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A simple(?) Analytical Geometry Question (Ellipse)The perimeter of the rectangle is $20$, diagonal is $8$ and side is $x$. Show that $x^2-10x+18=0$How to prove the quadrilateral formed by bisectors of a parallelogram is not always square?Find maximum width of a rectangle contained with another (diagonally)Prove this is a rectangleProve that the midpoints of the sides of a quadrilateral lie on a circle if and only if the quadrilateral is orthodiagonal.Area of a concave quadrilateralUnknown internal angles of a quadrilateral where its area and side lengths are knownFind the two missing angles in a quadrilateralGeometry find length given altitude - Is my understanding correct?

3

$begingroup$

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 2 hours ago

Blue

49.3k870157

asked 2 hours ago

Jack O.Jack O.

16

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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh. Just reread. The question is utter bullshit and completely wrong and the person who wrote the answer is a complete idiot. You are correct.
  $endgroup$
  – fleablood
  2 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
  2 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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 2 hours ago

Blue

49.3k870157

asked 2 hours ago

Jack O.Jack O.

16

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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh. Just reread. The question is utter bullshit and completely wrong and the person who wrote the answer is a complete idiot. You are correct.
  $endgroup$
  – fleablood
  2 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
  2 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 2 hours ago

Blue

49.3k870157

asked 2 hours ago

Jack O.Jack O.

16

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 2 hours ago

Blue

49.3k870157

asked 2 hours ago

Jack O.Jack O.

16

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 2 hours ago

Blue

49.3k870157

asked 2 hours ago

Jack O.Jack O.

16

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 2 hours ago

Blue

49.3k870157

edited 2 hours ago

Blue

49.3k870157

asked 2 hours ago

Jack O.Jack O.

16

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 2 hours ago

Jack O.Jack O.

16

3 Answers
3

active

oldest

votes

2

$begingroup$

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 2 hours ago

DeepakDeepak

17.7k11539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, thank you!
  $endgroup$
  – Jack O.
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Deepak
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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 2 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.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackO. See my answer. The correct answer would be "All sides must be known".
  $endgroup$
  – Deepak
  2 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
  1 hour ago
  

  
  
  
  

add a comment | 

0

$begingroup$

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 1 hour ago

user21820user21820

39.9k544159

$endgroup$

add a comment | 

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2

$begingroup$

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 2 hours ago

DeepakDeepak

17.7k11539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, thank you!
  $endgroup$
  – Jack O.
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Deepak
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 2 hours ago

DeepakDeepak

17.7k11539

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, thank you!
  $endgroup$
  – Jack O.
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Deepak
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

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 2 hours ago

DeepakDeepak

17.7k11539

$endgroup$

share|cite|improve this answer

answered 2 hours ago

DeepakDeepak

17.7k11539

answered 2 hours ago

DeepakDeepak

17.7k11539

answered 2 hours ago

DeepakDeepak

17.7k11539

1

$begingroup$

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 2 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.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackO. See my answer. The correct answer would be "All sides must be known".
  $endgroup$
  – Deepak
  2 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
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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 2 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.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackO. See my answer. The correct answer would be "All sides must be known".
  $endgroup$
  – Deepak
  2 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
  1 hour 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 2 hours ago

Robert IsraelRobert Israel

330k23219473

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Robert IsraelRobert Israel

330k23219473

answered 2 hours ago

Robert IsraelRobert Israel

330k23219473

answered 2 hours ago

Robert IsraelRobert Israel

330k23219473

0

$begingroup$

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 1 hour ago

user21820user21820

39.9k544159

$endgroup$

add a comment | 

0

$begingroup$

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 1 hour 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 1 hour ago

user21820user21820

39.9k544159

$endgroup$

share|cite|improve this answer

answered 1 hour ago

user21820user21820

39.9k544159

answered 1 hour ago

user21820user21820

39.9k544159

answered 1 hour ago

user21820user21820

39.9k544159