Did I arrive at this answer using sound reasoning or was it a fluke?
$begingroup$
Because E is the midpoint of AB and F is the midpoint of CD, I just added (25-12) to 25 to calculate that AD = 38. And because G is the midpoint AE and H is the midpoint of FD, I just subtracted (13/2) from 38 to calculate that GH = 31.5. This is the correct answer, but I'm not sure if I used sound reasoning.
geometry
asked Dec 20 '18 at 3:38
user27343user27343
343
$endgroup$
add a comment |
$begingroup$
Because E is the midpoint of AB and F is the midpoint of CD, I just added (25-12) to 25 to calculate that AD = 38. And because G is the midpoint AE and H is the midpoint of FD, I just subtracted (13/2) from 38 to calculate that GH = 31.5. This is the correct answer, but I'm not sure if I used sound reasoning.
geometry
asked Dec 20 '18 at 3:38
user27343user27343
343
$endgroup$
add a comment |
$begingroup$
Because E is the midpoint of AB and F is the midpoint of CD, I just added (25-12) to 25 to calculate that AD = 38. And because G is the midpoint AE and H is the midpoint of FD, I just subtracted (13/2) from 38 to calculate that GH = 31.5. This is the correct answer, but I'm not sure if I used sound reasoning.
geometry
asked Dec 20 '18 at 3:38
user27343user27343
343
$endgroup$
Because E is the midpoint of AB and F is the midpoint of CD, I just added (25-12) to 25 to calculate that AD = 38. And because G is the midpoint AE and H is the midpoint of FD, I just subtracted (13/2) from 38 to calculate that GH = 31.5. This is the correct answer, but I'm not sure if I used sound reasoning.
geometry
geometry
asked Dec 20 '18 at 3:38
user27343user27343
343
asked Dec 20 '18 at 3:38
user27343user27343
343
asked Dec 20 '18 at 3:38
user27343user27343
343
asked Dec 20 '18 at 3:38
user27343user27343
343
asked Dec 20 '18 at 3:38
user27343user27343
343
343
add a comment |
add a comment |
2 Answers
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$begingroup$
It is sound, but needs some justification. It is true that the length of chords parallel to BC or AD in your trapezoid are a linear function of the length from B to the endpoint or of the length from C to the endpoint. You can justify this by extending AB and DC until they meet at point Z. All the triangles with one vertex at Z and with the opposite side parallel to BC or AD are similar.
answered Dec 20 '18 at 3:45
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$begingroup$
Yes, I see that now. Thank you!
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
$begingroup$
The reasoning is basically sound. To help see this, try adding lines perpendicular lines from G, E, B, C, F and H to AD, say at points I, J, K, L, M and N. Now, due to the various lines which are parallel, the triangles AGI, AEJ and ABK are similar to each other, and the triangles CLD, FMD and HND are also similar to each other. You can then use the various midpoint values to determine the appropriate ratios of the various triangle side lines to each other and then use this to confirm what you determined.
answered Dec 20 '18 at 3:45
John OmielanJohn Omielan
4,1251215
$endgroup$
$begingroup$
Thank you for taking the time to answer my question.
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
It is sound, but needs some justification. It is true that the length of chords parallel to BC or AD in your trapezoid are a linear function of the length from B to the endpoint or of the length from C to the endpoint. You can justify this by extending AB and DC until they meet at point Z. All the triangles with one vertex at Z and with the opposite side parallel to BC or AD are similar.
answered Dec 20 '18 at 3:45
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$begingroup$
Yes, I see that now. Thank you!
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
$begingroup$
It is sound, but needs some justification. It is true that the length of chords parallel to BC or AD in your trapezoid are a linear function of the length from B to the endpoint or of the length from C to the endpoint. You can justify this by extending AB and DC until they meet at point Z. All the triangles with one vertex at Z and with the opposite side parallel to BC or AD are similar.
answered Dec 20 '18 at 3:45
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$begingroup$
Yes, I see that now. Thank you!
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
$begingroup$
It is sound, but needs some justification. It is true that the length of chords parallel to BC or AD in your trapezoid are a linear function of the length from B to the endpoint or of the length from C to the endpoint. You can justify this by extending AB and DC until they meet at point Z. All the triangles with one vertex at Z and with the opposite side parallel to BC or AD are similar.
answered Dec 20 '18 at 3:45
Ross MillikanRoss Millikan
300k24200374
$endgroup$
It is sound, but needs some justification. It is true that the length of chords parallel to BC or AD in your trapezoid are a linear function of the length from B to the endpoint or of the length from C to the endpoint. You can justify this by extending AB and DC until they meet at point Z. All the triangles with one vertex at Z and with the opposite side parallel to BC or AD are similar.
answered Dec 20 '18 at 3:45
Ross MillikanRoss Millikan
300k24200374
answered Dec 20 '18 at 3:45
Ross MillikanRoss Millikan
300k24200374
answered Dec 20 '18 at 3:45
Ross MillikanRoss Millikan
300k24200374
answered Dec 20 '18 at 3:45
Ross MillikanRoss Millikan
300k24200374
300k24200374
$begingroup$
Yes, I see that now. Thank you!
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
$begingroup$
Yes, I see that now. Thank you!
$endgroup$
– user27343
Dec 20 '18 at 4:01
$begingroup$
Yes, I see that now. Thank you!
$endgroup$
– user27343
Dec 20 '18 at 4:01
$begingroup$
Yes, I see that now. Thank you!
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
$begingroup$
The reasoning is basically sound. To help see this, try adding lines perpendicular lines from G, E, B, C, F and H to AD, say at points I, J, K, L, M and N. Now, due to the various lines which are parallel, the triangles AGI, AEJ and ABK are similar to each other, and the triangles CLD, FMD and HND are also similar to each other. You can then use the various midpoint values to determine the appropriate ratios of the various triangle side lines to each other and then use this to confirm what you determined.
answered Dec 20 '18 at 3:45
John OmielanJohn Omielan
4,1251215
$endgroup$
$begingroup$
Thank you for taking the time to answer my question.
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
$begingroup$
The reasoning is basically sound. To help see this, try adding lines perpendicular lines from G, E, B, C, F and H to AD, say at points I, J, K, L, M and N. Now, due to the various lines which are parallel, the triangles AGI, AEJ and ABK are similar to each other, and the triangles CLD, FMD and HND are also similar to each other. You can then use the various midpoint values to determine the appropriate ratios of the various triangle side lines to each other and then use this to confirm what you determined.
answered Dec 20 '18 at 3:45
John OmielanJohn Omielan
4,1251215
$endgroup$
$begingroup$
Thank you for taking the time to answer my question.
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
$begingroup$
The reasoning is basically sound. To help see this, try adding lines perpendicular lines from G, E, B, C, F and H to AD, say at points I, J, K, L, M and N. Now, due to the various lines which are parallel, the triangles AGI, AEJ and ABK are similar to each other, and the triangles CLD, FMD and HND are also similar to each other. You can then use the various midpoint values to determine the appropriate ratios of the various triangle side lines to each other and then use this to confirm what you determined.
answered Dec 20 '18 at 3:45
John OmielanJohn Omielan
4,1251215
$endgroup$
The reasoning is basically sound. To help see this, try adding lines perpendicular lines from G, E, B, C, F and H to AD, say at points I, J, K, L, M and N. Now, due to the various lines which are parallel, the triangles AGI, AEJ and ABK are similar to each other, and the triangles CLD, FMD and HND are also similar to each other. You can then use the various midpoint values to determine the appropriate ratios of the various triangle side lines to each other and then use this to confirm what you determined.
answered Dec 20 '18 at 3:45
John OmielanJohn Omielan
4,1251215
answered Dec 20 '18 at 3:45
John OmielanJohn Omielan
4,1251215
answered Dec 20 '18 at 3:45
John OmielanJohn Omielan
4,1251215
answered Dec 20 '18 at 3:45
John OmielanJohn Omielan
4,1251215
4,1251215
$begingroup$
Thank you for taking the time to answer my question.
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
$begingroup$
Thank you for taking the time to answer my question.
$endgroup$
– user27343
Dec 20 '18 at 4:01
$begingroup$
Thank you for taking the time to answer my question.
$endgroup$
– user27343
Dec 20 '18 at 4:01
$begingroup$
Thank you for taking the time to answer my question.
$endgroup$
– user27343
Dec 20 '18 at 4:01
add a comment |
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