Can we compute the area of a quadrilateral with one right angle when we only know the lengths of any three sides? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Maximize the Area of a Quadrilateral given Three SidesA simple(?) Analytical Geometry Question (Ellipse)The perimeter of the rectangle is $20$, diagonal is $8$ and side is $x$. Show that $x^2-10x+18=0$Solving right triangle given the area and one angleCan one deduce whether a given quantity is possible as the area of a triangle when supplied with the length of two of its sides?Prove that the midpoints of the sides of a quadrilateral lie on a circle if and only if the quadrilateral is orthodiagonal.Given three points, how can I tell if the angle is acute without using trigonometric functions?Area of a concave quadrilateralUnknown internal angles of a quadrilateral where its area and side lengths are knownSimilar quadrilateral to a given one with vertices of the same color
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Can we compute the area of a quadrilateral with one right angle when we only know the lengths of any three sides?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Maximize the Area of a Quadrilateral given Three SidesA simple(?) Analytical Geometry Question (Ellipse)The perimeter of the rectangle is $20$, diagonal is $8$ and side is $x$. Show that $x^2-10x+18=0$Solving right triangle given the area and one angleCan one deduce whether a given quantity is possible as the area of a triangle when supplied with the length of two of its sides?Prove that the midpoints of the sides of a quadrilateral lie on a circle if and only if the quadrilateral is orthodiagonal.Given three points, how can I tell if the angle is acute without using trigonometric functions?Area of a concave quadrilateralUnknown internal angles of a quadrilateral where its area and side lengths are knownSimilar quadrilateral to a given one with vertices of the same color
$begingroup$
I took an IQ test for fun recently, but I take issue with the answer to one of the questions. Here's the question:
My issue is that the explanation assumes angle DC is a right angle. Given that assumption, I can see the quadrilateral is indeed a rectangle and a right triangle and can follow their explanation. However, (from what I remember my high school geometry teacher telling me) even though an angle looks like a right angle, it shouldn't be assumed unless it is explicitly stated or you can prove it. To explain what I mean, if DC isn't a right angle and we exacerbated that difference, it would look like the following:
Thus, even being given A, B, C and D it seems like the area could not be calculated.
So my question is twofold:
- Is my criticism valid or am I just being too proud because I got a question wrong?
- Given my interpretation, DC is not a right angle, can this problem be solved?
geometry
edited Apr 3 at 9:09
Discrete lizard
14010
asked Apr 3 at 2:41
Jack O.Jack O.
1115
$endgroup$
add a comment |
$begingroup$
I took an IQ test for fun recently, but I take issue with the answer to one of the questions. Here's the question:
My issue is that the explanation assumes angle DC is a right angle. Given that assumption, I can see the quadrilateral is indeed a rectangle and a right triangle and can follow their explanation. However, (from what I remember my high school geometry teacher telling me) even though an angle looks like a right angle, it shouldn't be assumed unless it is explicitly stated or you can prove it. To explain what I mean, if DC isn't a right angle and we exacerbated that difference, it would look like the following:
Thus, even being given A, B, C and D it seems like the area could not be calculated.
So my question is twofold:
- Is my criticism valid or am I just being too proud because I got a question wrong?
- Given my interpretation, DC is not a right angle, can this problem be solved?
geometry
edited Apr 3 at 9:09
Discrete lizard
14010
asked Apr 3 at 2:41
Jack O.Jack O.
1115
$endgroup$
1
$begingroup$
You know it is a right angle because it has a large "90" on it. Now we can argue they never said why it has a "90" on it and as I am a nitpick I would agree with you... but... I think you and I would lose in any court.
$endgroup$
– fleablood
Apr 3 at 2:45
15
$begingroup$
Not that angle, the one below it.
$endgroup$
– Robert Israel
Apr 3 at 2:46
3
$begingroup$
" even though an angle looks like an angle, it shouldn't be assumed" but it doesn't even look like a right angle.
$endgroup$
– fleablood
Apr 3 at 2:56
7
$begingroup$
Where did you find that test? Online IQ tests are generally untrustworthy even before we get to this quadrilateral problem. Many don't even bother ending in an IQ estimate.
$endgroup$
– J.G.
Apr 3 at 19:09
$begingroup$
yes for Cyclic quadrilaterals. Brahmagupta the Indian mathematician of the 7th century CE knew it.
$endgroup$
– R.C.Cowsik
2 days ago
add a comment |
$begingroup$
I took an IQ test for fun recently, but I take issue with the answer to one of the questions. Here's the question:
My issue is that the explanation assumes angle DC is a right angle. Given that assumption, I can see the quadrilateral is indeed a rectangle and a right triangle and can follow their explanation. However, (from what I remember my high school geometry teacher telling me) even though an angle looks like a right angle, it shouldn't be assumed unless it is explicitly stated or you can prove it. To explain what I mean, if DC isn't a right angle and we exacerbated that difference, it would look like the following:
Thus, even being given A, B, C and D it seems like the area could not be calculated.
So my question is twofold:
- Is my criticism valid or am I just being too proud because I got a question wrong?
- Given my interpretation, DC is not a right angle, can this problem be solved?
geometry
edited Apr 3 at 9:09
Discrete lizard
14010
asked Apr 3 at 2:41
Jack O.Jack O.
1115
$endgroup$
I took an IQ test for fun recently, but I take issue with the answer to one of the questions. Here's the question:
My issue is that the explanation assumes angle DC is a right angle. Given that assumption, I can see the quadrilateral is indeed a rectangle and a right triangle and can follow their explanation. However, (from what I remember my high school geometry teacher telling me) even though an angle looks like a right angle, it shouldn't be assumed unless it is explicitly stated or you can prove it. To explain what I mean, if DC isn't a right angle and we exacerbated that difference, it would look like the following:
Thus, even being given A, B, C and D it seems like the area could not be calculated.
So my question is twofold:
- Is my criticism valid or am I just being too proud because I got a question wrong?
- Given my interpretation, DC is not a right angle, can this problem be solved?
geometry
geometry
edited Apr 3 at 9:09
Discrete lizard
14010
asked Apr 3 at 2:41
Jack O.Jack O.
1115
edited Apr 3 at 9:09
Discrete lizard
14010
asked Apr 3 at 2:41
Jack O.Jack O.
1115
edited Apr 3 at 9:09
Discrete lizard
14010
edited Apr 3 at 9:09
Discrete lizard
14010
edited Apr 3 at 9:09
Discrete lizard
14010
14010
asked Apr 3 at 2:41
Jack O.Jack O.
1115
asked Apr 3 at 2:41
Jack O.Jack O.
1115
asked Apr 3 at 2:41
Jack O.Jack O.
1115
1115
1
$begingroup$
You know it is a right angle because it has a large "90" on it. Now we can argue they never said why it has a "90" on it and as I am a nitpick I would agree with you... but... I think you and I would lose in any court.
$endgroup$
– fleablood
Apr 3 at 2:45
15
$begingroup$
Not that angle, the one below it.
$endgroup$
– Robert Israel
Apr 3 at 2:46
3
$begingroup$
" even though an angle looks like an angle, it shouldn't be assumed" but it doesn't even look like a right angle.
$endgroup$
– fleablood
Apr 3 at 2:56
7
$begingroup$
Where did you find that test? Online IQ tests are generally untrustworthy even before we get to this quadrilateral problem. Many don't even bother ending in an IQ estimate.
$endgroup$
– J.G.
Apr 3 at 19:09
$begingroup$
yes for Cyclic quadrilaterals. Brahmagupta the Indian mathematician of the 7th century CE knew it.
$endgroup$
– R.C.Cowsik
2 days ago
add a comment |
1
$begingroup$
You know it is a right angle because it has a large "90" on it. Now we can argue they never said why it has a "90" on it and as I am a nitpick I would agree with you... but... I think you and I would lose in any court.
$endgroup$
– fleablood
Apr 3 at 2:45
15
$begingroup$
Not that angle, the one below it.
$endgroup$
– Robert Israel
Apr 3 at 2:46
3
$begingroup$
" even though an angle looks like an angle, it shouldn't be assumed" but it doesn't even look like a right angle.
$endgroup$
– fleablood
Apr 3 at 2:56
7
$begingroup$
Where did you find that test? Online IQ tests are generally untrustworthy even before we get to this quadrilateral problem. Many don't even bother ending in an IQ estimate.
$endgroup$
– J.G.
Apr 3 at 19:09
$begingroup$
yes for Cyclic quadrilaterals. Brahmagupta the Indian mathematician of the 7th century CE knew it.
$endgroup$
– R.C.Cowsik
2 days ago
1
1
$begingroup$
You know it is a right angle because it has a large "90" on it. Now we can argue they never said why it has a "90" on it and as I am a nitpick I would agree with you... but... I think you and I would lose in any court.
$endgroup$
– fleablood
Apr 3 at 2:45
$begingroup$
You know it is a right angle because it has a large "90" on it. Now we can argue they never said why it has a "90" on it and as I am a nitpick I would agree with you... but... I think you and I would lose in any court.
$endgroup$
– fleablood
Apr 3 at 2:45
15
15
$begingroup$
Not that angle, the one below it.
$endgroup$
– Robert Israel
Apr 3 at 2:46
$begingroup$
Not that angle, the one below it.
$endgroup$
– Robert Israel
Apr 3 at 2:46
3
3
$begingroup$
" even though an angle looks like an angle, it shouldn't be assumed" but it doesn't even look like a right angle.
$endgroup$
– fleablood
Apr 3 at 2:56
$begingroup$
" even though an angle looks like an angle, it shouldn't be assumed" but it doesn't even look like a right angle.
$endgroup$
– fleablood
Apr 3 at 2:56
7
7
$begingroup$
Where did you find that test? Online IQ tests are generally untrustworthy even before we get to this quadrilateral problem. Many don't even bother ending in an IQ estimate.
$endgroup$
– J.G.
Apr 3 at 19:09
$begingroup$
Where did you find that test? Online IQ tests are generally untrustworthy even before we get to this quadrilateral problem. Many don't even bother ending in an IQ estimate.
$endgroup$
– J.G.
Apr 3 at 19:09
$begingroup$
yes for Cyclic quadrilaterals. Brahmagupta the Indian mathematician of the 7th century CE knew it.
$endgroup$
– R.C.Cowsik
2 days ago
$begingroup$
yes for Cyclic quadrilaterals. Brahmagupta the Indian mathematician of the 7th century CE knew it.
$endgroup$
– R.C.Cowsik
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You are right. The provided explanation is nonsensical. $DC$ cannot be assumed to be a right angle.
However, if you don't make that assumption, and take $BC$ as the only given right angle, the correct answer is "All four sides must be known."
The quadrilateral can be decomposed into two non-overlapping triangles. The first is a right angled triangle formed by sides $B$, $C$ and a hypotenuse, and its area is easy to determine. You can use Pythagoras' Theorem to find the hypotenuse of that right triangle formed by sides $B$ and $C$. That hypotenuse, together with sides $A$ and $D$ forms the other triangle. Its area can be computed using Heron's formula. Just sum the areas.
Note that I'm still making a tacit assumption that this is a convex quadrilateral. A concave quadrilateral is possible with the angle $AD$ being reflex, which is seemingly not representative of the diagram given, but is a possibility when one only has the information of a single right angle and four given sides. In the case of a concave quadrilateral, the area calculation that I detailed above would not be correct.
answered Apr 3 at 3:10
DeepakDeepak
18k11640
$endgroup$
1
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
Apr 3 at 3:21
$begingroup$
You're welcome.
$endgroup$
– Deepak
Apr 3 at 3:24
2
$begingroup$
+1 for Heron's formula - I learned something new today
$endgroup$
– slebetman
Apr 3 at 9:31
11
$begingroup$
@Deepak No, that's not correct. I took some time to double check. For example, see this image representing two quadrilaterals, both satisfying the conditions in the problem, both with the same side lengths, but having different areas - one is strictly smaller than the other.
$endgroup$
– David Z
Apr 3 at 10:02
4
$begingroup$
@Deepak Yes, that was exactly my point, that you can have a concave and convex quadrilateral with the same side lengths. Given that the question is about not assuming things that aren't explicitly stated, I think it may be worth at least mentioning that assumption.
$endgroup$
– David Z
Apr 3 at 10:41
|
show 5 more comments
$begingroup$
You are correct that the given solution is wrong. Worse still, even if you know that the angles between BC and CD are both right-angles, the purported answer is still wrong! This is because if you're given the lengths of A,B,C, it still does not uniquely determine D because we are not told that the angle between AB is less than $90°$.
In general, it is not enough even if you have all four side lengths. For example, consider convex quadrilateral $PQRS$ such that $PQ = 6$ and $QR = 9$ and $RS = 7$ and $SP = 8$. It is possible that $P,R$ are slightly less than $15$ apart, making $PQRS$ a very skinny quadrilateral whose area can be made arbitrarily close to zero. Alternatively, moving $P,R$ to a distance of $10$ makes $PQRS$ rather square-like with area clearly more than $48$. Bretschneider's formula gives the area for an arbitrary quadrilateral, and you can see from it as well that fixing the side lengths is not enough to determine the area, since it also varies with the sum of two opposite angles.
answered Apr 3 at 3:55
user21820user21820
40.1k544162
$endgroup$
add a comment |
$begingroup$
You are right: there is absolutely no indication that angle $DC$ is a right angle. If they wanted you to assume it was a right angle, they should have indicated that with another $90$. It really doesn't even look like a right angle (somebody had the bright idea of trying to render the picture in perspective, but we don't even know where the horizon is supposed to be).
answered Apr 3 at 2:46
Robert IsraelRobert Israel
331k23221477
$endgroup$
$begingroup$
That's what I thought. It should explicitly state if any angles are right. However my second question remains, given DC is ambiguous, is this question solvable? I don't think there would be enough information to solve in this case.
$endgroup$
– Jack O.
Apr 3 at 2:52
$begingroup$
@JackO. See my answer. The correct answer would be "All sides must be known".
$endgroup$
– Deepak
Apr 3 at 3:11
$begingroup$
If we know all four lengths and assume no angle is more than 180, then I think there is only one quadrilateral so the area will be unique. I think. But you need all four. If you only three the fourth can be many lengths if the third one "swings".
$endgroup$
– fleablood
Apr 3 at 3:15
$begingroup$
@fleablood: No, in general there are infinitely many quadrilaterals with the same sides in the same order. See Bretschneider's formula for area of a quadrilateral given all sides and 2 opposite angles.
$endgroup$
– user21820
Apr 3 at 12:04
$begingroup$
The rectangle appears to be drawn with "thickness", and if we assume that the thickness is perpendicular to the other sides, we might be able to figure out where the horizon is ... but I'm not sure that there is actually any consistent solution; the other two corners don't have thickness, which imposes constraints on the horizon (it must be such that the other "side" of the thickness for those corners is behind the corners) that may not be satisfiable.
$endgroup$
– Acccumulation
Apr 3 at 15:46
|
show 4 more comments
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are right. The provided explanation is nonsensical. $DC$ cannot be assumed to be a right angle.
However, if you don't make that assumption, and take $BC$ as the only given right angle, the correct answer is "All four sides must be known."
The quadrilateral can be decomposed into two non-overlapping triangles. The first is a right angled triangle formed by sides $B$, $C$ and a hypotenuse, and its area is easy to determine. You can use Pythagoras' Theorem to find the hypotenuse of that right triangle formed by sides $B$ and $C$. That hypotenuse, together with sides $A$ and $D$ forms the other triangle. Its area can be computed using Heron's formula. Just sum the areas.
Note that I'm still making a tacit assumption that this is a convex quadrilateral. A concave quadrilateral is possible with the angle $AD$ being reflex, which is seemingly not representative of the diagram given, but is a possibility when one only has the information of a single right angle and four given sides. In the case of a concave quadrilateral, the area calculation that I detailed above would not be correct.
answered Apr 3 at 3:10
DeepakDeepak
18k11640
$endgroup$
1
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
Apr 3 at 3:21
$begingroup$
You're welcome.
$endgroup$
– Deepak
Apr 3 at 3:24
2
$begingroup$
+1 for Heron's formula - I learned something new today
$endgroup$
– slebetman
Apr 3 at 9:31
11
$begingroup$
@Deepak No, that's not correct. I took some time to double check. For example, see this image representing two quadrilaterals, both satisfying the conditions in the problem, both with the same side lengths, but having different areas - one is strictly smaller than the other.
$endgroup$
– David Z
Apr 3 at 10:02
4
$begingroup$
@Deepak Yes, that was exactly my point, that you can have a concave and convex quadrilateral with the same side lengths. Given that the question is about not assuming things that aren't explicitly stated, I think it may be worth at least mentioning that assumption.
$endgroup$
– David Z
Apr 3 at 10:41
|
show 5 more comments
$begingroup$
You are right. The provided explanation is nonsensical. $DC$ cannot be assumed to be a right angle.
However, if you don't make that assumption, and take $BC$ as the only given right angle, the correct answer is "All four sides must be known."
The quadrilateral can be decomposed into two non-overlapping triangles. The first is a right angled triangle formed by sides $B$, $C$ and a hypotenuse, and its area is easy to determine. You can use Pythagoras' Theorem to find the hypotenuse of that right triangle formed by sides $B$ and $C$. That hypotenuse, together with sides $A$ and $D$ forms the other triangle. Its area can be computed using Heron's formula. Just sum the areas.
Note that I'm still making a tacit assumption that this is a convex quadrilateral. A concave quadrilateral is possible with the angle $AD$ being reflex, which is seemingly not representative of the diagram given, but is a possibility when one only has the information of a single right angle and four given sides. In the case of a concave quadrilateral, the area calculation that I detailed above would not be correct.
answered Apr 3 at 3:10
DeepakDeepak
18k11640
$endgroup$
1
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
Apr 3 at 3:21
$begingroup$
You're welcome.
$endgroup$
– Deepak
Apr 3 at 3:24
2
$begingroup$
+1 for Heron's formula - I learned something new today
$endgroup$
– slebetman
Apr 3 at 9:31
11
$begingroup$
@Deepak No, that's not correct. I took some time to double check. For example, see this image representing two quadrilaterals, both satisfying the conditions in the problem, both with the same side lengths, but having different areas - one is strictly smaller than the other.
$endgroup$
– David Z
Apr 3 at 10:02
4
$begingroup$
@Deepak Yes, that was exactly my point, that you can have a concave and convex quadrilateral with the same side lengths. Given that the question is about not assuming things that aren't explicitly stated, I think it may be worth at least mentioning that assumption.
$endgroup$
– David Z
Apr 3 at 10:41
|
show 5 more comments
$begingroup$
You are right. The provided explanation is nonsensical. $DC$ cannot be assumed to be a right angle.
However, if you don't make that assumption, and take $BC$ as the only given right angle, the correct answer is "All four sides must be known."
The quadrilateral can be decomposed into two non-overlapping triangles. The first is a right angled triangle formed by sides $B$, $C$ and a hypotenuse, and its area is easy to determine. You can use Pythagoras' Theorem to find the hypotenuse of that right triangle formed by sides $B$ and $C$. That hypotenuse, together with sides $A$ and $D$ forms the other triangle. Its area can be computed using Heron's formula. Just sum the areas.
Note that I'm still making a tacit assumption that this is a convex quadrilateral. A concave quadrilateral is possible with the angle $AD$ being reflex, which is seemingly not representative of the diagram given, but is a possibility when one only has the information of a single right angle and four given sides. In the case of a concave quadrilateral, the area calculation that I detailed above would not be correct.
answered Apr 3 at 3:10
DeepakDeepak
18k11640
$endgroup$
You are right. The provided explanation is nonsensical. $DC$ cannot be assumed to be a right angle.
However, if you don't make that assumption, and take $BC$ as the only given right angle, the correct answer is "All four sides must be known."
The quadrilateral can be decomposed into two non-overlapping triangles. The first is a right angled triangle formed by sides $B$, $C$ and a hypotenuse, and its area is easy to determine. You can use Pythagoras' Theorem to find the hypotenuse of that right triangle formed by sides $B$ and $C$. That hypotenuse, together with sides $A$ and $D$ forms the other triangle. Its area can be computed using Heron's formula. Just sum the areas.
Note that I'm still making a tacit assumption that this is a convex quadrilateral. A concave quadrilateral is possible with the angle $AD$ being reflex, which is seemingly not representative of the diagram given, but is a possibility when one only has the information of a single right angle and four given sides. In the case of a concave quadrilateral, the area calculation that I detailed above would not be correct.
answered Apr 3 at 3:10
DeepakDeepak
18k11640
edited Apr 3 at 11:08
answered Apr 3 at 3:10
DeepakDeepak
18k11640
answered Apr 3 at 3:10
DeepakDeepak
18k11640
answered Apr 3 at 3:10
DeepakDeepak
18k11640
18k11640
1
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
Apr 3 at 3:21
$begingroup$
You're welcome.
$endgroup$
– Deepak
Apr 3 at 3:24
2
$begingroup$
+1 for Heron's formula - I learned something new today
$endgroup$
– slebetman
Apr 3 at 9:31
11
$begingroup$
@Deepak No, that's not correct. I took some time to double check. For example, see this image representing two quadrilaterals, both satisfying the conditions in the problem, both with the same side lengths, but having different areas - one is strictly smaller than the other.
$endgroup$
– David Z
Apr 3 at 10:02
4
$begingroup$
@Deepak Yes, that was exactly my point, that you can have a concave and convex quadrilateral with the same side lengths. Given that the question is about not assuming things that aren't explicitly stated, I think it may be worth at least mentioning that assumption.
$endgroup$
– David Z
Apr 3 at 10:41
|
show 5 more comments
1
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
Apr 3 at 3:21
$begingroup$
You're welcome.
$endgroup$
– Deepak
Apr 3 at 3:24
2
$begingroup$
+1 for Heron's formula - I learned something new today
$endgroup$
– slebetman
Apr 3 at 9:31
11
$begingroup$
@Deepak No, that's not correct. I took some time to double check. For example, see this image representing two quadrilaterals, both satisfying the conditions in the problem, both with the same side lengths, but having different areas - one is strictly smaller than the other.
$endgroup$
– David Z
Apr 3 at 10:02
4
$begingroup$
@Deepak Yes, that was exactly my point, that you can have a concave and convex quadrilateral with the same side lengths. Given that the question is about not assuming things that aren't explicitly stated, I think it may be worth at least mentioning that assumption.
$endgroup$
– David Z
Apr 3 at 10:41
1
1
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
Apr 3 at 3:21
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
Apr 3 at 3:21
$begingroup$
You're welcome.
$endgroup$
– Deepak
Apr 3 at 3:24
$begingroup$
You're welcome.
$endgroup$
– Deepak
Apr 3 at 3:24
2
2
$begingroup$
+1 for Heron's formula - I learned something new today
$endgroup$
– slebetman
Apr 3 at 9:31
$begingroup$
+1 for Heron's formula - I learned something new today
$endgroup$
– slebetman
Apr 3 at 9:31
11
11
$begingroup$
@Deepak No, that's not correct. I took some time to double check. For example, see this image representing two quadrilaterals, both satisfying the conditions in the problem, both with the same side lengths, but having different areas - one is strictly smaller than the other.
$endgroup$
– David Z
Apr 3 at 10:02
$begingroup$
@Deepak No, that's not correct. I took some time to double check. For example, see this image representing two quadrilaterals, both satisfying the conditions in the problem, both with the same side lengths, but having different areas - one is strictly smaller than the other.
$endgroup$
– David Z
Apr 3 at 10:02
4
4
$begingroup$
@Deepak Yes, that was exactly my point, that you can have a concave and convex quadrilateral with the same side lengths. Given that the question is about not assuming things that aren't explicitly stated, I think it may be worth at least mentioning that assumption.
$endgroup$
– David Z
Apr 3 at 10:41
$begingroup$
@Deepak Yes, that was exactly my point, that you can have a concave and convex quadrilateral with the same side lengths. Given that the question is about not assuming things that aren't explicitly stated, I think it may be worth at least mentioning that assumption.
$endgroup$
– David Z
Apr 3 at 10:41
|
show 5 more comments
$begingroup$
You are correct that the given solution is wrong. Worse still, even if you know that the angles between BC and CD are both right-angles, the purported answer is still wrong! This is because if you're given the lengths of A,B,C, it still does not uniquely determine D because we are not told that the angle between AB is less than $90°$.
In general, it is not enough even if you have all four side lengths. For example, consider convex quadrilateral $PQRS$ such that $PQ = 6$ and $QR = 9$ and $RS = 7$ and $SP = 8$. It is possible that $P,R$ are slightly less than $15$ apart, making $PQRS$ a very skinny quadrilateral whose area can be made arbitrarily close to zero. Alternatively, moving $P,R$ to a distance of $10$ makes $PQRS$ rather square-like with area clearly more than $48$. Bretschneider's formula gives the area for an arbitrary quadrilateral, and you can see from it as well that fixing the side lengths is not enough to determine the area, since it also varies with the sum of two opposite angles.
answered Apr 3 at 3:55
user21820user21820
40.1k544162
$endgroup$
add a comment |
$begingroup$
You are correct that the given solution is wrong. Worse still, even if you know that the angles between BC and CD are both right-angles, the purported answer is still wrong! This is because if you're given the lengths of A,B,C, it still does not uniquely determine D because we are not told that the angle between AB is less than $90°$.
In general, it is not enough even if you have all four side lengths. For example, consider convex quadrilateral $PQRS$ such that $PQ = 6$ and $QR = 9$ and $RS = 7$ and $SP = 8$. It is possible that $P,R$ are slightly less than $15$ apart, making $PQRS$ a very skinny quadrilateral whose area can be made arbitrarily close to zero. Alternatively, moving $P,R$ to a distance of $10$ makes $PQRS$ rather square-like with area clearly more than $48$. Bretschneider's formula gives the area for an arbitrary quadrilateral, and you can see from it as well that fixing the side lengths is not enough to determine the area, since it also varies with the sum of two opposite angles.
answered Apr 3 at 3:55
user21820user21820
40.1k544162
$endgroup$
add a comment |
$begingroup$
You are correct that the given solution is wrong. Worse still, even if you know that the angles between BC and CD are both right-angles, the purported answer is still wrong! This is because if you're given the lengths of A,B,C, it still does not uniquely determine D because we are not told that the angle between AB is less than $90°$.
In general, it is not enough even if you have all four side lengths. For example, consider convex quadrilateral $PQRS$ such that $PQ = 6$ and $QR = 9$ and $RS = 7$ and $SP = 8$. It is possible that $P,R$ are slightly less than $15$ apart, making $PQRS$ a very skinny quadrilateral whose area can be made arbitrarily close to zero. Alternatively, moving $P,R$ to a distance of $10$ makes $PQRS$ rather square-like with area clearly more than $48$. Bretschneider's formula gives the area for an arbitrary quadrilateral, and you can see from it as well that fixing the side lengths is not enough to determine the area, since it also varies with the sum of two opposite angles.
answered Apr 3 at 3:55
user21820user21820
40.1k544162
$endgroup$
You are correct that the given solution is wrong. Worse still, even if you know that the angles between BC and CD are both right-angles, the purported answer is still wrong! This is because if you're given the lengths of A,B,C, it still does not uniquely determine D because we are not told that the angle between AB is less than $90°$.
In general, it is not enough even if you have all four side lengths. For example, consider convex quadrilateral $PQRS$ such that $PQ = 6$ and $QR = 9$ and $RS = 7$ and $SP = 8$. It is possible that $P,R$ are slightly less than $15$ apart, making $PQRS$ a very skinny quadrilateral whose area can be made arbitrarily close to zero. Alternatively, moving $P,R$ to a distance of $10$ makes $PQRS$ rather square-like with area clearly more than $48$. Bretschneider's formula gives the area for an arbitrary quadrilateral, and you can see from it as well that fixing the side lengths is not enough to determine the area, since it also varies with the sum of two opposite angles.
answered Apr 3 at 3:55
user21820user21820
40.1k544162
edited Apr 3 at 12:29
answered Apr 3 at 3:55
user21820user21820
40.1k544162
answered Apr 3 at 3:55
user21820user21820
40.1k544162
answered Apr 3 at 3:55
user21820user21820
40.1k544162
40.1k544162
add a comment |
add a comment |
$begingroup$
You are right: there is absolutely no indication that angle $DC$ is a right angle. If they wanted you to assume it was a right angle, they should have indicated that with another $90$. It really doesn't even look like a right angle (somebody had the bright idea of trying to render the picture in perspective, but we don't even know where the horizon is supposed to be).
answered Apr 3 at 2:46
Robert IsraelRobert Israel
331k23221477
$endgroup$
$begingroup$
That's what I thought. It should explicitly state if any angles are right. However my second question remains, given DC is ambiguous, is this question solvable? I don't think there would be enough information to solve in this case.
$endgroup$
– Jack O.
Apr 3 at 2:52
$begingroup$
@JackO. See my answer. The correct answer would be "All sides must be known".
$endgroup$
– Deepak
Apr 3 at 3:11
$begingroup$
If we know all four lengths and assume no angle is more than 180, then I think there is only one quadrilateral so the area will be unique. I think. But you need all four. If you only three the fourth can be many lengths if the third one "swings".
$endgroup$
– fleablood
Apr 3 at 3:15
$begingroup$
@fleablood: No, in general there are infinitely many quadrilaterals with the same sides in the same order. See Bretschneider's formula for area of a quadrilateral given all sides and 2 opposite angles.
$endgroup$
– user21820
Apr 3 at 12:04
$begingroup$
The rectangle appears to be drawn with "thickness", and if we assume that the thickness is perpendicular to the other sides, we might be able to figure out where the horizon is ... but I'm not sure that there is actually any consistent solution; the other two corners don't have thickness, which imposes constraints on the horizon (it must be such that the other "side" of the thickness for those corners is behind the corners) that may not be satisfiable.
$endgroup$
– Acccumulation
Apr 3 at 15:46
|
show 4 more comments
$begingroup$
You are right: there is absolutely no indication that angle $DC$ is a right angle. If they wanted you to assume it was a right angle, they should have indicated that with another $90$. It really doesn't even look like a right angle (somebody had the bright idea of trying to render the picture in perspective, but we don't even know where the horizon is supposed to be).
answered Apr 3 at 2:46
Robert IsraelRobert Israel
331k23221477
$endgroup$
$begingroup$
That's what I thought. It should explicitly state if any angles are right. However my second question remains, given DC is ambiguous, is this question solvable? I don't think there would be enough information to solve in this case.
$endgroup$
– Jack O.
Apr 3 at 2:52
$begingroup$
@JackO. See my answer. The correct answer would be "All sides must be known".
$endgroup$
– Deepak
Apr 3 at 3:11
$begingroup$
If we know all four lengths and assume no angle is more than 180, then I think there is only one quadrilateral so the area will be unique. I think. But you need all four. If you only three the fourth can be many lengths if the third one "swings".
$endgroup$
– fleablood
Apr 3 at 3:15
$begingroup$
@fleablood: No, in general there are infinitely many quadrilaterals with the same sides in the same order. See Bretschneider's formula for area of a quadrilateral given all sides and 2 opposite angles.
$endgroup$
– user21820
Apr 3 at 12:04
$begingroup$
The rectangle appears to be drawn with "thickness", and if we assume that the thickness is perpendicular to the other sides, we might be able to figure out where the horizon is ... but I'm not sure that there is actually any consistent solution; the other two corners don't have thickness, which imposes constraints on the horizon (it must be such that the other "side" of the thickness for those corners is behind the corners) that may not be satisfiable.
$endgroup$
– Acccumulation
Apr 3 at 15:46
|
show 4 more comments
$begingroup$
You are right: there is absolutely no indication that angle $DC$ is a right angle. If they wanted you to assume it was a right angle, they should have indicated that with another $90$. It really doesn't even look like a right angle (somebody had the bright idea of trying to render the picture in perspective, but we don't even know where the horizon is supposed to be).
answered Apr 3 at 2:46
Robert IsraelRobert Israel
331k23221477
$endgroup$
You are right: there is absolutely no indication that angle $DC$ is a right angle. If they wanted you to assume it was a right angle, they should have indicated that with another $90$. It really doesn't even look like a right angle (somebody had the bright idea of trying to render the picture in perspective, but we don't even know where the horizon is supposed to be).
answered Apr 3 at 2:46
Robert IsraelRobert Israel
331k23221477
answered Apr 3 at 2:46
Robert IsraelRobert Israel
331k23221477
answered Apr 3 at 2:46
Robert IsraelRobert Israel
331k23221477
answered Apr 3 at 2:46
Robert IsraelRobert Israel
331k23221477
331k23221477
$begingroup$
That's what I thought. It should explicitly state if any angles are right. However my second question remains, given DC is ambiguous, is this question solvable? I don't think there would be enough information to solve in this case.
$endgroup$
– Jack O.
Apr 3 at 2:52
$begingroup$
@JackO. See my answer. The correct answer would be "All sides must be known".
$endgroup$
– Deepak
Apr 3 at 3:11
$begingroup$
If we know all four lengths and assume no angle is more than 180, then I think there is only one quadrilateral so the area will be unique. I think. But you need all four. If you only three the fourth can be many lengths if the third one "swings".
$endgroup$
– fleablood
Apr 3 at 3:15
$begingroup$
@fleablood: No, in general there are infinitely many quadrilaterals with the same sides in the same order. See Bretschneider's formula for area of a quadrilateral given all sides and 2 opposite angles.
$endgroup$
– user21820
Apr 3 at 12:04
$begingroup$
The rectangle appears to be drawn with "thickness", and if we assume that the thickness is perpendicular to the other sides, we might be able to figure out where the horizon is ... but I'm not sure that there is actually any consistent solution; the other two corners don't have thickness, which imposes constraints on the horizon (it must be such that the other "side" of the thickness for those corners is behind the corners) that may not be satisfiable.
$endgroup$
– Acccumulation
Apr 3 at 15:46
|
show 4 more comments
$begingroup$
That's what I thought. It should explicitly state if any angles are right. However my second question remains, given DC is ambiguous, is this question solvable? I don't think there would be enough information to solve in this case.
$endgroup$
– Jack O.
Apr 3 at 2:52
$begingroup$
@JackO. See my answer. The correct answer would be "All sides must be known".
$endgroup$
– Deepak
Apr 3 at 3:11
$begingroup$
If we know all four lengths and assume no angle is more than 180, then I think there is only one quadrilateral so the area will be unique. I think. But you need all four. If you only three the fourth can be many lengths if the third one "swings".
$endgroup$
– fleablood
Apr 3 at 3:15
$begingroup$
@fleablood: No, in general there are infinitely many quadrilaterals with the same sides in the same order. See Bretschneider's formula for area of a quadrilateral given all sides and 2 opposite angles.
$endgroup$
– user21820
Apr 3 at 12:04
$begingroup$
The rectangle appears to be drawn with "thickness", and if we assume that the thickness is perpendicular to the other sides, we might be able to figure out where the horizon is ... but I'm not sure that there is actually any consistent solution; the other two corners don't have thickness, which imposes constraints on the horizon (it must be such that the other "side" of the thickness for those corners is behind the corners) that may not be satisfiable.
$endgroup$
– Acccumulation
Apr 3 at 15:46
$begingroup$
That's what I thought. It should explicitly state if any angles are right. However my second question remains, given DC is ambiguous, is this question solvable? I don't think there would be enough information to solve in this case.
$endgroup$
– Jack O.
Apr 3 at 2:52
$begingroup$
That's what I thought. It should explicitly state if any angles are right. However my second question remains, given DC is ambiguous, is this question solvable? I don't think there would be enough information to solve in this case.
$endgroup$
– Jack O.
Apr 3 at 2:52
$begingroup$
@JackO. See my answer. The correct answer would be "All sides must be known".
$endgroup$
– Deepak
Apr 3 at 3:11
$begingroup$
@JackO. See my answer. The correct answer would be "All sides must be known".
$endgroup$
– Deepak
Apr 3 at 3:11
$begingroup$
If we know all four lengths and assume no angle is more than 180, then I think there is only one quadrilateral so the area will be unique. I think. But you need all four. If you only three the fourth can be many lengths if the third one "swings".
$endgroup$
– fleablood
Apr 3 at 3:15
$begingroup$
If we know all four lengths and assume no angle is more than 180, then I think there is only one quadrilateral so the area will be unique. I think. But you need all four. If you only three the fourth can be many lengths if the third one "swings".
$endgroup$
– fleablood
Apr 3 at 3:15
$begingroup$
@fleablood: No, in general there are infinitely many quadrilaterals with the same sides in the same order. See Bretschneider's formula for area of a quadrilateral given all sides and 2 opposite angles.
$endgroup$
– user21820
Apr 3 at 12:04
$begingroup$
@fleablood: No, in general there are infinitely many quadrilaterals with the same sides in the same order. See Bretschneider's formula for area of a quadrilateral given all sides and 2 opposite angles.
$endgroup$
– user21820
Apr 3 at 12:04
$begingroup$
The rectangle appears to be drawn with "thickness", and if we assume that the thickness is perpendicular to the other sides, we might be able to figure out where the horizon is ... but I'm not sure that there is actually any consistent solution; the other two corners don't have thickness, which imposes constraints on the horizon (it must be such that the other "side" of the thickness for those corners is behind the corners) that may not be satisfiable.
$endgroup$
– Acccumulation
Apr 3 at 15:46
$begingroup$
The rectangle appears to be drawn with "thickness", and if we assume that the thickness is perpendicular to the other sides, we might be able to figure out where the horizon is ... but I'm not sure that there is actually any consistent solution; the other two corners don't have thickness, which imposes constraints on the horizon (it must be such that the other "side" of the thickness for those corners is behind the corners) that may not be satisfiable.
$endgroup$
– Acccumulation
Apr 3 at 15:46
|
show 4 more comments
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1
$begingroup$
You know it is a right angle because it has a large "90" on it. Now we can argue they never said why it has a "90" on it and as I am a nitpick I would agree with you... but... I think you and I would lose in any court.
$endgroup$
– fleablood
Apr 3 at 2:45
15
$begingroup$
Not that angle, the one below it.
$endgroup$
– Robert Israel
Apr 3 at 2:46
3
$begingroup$
" even though an angle looks like an angle, it shouldn't be assumed" but it doesn't even look like a right angle.
$endgroup$
– fleablood
Apr 3 at 2:56
7
$begingroup$
Where did you find that test? Online IQ tests are generally untrustworthy even before we get to this quadrilateral problem. Many don't even bother ending in an IQ estimate.
$endgroup$
– J.G.
Apr 3 at 19:09
$begingroup$
yes for Cyclic quadrilaterals. Brahmagupta the Indian mathematician of the 7th century CE knew it.
$endgroup$
– R.C.Cowsik
2 days ago