Find the length of the segment of the straight line connecting the midpoints of its diagonals.
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The bases of a trapezoid are $a,b$ long.Find the length of the segment of the straight line connecting the midpoints of its diagonals.
I dont know how to start this question.Please give me some hints.The answer given is $frac{1}{2}|a-b|$.
geometry
asked Dec 18 '18 at 15:14
user984325user984325
246112
$endgroup$
add a comment |
$begingroup$
The bases of a trapezoid are $a,b$ long.Find the length of the segment of the straight line connecting the midpoints of its diagonals.
I dont know how to start this question.Please give me some hints.The answer given is $frac{1}{2}|a-b|$.
geometry
asked Dec 18 '18 at 15:14
user984325user984325
246112
$endgroup$
add a comment |
$begingroup$
The bases of a trapezoid are $a,b$ long.Find the length of the segment of the straight line connecting the midpoints of its diagonals.
I dont know how to start this question.Please give me some hints.The answer given is $frac{1}{2}|a-b|$.
geometry
asked Dec 18 '18 at 15:14
user984325user984325
246112
$endgroup$
The bases of a trapezoid are $a,b$ long.Find the length of the segment of the straight line connecting the midpoints of its diagonals.
I dont know how to start this question.Please give me some hints.The answer given is $frac{1}{2}|a-b|$.
geometry
geometry
asked Dec 18 '18 at 15:14
user984325user984325
246112
asked Dec 18 '18 at 15:14
user984325user984325
246112
asked Dec 18 '18 at 15:14
user984325user984325
246112
asked Dec 18 '18 at 15:14
user984325user984325
246112
asked Dec 18 '18 at 15:14
user984325user984325
246112
246112
add a comment |
add a comment |
2 Answers
2
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votes
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Assume the parallels horizontal. Then one parallel edge is $[u,u+a]$ on some level $y=m+{hover2}$, and the other parallel edge is $[v,v+b]$ on the level $y=m-{hover2}$, whereby $h$ denotes the height of the trapezoid. Now compute the midpoints of the two diagonals.
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
175k8115327
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add a comment |
$begingroup$
Of course, this is just a mean geometry sending a line of length $a$ to one of $-b$, so the signed length is $frac{1}{2}(a-b)$, and unsigned is $frac{1}{2}arrowvert a-barrowvert$, as desired.
answered Dec 18 '18 at 15:17
william122william122
54912
$endgroup$
$begingroup$
I did not understand your method.
$endgroup$
– user984325
Dec 18 '18 at 15:32
add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Assume the parallels horizontal. Then one parallel edge is $[u,u+a]$ on some level $y=m+{hover2}$, and the other parallel edge is $[v,v+b]$ on the level $y=m-{hover2}$, whereby $h$ denotes the height of the trapezoid. Now compute the midpoints of the two diagonals.
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
Assume the parallels horizontal. Then one parallel edge is $[u,u+a]$ on some level $y=m+{hover2}$, and the other parallel edge is $[v,v+b]$ on the level $y=m-{hover2}$, whereby $h$ denotes the height of the trapezoid. Now compute the midpoints of the two diagonals.
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
Assume the parallels horizontal. Then one parallel edge is $[u,u+a]$ on some level $y=m+{hover2}$, and the other parallel edge is $[v,v+b]$ on the level $y=m-{hover2}$, whereby $h$ denotes the height of the trapezoid. Now compute the midpoints of the two diagonals.
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
175k8115327
$endgroup$
Assume the parallels horizontal. Then one parallel edge is $[u,u+a]$ on some level $y=m+{hover2}$, and the other parallel edge is $[v,v+b]$ on the level $y=m-{hover2}$, whereby $h$ denotes the height of the trapezoid. Now compute the midpoints of the two diagonals.
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
175k8115327
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
175k8115327
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
175k8115327
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
175k8115327
175k8115327
add a comment |
add a comment |
$begingroup$
Of course, this is just a mean geometry sending a line of length $a$ to one of $-b$, so the signed length is $frac{1}{2}(a-b)$, and unsigned is $frac{1}{2}arrowvert a-barrowvert$, as desired.
answered Dec 18 '18 at 15:17
william122william122
54912
$endgroup$
$begingroup$
I did not understand your method.
$endgroup$
– user984325
Dec 18 '18 at 15:32
add a comment |
$begingroup$
Of course, this is just a mean geometry sending a line of length $a$ to one of $-b$, so the signed length is $frac{1}{2}(a-b)$, and unsigned is $frac{1}{2}arrowvert a-barrowvert$, as desired.
answered Dec 18 '18 at 15:17
william122william122
54912
$endgroup$
$begingroup$
I did not understand your method.
$endgroup$
– user984325
Dec 18 '18 at 15:32
add a comment |
$begingroup$
Of course, this is just a mean geometry sending a line of length $a$ to one of $-b$, so the signed length is $frac{1}{2}(a-b)$, and unsigned is $frac{1}{2}arrowvert a-barrowvert$, as desired.
answered Dec 18 '18 at 15:17
william122william122
54912
$endgroup$
Of course, this is just a mean geometry sending a line of length $a$ to one of $-b$, so the signed length is $frac{1}{2}(a-b)$, and unsigned is $frac{1}{2}arrowvert a-barrowvert$, as desired.
answered Dec 18 '18 at 15:17
william122william122
54912
answered Dec 18 '18 at 15:17
william122william122
54912
answered Dec 18 '18 at 15:17
william122william122
54912
answered Dec 18 '18 at 15:17
william122william122
54912
54912
$begingroup$
I did not understand your method.
$endgroup$
– user984325
Dec 18 '18 at 15:32
add a comment |
$begingroup$
I did not understand your method.
$endgroup$
– user984325
Dec 18 '18 at 15:32
$begingroup$
I did not understand your method.
$endgroup$
– user984325
Dec 18 '18 at 15:32
$begingroup$
I did not understand your method.
$endgroup$
– user984325
Dec 18 '18 at 15:32
add a comment |
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