Is this answer explanation correct?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?
Theorists sure want true answers to this!
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Is this answer explanation correct?
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?
$begingroup$
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 1 hour 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.
$endgroup$
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 1 hour 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:
- 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 1 hour 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.
$endgroup$
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 1 hour 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 1 hour 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 1 hour ago
Blue
49.3k870157
edited 1 hour ago
Blue
49.3k870157
edited 1 hour ago
Blue
49.3k870157
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
asked 2 hours ago
Jack O.Jack O.
16
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.
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.
Jack O. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
1
$begingroup$
Not that angle, the one below it.
$endgroup$
– Robert Israel
2 hours ago
$begingroup$
Not that angle, the one below it.
$endgroup$
– Robert Israel
2 hours ago
1
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
$begingroup$
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
$begingroup$
" 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 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.
answered 1 hour 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: 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 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
1 hour 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 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 $90°$.
answered 1 hour ago
user21820user21820
39.9k544159
$endgroup$
add a comment |
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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.
answered 1 hour 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.
answered 1 hour 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.
answered 1 hour ago
DeepakDeepak
17.7k11539
$endgroup$
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.
answered 1 hour ago
DeepakDeepak
17.7k11539
answered 1 hour ago
DeepakDeepak
17.7k11539
answered 1 hour ago
DeepakDeepak
17.7k11539
answered 1 hour ago
DeepakDeepak
17.7k11539
17.7k11539
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Deepak
1 hour ago
add a comment |
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Deepak
1 hour ago
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
1 hour ago
$begingroup$
Perfect, thank you!
$endgroup$
– Jack O.
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Deepak
1 hour ago
$begingroup$
You're welcome.
$endgroup$
– Deepak
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).
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
1 hour 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).
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
1 hour 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).
answered 2 hours ago
Robert IsraelRobert Israel
330k23219473
$endgroup$
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 2 hours ago
Robert IsraelRobert Israel
330k23219473
answered 2 hours ago
Robert IsraelRobert Israel
330k23219473
answered 2 hours ago
Robert IsraelRobert Israel
330k23219473
answered 2 hours ago
Robert IsraelRobert Israel
330k23219473
330k23219473
$begingroup$
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
1 hour 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 |
$begingroup$
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
1 hour 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
$begingroup$
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
$begingroup$
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
1 hour ago
$begingroup$
@JackO. See my answer. The correct answer would be "All sides must be known".
$endgroup$
– Deepak
1 hour 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
$begingroup$
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 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 $90°$.
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 $90°$.
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 $90°$.
answered 1 hour ago
user21820user21820
39.9k544159
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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 $90°$.
answered 1 hour ago
user21820user21820
39.9k544159
answered 1 hour ago
user21820user21820
39.9k544159
answered 1 hour ago
user21820user21820
39.9k544159
answered 1 hour ago
user21820user21820
39.9k544159
39.9k544159
add a comment |
add a comment |
Jack O. is a new contributor. Be nice, and check out our Code of Conduct.
Jack O. is a new contributor. Be nice, and check out our Code of Conduct.
Jack O. is a new contributor. Be nice, and check out our Code of Conduct.
Jack O. is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
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.
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– fleablood
2 hours ago
1
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Not that angle, the one below it.
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– Robert Israel
2 hours ago
1
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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.
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– fleablood
2 hours ago