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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?

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

6

$begingroup$

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:

1. Is my criticism valid or am I just being too proud because I got a question wrong?


2. Given my interpretation, DC is not a right angle, can this problem be solved?

geometry

share|cite|improve this question

edited 1 hour ago

Discrete lizard

14010

asked 8 hours ago

Jack O.Jack O.

313

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not that angle, the one below it.
  $endgroup$
  – Robert Israel
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  " 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
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

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:

1. Is my criticism valid or am I just being too proud because I got a question wrong?


2. Given my interpretation, DC is not a right angle, can this problem be solved?

geometry

share|cite|improve this question

edited 1 hour ago

Discrete lizard

14010

asked 8 hours ago

Jack O.Jack O.

313

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not that angle, the one below it.
  $endgroup$
  – Robert Israel
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  " 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
  8 hours 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:

1. Is my criticism valid or am I just being too proud because I got a question wrong?


2. Given my interpretation, DC is not a right angle, can this problem be solved?

geometry

share|cite|improve this question

edited 1 hour ago

Discrete lizard

14010

asked 8 hours ago

Jack O.Jack O.

313

New contributor

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

$endgroup$

share|cite|improve this question

edited 1 hour ago

Discrete lizard

14010

asked 8 hours ago

Jack O.Jack O.

313

New contributor

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

share|cite|improve this question

edited 1 hour ago

Discrete lizard

14010

asked 8 hours ago

Jack O.Jack O.

313

New contributor

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

edited 1 hour ago

Discrete lizard

14010

edited 1 hour ago

Discrete lizard

14010

asked 8 hours ago

Jack O.Jack O.

313

New contributor

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

asked 8 hours ago

Jack O.Jack O.

313

3 Answers
3

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9

$begingroup$

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.

share|cite|improve this answer

answered 7 hours ago

DeepakDeepak

17.8k11539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, thank you!
  $endgroup$
  – Jack O.
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Deepak
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for Heron's formula - I learned something new today
  $endgroup$
  – slebetman
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  25 mins ago
  

  
  
  
  

 | 
show 4 more comments

7

$begingroup$

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 $90°$.

share|cite|improve this answer

answered 7 hours ago

user21820user21820

39.9k544159

$endgroup$

add a comment | 

4

$begingroup$

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).

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackO. See my answer. The correct answer would be "All sides must be known".
  $endgroup$
  – Deepak
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

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9

$begingroup$

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.

share|cite|improve this answer

answered 7 hours ago

DeepakDeepak

17.8k11539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, thank you!
  $endgroup$
  – Jack O.
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Deepak
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for Heron's formula - I learned something new today
  $endgroup$
  – slebetman
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  25 mins ago
  

  
  
  
  

 | 
show 4 more comments

9

$begingroup$

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.

share|cite|improve this answer

answered 7 hours ago

DeepakDeepak

17.8k11539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, thank you!
  $endgroup$
  – Jack O.
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Deepak
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for Heron's formula - I learned something new today
  $endgroup$
  – slebetman
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  25 mins ago
  

  
  
  
  

 | 
show 4 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.

share|cite|improve this answer

answered 7 hours ago

DeepakDeepak

17.8k11539

$endgroup$

share|cite|improve this answer

answered 7 hours ago

DeepakDeepak

17.8k11539

answered 7 hours ago

DeepakDeepak

17.8k11539

answered 7 hours ago

DeepakDeepak

17.8k11539

7

$begingroup$

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 $90°$.

share|cite|improve this answer

answered 7 hours ago

user21820user21820

39.9k544159

$endgroup$

add a comment | 

7

$begingroup$

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 $90°$.

share|cite|improve this answer

answered 7 hours ago

user21820user21820

39.9k544159

$endgroup$

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 $90°$.

share|cite|improve this answer

answered 7 hours ago

user21820user21820

39.9k544159

$endgroup$

share|cite|improve this answer

answered 7 hours ago

user21820user21820

39.9k544159

answered 7 hours ago

user21820user21820

39.9k544159

answered 7 hours ago

user21820user21820

39.9k544159

4

$begingroup$

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).

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackO. See my answer. The correct answer would be "All sides must be known".
  $endgroup$
  – Deepak
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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).

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

330k23219473

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackO. See my answer. The correct answer would be "All sides must be known".
  $endgroup$
  – Deepak
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

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).

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

330k23219473

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

330k23219473

answered 8 hours ago

Robert IsraelRobert Israel

330k23219473

answered 8 hours ago

Robert IsraelRobert Israel

330k23219473