positive.
So we have a negative number that must be greater than a positive number. A contradiction.
add a comment |
up vote
0
down vote
A wall has a thickness.
Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.
Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.
The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.
We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.
This t can be used to calculate the area of the wall as (72 - 4t) * t.
Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.
add a comment |
11 Answers
11
active
oldest
votes
11 Answers
11
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
69
down vote
accepted
The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
add a comment |
up vote
69
down vote
accepted
The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
add a comment |
up vote
69
down vote
accepted
up vote
69
down vote
accepted
The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
The area of the square built on the diagonal must be at least twice the area of the rectangle:
$hskip 4 cm$
answered 2 days ago
Théophile
19.1k12944
19.1k12944
add a comment |
add a comment |
up vote
65
down vote
Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
+1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
– mckenzm
yesterday
1
Here is a more dynamic, animated version of the same picture.
– Xander Henderson
yesterday
add a comment |
up vote
65
down vote
Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
+1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
– mckenzm
yesterday
1
Here is a more dynamic, animated version of the same picture.
– Xander Henderson
yesterday
add a comment |
up vote
65
down vote
up vote
65
down vote
Another proof without words, at the suggestion of Semiclassical:
The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.
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 2 days ago
community wiki
Xander Henderson
+1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
– mckenzm
yesterday
1
Here is a more dynamic, animated version of the same picture.
– Xander Henderson
yesterday
add a comment |
+1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
– mckenzm
yesterday
1
Here is a more dynamic, animated version of the same picture.
– Xander Henderson
yesterday
+1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
– mckenzm
yesterday
+1 It is often useful to consider the extremes, similar to "annulus" type problems. there is no solution for the perimeter between l=0 and l=18, therefore the perimeter or the diagonal has been misstated. A graph of the length vs perimeter would show the same.
– mckenzm
yesterday
1
1
Here is a more dynamic, animated version of the same picture.
– Xander Henderson
yesterday
Here is a more dynamic, animated version of the same picture.
– Xander Henderson
yesterday
add a comment |
up vote
23
down vote
A simple explanation without proof or pictures:
The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.
Great explanation, but that “it's” is jarring...
– DaG
10 hours ago
Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
– AlexanderJ93
10 hours ago
+1 Thank you for the one-line proof!
– DaG
9 hours ago
add a comment |
up vote
23
down vote
A simple explanation without proof or pictures:
The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.
Great explanation, but that “it's” is jarring...
– DaG
10 hours ago
Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
– AlexanderJ93
10 hours ago
+1 Thank you for the one-line proof!
– DaG
9 hours ago
add a comment |
up vote
23
down vote
up vote
23
down vote
A simple explanation without proof or pictures:
The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.
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 10 hours ago
answered yesterday
AlexanderJ93
5,127422
5,127422
Great explanation, but that “it's” is jarring...
– DaG
10 hours ago
Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
– AlexanderJ93
10 hours ago
+1 Thank you for the one-line proof!
– DaG
9 hours ago
add a comment |
Great explanation, but that “it's” is jarring...
– DaG
10 hours ago
Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
– AlexanderJ93
10 hours ago
+1 Thank you for the one-line proof!
– DaG
9 hours ago
Great explanation, but that “it's” is jarring...
– DaG
10 hours ago
Great explanation, but that “it's” is jarring...
– DaG
10 hours ago
Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
– AlexanderJ93
10 hours ago
Sorry! This is my number 1 typo, and I miss it all the time in proofreading. Fixed now, thanks!
– AlexanderJ93
10 hours ago
+1 Thank you for the one-line proof!
– DaG
9 hours ago
+1 Thank you for the one-line proof!
– DaG
9 hours ago
add a comment |
up vote
20
down vote
In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.
add a comment |
up vote
20
down vote
In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.
add a comment |
up vote
20
down vote
up vote
20
down vote
In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2ge 2ab$ if orthogonal sides' lengths are $a,,b$. Why? Because the difference is $(a-b)^2ge 0$.
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 2 days ago
J.G.
17.9k11830
17.9k11830
add a comment |
add a comment |
up vote
8
down vote
You can prove that no such rectangle exists as follows:
Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$
Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.
The answer given, though arithmetically correct does not represent a real wall.
I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.
To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
– Ilmari Karonen
yesterday
1
@IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
– Mark Bennet
yesterday
add a comment |
up vote
8
down vote
You can prove that no such rectangle exists as follows:
Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$
Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.
The answer given, though arithmetically correct does not represent a real wall.
I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.
To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
– Ilmari Karonen
yesterday
1
@IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
– Mark Bennet
yesterday
add a comment |
up vote
8
down vote
up vote
8
down vote
You can prove that no such rectangle exists as follows:
Let $l ge b$ (one side of the rectangle has to be the longest). Since $2l+2b=72$ you have $2lge l+b= 36$ so $l ge 18$
Then the diagonal of a right-angled triangle is the longest side so $dgt lge 18$ for a non-degenerate triangle, and the only degenerate case which arises is with $l=36, b=0, d=36$.
The answer given, though arithmetically correct does not represent a real wall.
I am not sure what the question means, though, as it is curiously phrased. The question asks for "the area of the wall" and not "the area of the rectangle bounded by the wall" and had the answer not been set out, I might have been thinking of a wall of uniform thickness and external perimeter $72$ and an internal diagonal of $18$ to make any sense of it.
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 2 days ago
Mark Bennet
79.5k978177
79.5k978177
To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
– Ilmari Karonen
yesterday
1
@IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
– Mark Bennet
yesterday
add a comment |
To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
– Ilmari Karonen
yesterday
1
@IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
– Mark Bennet
yesterday
To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
– Ilmari Karonen
yesterday
To address your comment at the end, I'm pretty sure they mean "a wall" as in "one of the (usually) four walls of a room", not as in "a wall surrounding an area of land". So the wall itself is a rectangular surface, with width $ell$ and height $b$.
– Ilmari Karonen
yesterday
1
1
@IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
– Mark Bennet
yesterday
@IlmariKaronen Just goes to show how careful one has to be reading and interpreting (and setting) questions.
– Mark Bennet
yesterday
add a comment |
up vote
4
down vote
No. As others have said.
What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.
If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.
If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:
Total perimeter: 70
Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)
This should now give the solution of:
- $I^2 + B^2 = 25^2 = 625 $
- $2I + 2B = 70 $
- $I + B = 35 $
- $I^2 + 2IB + B^2 = 1,225 $
- $2IB = 600 $
$IB = 300$ ,
which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)
Hope that helps!
-Van
New contributor
add a comment |
up vote
4
down vote
No. As others have said.
What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.
If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.
If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:
Total perimeter: 70
Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)
This should now give the solution of:
- $I^2 + B^2 = 25^2 = 625 $
- $2I + 2B = 70 $
- $I + B = 35 $
- $I^2 + 2IB + B^2 = 1,225 $
- $2IB = 600 $
$IB = 300$ ,
which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)
Hope that helps!
-Van
New contributor
add a comment |
up vote
4
down vote
up vote
4
down vote
No. As others have said.
What this looks like to me (as someone who has taught HS Chem & Physics for years and has helped write middle school math content) is a question written trying to get someone to put together the solution as shown, but without checking whether the numbers given make any real-world sense. I have certainly made this mistake myself, even though I try really hard to catch it.
If this is a standardized test question (or one from a textbook, practice book, online resource, etc.), fair play, we've caught a poorly written question.
If this is a question you are writing yourself, and you want to improve it, you could change the parameters this way:
Total perimeter: 70
Diagonal: 25 (I don't think you'll find any nice whole numbers - aka pythagorean triples - using a perimeter of 72.)
This should now give the solution of:
- $I^2 + B^2 = 25^2 = 625 $
- $2I + 2B = 70 $
- $I + B = 35 $
- $I^2 + 2IB + B^2 = 1,225 $
- $2IB = 600 $
$IB = 300$ ,
which makes sense, given that I used a (3,4,5) right triangle (scaled by 5) in my setup. (Which means that I = 15 and B = 20, for a hypotenuse of 25.)
Hope that helps!
-Van
New contributor
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
New contributor
New contributor
answered yesterday
Van
413
413
New contributor
New contributor
add a comment |
add a comment |
up vote
2
down vote
No. Using Pythagoras and a simple inequality we get
$$d^2=a^2+b^2geq 2abgeq ab$$
If $a,b$ are the sides and $d$ the diagonal
add a comment |
up vote
2
down vote
No. Using Pythagoras and a simple inequality we get
$$d^2=a^2+b^2geq 2abgeq ab$$
If $a,b$ are the sides and $d$ the diagonal
add a comment |
up vote
2
down vote
up vote
2
down vote
No. Using Pythagoras and a simple inequality we get
$$d^2=a^2+b^2geq 2abgeq ab$$
If $a,b$ are the sides and $d$ the diagonal
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 2 days ago
b00n heT
10.1k12134
10.1k12134
add a comment |
add a comment |
up vote
2
down vote
No, use the Pythagorean Theorem.
$$c^2 = a^2+b^2$$
$c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).
Recall for any real number, its square must be non-negative.
$$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$
The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.
Now, to find the area itself.
For the diagonal:
$$c^2 = a^2+b^2$$
$$implies 18^2 = a^2+b^2$$
$$color{blue}{324 = a^2+b^2} tag{1}$$
For the perimeter:
$$2(a+b) = 72$$
$$a+b = 36$$
Now, define one variable in terms of the other.
$$color{purple}{a = 36-b} tag{2}$$
Combine $(1)$ and $(2)$.
$$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$
$$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$
But $$Delta = b^2-4ac$$
$$Delta = 72^2-4(2)(972) = -2592$$
$$implies Delta < 0$$
Thus, there is no solution. (No such rectangle exists.)
add a comment |
up vote
2
down vote
No, use the Pythagorean Theorem.
$$c^2 = a^2+b^2$$
$c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).
Recall for any real number, its square must be non-negative.
$$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$
The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.
Now, to find the area itself.
For the diagonal:
$$c^2 = a^2+b^2$$
$$implies 18^2 = a^2+b^2$$
$$color{blue}{324 = a^2+b^2} tag{1}$$
For the perimeter:
$$2(a+b) = 72$$
$$a+b = 36$$
Now, define one variable in terms of the other.
$$color{purple}{a = 36-b} tag{2}$$
Combine $(1)$ and $(2)$.
$$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$
$$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$
But $$Delta = b^2-4ac$$
$$Delta = 72^2-4(2)(972) = -2592$$
$$implies Delta < 0$$
Thus, there is no solution. (No such rectangle exists.)
add a comment |
up vote
2
down vote
up vote
2
down vote
No, use the Pythagorean Theorem.
$$c^2 = a^2+b^2$$
$c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).
Recall for any real number, its square must be non-negative.
$$(a-b)^2 geq 0 implies a^2-2ab+b^2 geq 0 implies color{blue}{a^2+b^2 geq 2ab}$$
The area of the rectangle is $ab$, but $c^2 geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.
Now, to find the area itself.
For the diagonal:
$$c^2 = a^2+b^2$$
$$implies 18^2 = a^2+b^2$$
$$color{blue}{324 = a^2+b^2} tag{1}$$
For the perimeter:
$$2(a+b) = 72$$
$$a+b = 36$$
Now, define one variable in terms of the other.
$$color{purple}{a = 36-b} tag{2}$$
Combine $(1)$ and $(2)$.
$$324 = a^2+b^2 implies 324 = (36-b)^2+b^2$$
$$324 = 36^2-2(36)b+b^2+b^2 implies 324 = 1296-72b+2b^2 implies 2b^2-72b+972 = 0$$
But $$Delta = b^2-4ac$$
$$Delta = 72^2-4(2)(972) = -2592$$
$$implies Delta < 0$$
Thus, there is no solution. (No such rectangle exists.)
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 2 days ago
answered 2 days ago
KM101
1,765313
1,765313
add a comment |
add a comment |
up vote
2
down vote
Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
add a comment |
up vote
2
down vote
Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
add a comment |
up vote
2
down vote
up vote
2
down vote
Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
Another PWW (noted by AlexanderJ93 and others):
$hspace{5cm}$
answered 9 hours ago
farruhota
17.4k2736
17.4k2736
add a comment |
add a comment |
up vote
0
down vote
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
Can $A>d^2$?
Can $lw>l^2+w^2$?
$-lw>l^2-2wl+w^2=(l-w)^2$
Width and length are necessarily positive. The square of their difference also must be positive.
So we have a negative number that must be greater than a positive number. A contradiction.
add a comment |
up vote
0
down vote
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
Can $A>d^2$?
Can $lw>l^2+w^2$?
$-lw>l^2-2wl+w^2=(l-w)^2$
Width and length are necessarily positive. The square of their difference also must be positive.
So we have a negative number that must be greater than a positive number. A contradiction.
add a comment |
up vote
0
down vote
up vote
0
down vote
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
Can $A>d^2$?
Can $lw>l^2+w^2$?
$-lw>l^2-2wl+w^2=(l-w)^2$
Width and length are necessarily positive. The square of their difference also must be positive.
So we have a negative number that must be greater than a positive number. A contradiction.
$A=lw$
$P=2(l+w)$
$d=sqrt{l^2+w^2}$
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 yesterday
TurlocTheRed
55819
55819
add a comment |
add a comment |
up vote
0
down vote
A wall has a thickness.
Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.
Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.
The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.
We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.
This t can be used to calculate the area of the wall as (72 - 4t) * t.
Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.
add a comment |
up vote
0
down vote
A wall has a thickness.
Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.
Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.
The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.
We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.
This t can be used to calculate the area of the wall as (72 - 4t) * t.
Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.
add a comment |
up vote
0
down vote
up vote
0
down vote
A wall has a thickness.
Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.
Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.
The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.
We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.
This t can be used to calculate the area of the wall as (72 - 4t) * t.
Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.
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 20 hours ago
gnasher729
5,9511028
5,9511028
add a comment |
add a comment |
user17838 is a new contributor. Be nice, and check out our Code of Conduct.
user17838 is a new contributor. Be nice, and check out our Code of Conduct.
user17838 is a new contributor. Be nice, and check out our Code of Conduct.
user17838 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000024%2fcan-area-of-rectangle-be-greater-than-the-square-of-its-diagonal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
21
The given solution is preposterous. (Note that for a rectangle with fixed perimeter, the diagonal is shortest when the rectangle is a square. Therefore the diagonal must be at least $18sqrt2 textrm m$ long.) On the bright side, perhaps you are better off not working for a company that won't listen to reason.
– Théophile
2 days ago
8
The area of the wall is $144h$ m${}^2$, where $h$ is the height of the wall. (Note, trick question alert - the wall has two sides so the answer is not $72h$.) Oh, you mean you want the area enclosed by the wall? Then why didn't you say so???
– alephzero
yesterday
6
In fairness, they do say "if the length of the diagonal is 18...". Given that this premise P is false (the diagonal cannot be 18!), any statement of the form "if P then X" is true, so they are right that the area may as well be 486!
– amalloy
yesterday
7
The problem with the question is that if you solve for the side lengths you get complex numbers.
– 1123581321
yesterday
8
@1123581321 Specifically, $l$ and $b$ are $18pm9isqrt2$.
– Teepeemm
yesterday