Find the length of the segment of the straight line connecting the midpoints of its diagonals.
$begingroup$
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
$endgroup$
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
active
oldest
votes
$begingroup$
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
174k8115327
$endgroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045272%2ffind-the-length-of-the-segment-of-the-straight-line-connecting-the-midpoints-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
174k8115327
$endgroup$
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
174k8115327
$endgroup$
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
174k8115327
$endgroup$
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
174k8115327
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
174k8115327
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
174k8115327
answered Dec 18 '18 at 16:03
Christian BlatterChristian Blatter
174k8115327
174k8115327
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045272%2ffind-the-length-of-the-segment-of-the-straight-line-connecting-the-midpoints-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
