Mixing
Geometry . Finding height of trapezoid.
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Find the height of the isoceles trapezoid given base AB=24 and DC=6. The diagonals are perpendicular to each other .
geometry
asked Feb 2 at 17:12
samuel darksamuel dark
1
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add a comment |
$begingroup$
Find the height of the isoceles trapezoid given base AB=24 and DC=6. The diagonals are perpendicular to each other .
geometry
asked Feb 2 at 17:12
samuel darksamuel dark
1
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1
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Hi, and welcome to MSE. If you are able to show any attempts or effort into doing this problem, more people are willing to answer your question. You should also format your question using MathJax.
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– Kyky
Feb 2 at 17:16
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I don't believe there is a unique answer. Two cases. In both start with the long base at the bottom and one side perpendicular to the two bases. For case one at some height the diagonals will be perpendicular. For case two, start with the two bases close together and slide the top base sideways keeping the distance between the bases constant. At some point the diagonals will be perpendicular. For these cases the heights will be greatly different.
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– herb steinberg
Feb 2 at 17:34
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@herbsteinberg the trapezoid is isosceles.
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– user376343
Feb 2 at 18:05
add a comment |
$begingroup$
Find the height of the isoceles trapezoid given base AB=24 and DC=6. The diagonals are perpendicular to each other .
geometry
asked Feb 2 at 17:12
samuel darksamuel dark
1
$endgroup$
Find the height of the isoceles trapezoid given base AB=24 and DC=6. The diagonals are perpendicular to each other .
geometry
geometry
asked Feb 2 at 17:12
samuel darksamuel dark
1
asked Feb 2 at 17:12
samuel darksamuel dark
1
asked Feb 2 at 17:12
samuel darksamuel dark
1
asked Feb 2 at 17:12
samuel darksamuel dark
1
asked Feb 2 at 17:12
samuel darksamuel dark
1
1
1
$begingroup$
Hi, and welcome to MSE. If you are able to show any attempts or effort into doing this problem, more people are willing to answer your question. You should also format your question using MathJax.
$endgroup$
– Kyky
Feb 2 at 17:16
$begingroup$
I don't believe there is a unique answer. Two cases. In both start with the long base at the bottom and one side perpendicular to the two bases. For case one at some height the diagonals will be perpendicular. For case two, start with the two bases close together and slide the top base sideways keeping the distance between the bases constant. At some point the diagonals will be perpendicular. For these cases the heights will be greatly different.
$endgroup$
– herb steinberg
Feb 2 at 17:34
$begingroup$
@herbsteinberg the trapezoid is isosceles.
$endgroup$
– user376343
Feb 2 at 18:05
add a comment |
1
$begingroup$
Hi, and welcome to MSE. If you are able to show any attempts or effort into doing this problem, more people are willing to answer your question. You should also format your question using MathJax.
$endgroup$
– Kyky
Feb 2 at 17:16
$begingroup$
I don't believe there is a unique answer. Two cases. In both start with the long base at the bottom and one side perpendicular to the two bases. For case one at some height the diagonals will be perpendicular. For case two, start with the two bases close together and slide the top base sideways keeping the distance between the bases constant. At some point the diagonals will be perpendicular. For these cases the heights will be greatly different.
$endgroup$
– herb steinberg
Feb 2 at 17:34
$begingroup$
@herbsteinberg the trapezoid is isosceles.
$endgroup$
– user376343
Feb 2 at 18:05
1
1
$begingroup$
Hi, and welcome to MSE. If you are able to show any attempts or effort into doing this problem, more people are willing to answer your question. You should also format your question using MathJax.
$endgroup$
– Kyky
Feb 2 at 17:16
$begingroup$
Hi, and welcome to MSE. If you are able to show any attempts or effort into doing this problem, more people are willing to answer your question. You should also format your question using MathJax.
$endgroup$
– Kyky
Feb 2 at 17:16
$begingroup$
I don't believe there is a unique answer. Two cases. In both start with the long base at the bottom and one side perpendicular to the two bases. For case one at some height the diagonals will be perpendicular. For case two, start with the two bases close together and slide the top base sideways keeping the distance between the bases constant. At some point the diagonals will be perpendicular. For these cases the heights will be greatly different.
$endgroup$
– herb steinberg
Feb 2 at 17:34
$begingroup$
I don't believe there is a unique answer. Two cases. In both start with the long base at the bottom and one side perpendicular to the two bases. For case one at some height the diagonals will be perpendicular. For case two, start with the two bases close together and slide the top base sideways keeping the distance between the bases constant. At some point the diagonals will be perpendicular. For these cases the heights will be greatly different.
$endgroup$
– herb steinberg
Feb 2 at 17:34
$begingroup$
@herbsteinberg the trapezoid is isosceles.
$endgroup$
– user376343
Feb 2 at 18:05
$begingroup$
@herbsteinberg the trapezoid is isosceles.
$endgroup$
– user376343
Feb 2 at 18:05
add a comment |
2 Answers
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If O is the intersection of the diagonals then the triangle ABO is isoscel and with a right angle. If you take OM perpendicular on AB than OM=AB/2. The same way you can find ON perpendicular on CD. Sum ON and OM and you will find the height.
answered Feb 2 at 17:16
mathlearningmathlearning
1887
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add a comment |
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Can you finish it with the use of this picture?
answered Feb 2 at 18:04
user376343user376343
3,9834829
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2 Answers
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active
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2 Answers
2
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$begingroup$
If O is the intersection of the diagonals then the triangle ABO is isoscel and with a right angle. If you take OM perpendicular on AB than OM=AB/2. The same way you can find ON perpendicular on CD. Sum ON and OM and you will find the height.
answered Feb 2 at 17:16
mathlearningmathlearning
1887
$endgroup$
add a comment |
$begingroup$
If O is the intersection of the diagonals then the triangle ABO is isoscel and with a right angle. If you take OM perpendicular on AB than OM=AB/2. The same way you can find ON perpendicular on CD. Sum ON and OM and you will find the height.
answered Feb 2 at 17:16
mathlearningmathlearning
1887
$endgroup$
add a comment |
$begingroup$
If O is the intersection of the diagonals then the triangle ABO is isoscel and with a right angle. If you take OM perpendicular on AB than OM=AB/2. The same way you can find ON perpendicular on CD. Sum ON and OM and you will find the height.
answered Feb 2 at 17:16
mathlearningmathlearning
1887
$endgroup$
If O is the intersection of the diagonals then the triangle ABO is isoscel and with a right angle. If you take OM perpendicular on AB than OM=AB/2. The same way you can find ON perpendicular on CD. Sum ON and OM and you will find the height.
answered Feb 2 at 17:16
mathlearningmathlearning
1887
answered Feb 2 at 17:16
mathlearningmathlearning
1887
answered Feb 2 at 17:16
mathlearningmathlearning
1887
answered Feb 2 at 17:16
mathlearningmathlearning
1887
1887
add a comment |
add a comment |
$begingroup$
Can you finish it with the use of this picture?
answered Feb 2 at 18:04
user376343user376343
3,9834829
$endgroup$
add a comment |
$begingroup$
Can you finish it with the use of this picture?
answered Feb 2 at 18:04
user376343user376343
3,9834829
$endgroup$
add a comment |
$begingroup$
Can you finish it with the use of this picture?
answered Feb 2 at 18:04
user376343user376343
3,9834829
$endgroup$
Can you finish it with the use of this picture?
answered Feb 2 at 18:04
user376343user376343
3,9834829
answered Feb 2 at 18:04
user376343user376343
3,9834829
answered Feb 2 at 18:04
user376343user376343
3,9834829
answered Feb 2 at 18:04
user376343user376343
3,9834829
3,9834829
add a comment |
add a comment |
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$begingroup$
Hi, and welcome to MSE. If you are able to show any attempts or effort into doing this problem, more people are willing to answer your question. You should also format your question using MathJax.
$endgroup$
– Kyky
Feb 2 at 17:16
$begingroup$
I don't believe there is a unique answer. Two cases. In both start with the long base at the bottom and one side perpendicular to the two bases. For case one at some height the diagonals will be perpendicular. For case two, start with the two bases close together and slide the top base sideways keeping the distance between the bases constant. At some point the diagonals will be perpendicular. For these cases the heights will be greatly different.
$endgroup$
– herb steinberg
Feb 2 at 17:34
$begingroup$
@herbsteinberg the trapezoid is isosceles.
$endgroup$
– user376343
Feb 2 at 18:05