trapezoids similarity












1












$begingroup$


I found this geometry problem in an IGCSE text. Can't seem to find the missing lengths that they are asking for. I got an estimate for f (f=2.6666... cm) by drawing a line parallel to the left side of the trapezium that intersects the top right-hand vertex and forms two similar triangles.



is it just possible that some data is missing in this problem?



Screenshot of the problem with accompanying diagram:
enter image description here






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Apologies for not mentioning this before. The info in that screen shot was all that was provided.
    $endgroup$
    – Indula
    Dec 21 '18 at 9:18






  • 1




    $begingroup$
    First thoughts: Looking at the angle that the slanted edge makes with the vertical edge, surely it must be true that $$ frac{4}{3} = frac{f}{2} $$ Do you agree?
    $endgroup$
    – Matti P.
    Dec 21 '18 at 9:23










  • $begingroup$
    @Indula Similar quadrilaterals work in much the same way as similar triangles
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 9:28






  • 1




    $begingroup$
    Yes, you could call it the '$AAAA$' criterion for quadrilaterals. Anywho, you could have a solution using triangles.
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 9:50








  • 3




    $begingroup$
    The two trapezoids are NOT similar. Having the same angles does not mean they are merely scaled up versions of one another. For example rectangles are not all similar - they are only similar if they have the same aspect ratio. However, in this problem, if you extend the sloped sides of the trapezoid till they meet, you can use three triangles, and triangles are similar if they have the same angles.
    $endgroup$
    – Jaap Scherphuis
    Dec 21 '18 at 10:38


















1












$begingroup$


I found this geometry problem in an IGCSE text. Can't seem to find the missing lengths that they are asking for. I got an estimate for f (f=2.6666... cm) by drawing a line parallel to the left side of the trapezium that intersects the top right-hand vertex and forms two similar triangles.



is it just possible that some data is missing in this problem?



Screenshot of the problem with accompanying diagram:
enter image description here






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Apologies for not mentioning this before. The info in that screen shot was all that was provided.
    $endgroup$
    – Indula
    Dec 21 '18 at 9:18






  • 1




    $begingroup$
    First thoughts: Looking at the angle that the slanted edge makes with the vertical edge, surely it must be true that $$ frac{4}{3} = frac{f}{2} $$ Do you agree?
    $endgroup$
    – Matti P.
    Dec 21 '18 at 9:23










  • $begingroup$
    @Indula Similar quadrilaterals work in much the same way as similar triangles
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 9:28






  • 1




    $begingroup$
    Yes, you could call it the '$AAAA$' criterion for quadrilaterals. Anywho, you could have a solution using triangles.
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 9:50








  • 3




    $begingroup$
    The two trapezoids are NOT similar. Having the same angles does not mean they are merely scaled up versions of one another. For example rectangles are not all similar - they are only similar if they have the same aspect ratio. However, in this problem, if you extend the sloped sides of the trapezoid till they meet, you can use three triangles, and triangles are similar if they have the same angles.
    $endgroup$
    – Jaap Scherphuis
    Dec 21 '18 at 10:38
















1












1








1





$begingroup$


I found this geometry problem in an IGCSE text. Can't seem to find the missing lengths that they are asking for. I got an estimate for f (f=2.6666... cm) by drawing a line parallel to the left side of the trapezium that intersects the top right-hand vertex and forms two similar triangles.



is it just possible that some data is missing in this problem?



Screenshot of the problem with accompanying diagram:
enter image description here






geometry







share|cite|improve this question











$endgroup$




I found this geometry problem in an IGCSE text. Can't seem to find the missing lengths that they are asking for. I got an estimate for f (f=2.6666... cm) by drawing a line parallel to the left side of the trapezium that intersects the top right-hand vertex and forms two similar triangles.



is it just possible that some data is missing in this problem?



Screenshot of the problem with accompanying diagram:
enter image description here






geometry




geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 15:41







Indula

















asked Dec 21 '18 at 9:16









IndulaIndula

506




506












  • $begingroup$
    Apologies for not mentioning this before. The info in that screen shot was all that was provided.
    $endgroup$
    – Indula
    Dec 21 '18 at 9:18






  • 1




    $begingroup$
    First thoughts: Looking at the angle that the slanted edge makes with the vertical edge, surely it must be true that $$ frac{4}{3} = frac{f}{2} $$ Do you agree?
    $endgroup$
    – Matti P.
    Dec 21 '18 at 9:23










  • $begingroup$
    @Indula Similar quadrilaterals work in much the same way as similar triangles
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 9:28






  • 1




    $begingroup$
    Yes, you could call it the '$AAAA$' criterion for quadrilaterals. Anywho, you could have a solution using triangles.
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 9:50








  • 3




    $begingroup$
    The two trapezoids are NOT similar. Having the same angles does not mean they are merely scaled up versions of one another. For example rectangles are not all similar - they are only similar if they have the same aspect ratio. However, in this problem, if you extend the sloped sides of the trapezoid till they meet, you can use three triangles, and triangles are similar if they have the same angles.
    $endgroup$
    – Jaap Scherphuis
    Dec 21 '18 at 10:38




















  • $begingroup$
    Apologies for not mentioning this before. The info in that screen shot was all that was provided.
    $endgroup$
    – Indula
    Dec 21 '18 at 9:18






  • 1




    $begingroup$
    First thoughts: Looking at the angle that the slanted edge makes with the vertical edge, surely it must be true that $$ frac{4}{3} = frac{f}{2} $$ Do you agree?
    $endgroup$
    – Matti P.
    Dec 21 '18 at 9:23










  • $begingroup$
    @Indula Similar quadrilaterals work in much the same way as similar triangles
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 9:28






  • 1




    $begingroup$
    Yes, you could call it the '$AAAA$' criterion for quadrilaterals. Anywho, you could have a solution using triangles.
    $endgroup$
    – Shubham Johri
    Dec 21 '18 at 9:50








  • 3




    $begingroup$
    The two trapezoids are NOT similar. Having the same angles does not mean they are merely scaled up versions of one another. For example rectangles are not all similar - they are only similar if they have the same aspect ratio. However, in this problem, if you extend the sloped sides of the trapezoid till they meet, you can use three triangles, and triangles are similar if they have the same angles.
    $endgroup$
    – Jaap Scherphuis
    Dec 21 '18 at 10:38


















$begingroup$
Apologies for not mentioning this before. The info in that screen shot was all that was provided.
$endgroup$
– Indula
Dec 21 '18 at 9:18




$begingroup$
Apologies for not mentioning this before. The info in that screen shot was all that was provided.
$endgroup$
– Indula
Dec 21 '18 at 9:18




1




1




$begingroup$
First thoughts: Looking at the angle that the slanted edge makes with the vertical edge, surely it must be true that $$ frac{4}{3} = frac{f}{2} $$ Do you agree?
$endgroup$
– Matti P.
Dec 21 '18 at 9:23




$begingroup$
First thoughts: Looking at the angle that the slanted edge makes with the vertical edge, surely it must be true that $$ frac{4}{3} = frac{f}{2} $$ Do you agree?
$endgroup$
– Matti P.
Dec 21 '18 at 9:23












$begingroup$
@Indula Similar quadrilaterals work in much the same way as similar triangles
$endgroup$
– Shubham Johri
Dec 21 '18 at 9:28




$begingroup$
@Indula Similar quadrilaterals work in much the same way as similar triangles
$endgroup$
– Shubham Johri
Dec 21 '18 at 9:28




1




1




$begingroup$
Yes, you could call it the '$AAAA$' criterion for quadrilaterals. Anywho, you could have a solution using triangles.
$endgroup$
– Shubham Johri
Dec 21 '18 at 9:50






$begingroup$
Yes, you could call it the '$AAAA$' criterion for quadrilaterals. Anywho, you could have a solution using triangles.
$endgroup$
– Shubham Johri
Dec 21 '18 at 9:50






3




3




$begingroup$
The two trapezoids are NOT similar. Having the same angles does not mean they are merely scaled up versions of one another. For example rectangles are not all similar - they are only similar if they have the same aspect ratio. However, in this problem, if you extend the sloped sides of the trapezoid till they meet, you can use three triangles, and triangles are similar if they have the same angles.
$endgroup$
– Jaap Scherphuis
Dec 21 '18 at 10:38






$begingroup$
The two trapezoids are NOT similar. Having the same angles does not mean they are merely scaled up versions of one another. For example rectangles are not all similar - they are only similar if they have the same aspect ratio. However, in this problem, if you extend the sloped sides of the trapezoid till they meet, you can use three triangles, and triangles are similar if they have the same angles.
$endgroup$
– Jaap Scherphuis
Dec 21 '18 at 10:38












2 Answers
2






active

oldest

votes


















1












$begingroup$

You wrote that you drew a line parallel to the left side from the top-right corner down to the base. That line will also be split into lengths of exactly $3$ and $2$, just like the left side. This is because you have parallelograms on the left. This mean that the ratios on the triangles on the right is $3:(2+3)$, and this gives you that the right hand side of the trapezoid is $4*5/3=20/3$, making $f=20/3 - 4 = 8/3$.



To find $e$ you will have to make some assumption. From the right hand end of the horizontal line of length $6$, you can draw a line down to the base, parallel to the left side of the trapezoid. To the left of that line you have a parallelogram, so that line is of length $2$, and the part of the base to the left of it is of length $6$. So to find $e$, you need the part of the base to the right of it. That unknown part is the base of a triangle with sides $2$ and $f=8/3$. Unfortunately, it is impossible to find that third length unless we know one of the angles of that triangle.



To illustrate this, I have drawn the diagram in two ways:



enter image description here



You can clearly see the length of the base $e$ differs, even though the given lengths and the length of $f$ all remain unchanged.



Presumably the left side is supposed to be perpendicular to the base. In that case you can use Pythagoras to find out the exact length.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks so much! There was a typo in the very first post I made which indicated f=1.6666... which should have been f=8/3=2.666... Thanks for verifying my method for finding the length f. I was thinking about assuming that the trapezium is right-angled (it looks like that after all) and using the Pythagorean theorem to find the length e but was hesitant about explicitly adopting that assumption seeing as they haven't indicated that explicitly.
    $endgroup$
    – Indula
    Dec 21 '18 at 15:15










  • $begingroup$
    But on another note, if it hadn't been right-angled, and instead had been something akin to what you've sketched in your first diagram, is it possible to find the length, even if we knew the degrees of the angle?
    $endgroup$
    – Indula
    Dec 21 '18 at 15:17






  • 1




    $begingroup$
    Yes, in that case you will have to use the law of cosines, which can be considered an extension of the Pythagorean theorem that works for an arbitrary angle, not just a right angle.
    $endgroup$
    – Jaap Scherphuis
    Dec 21 '18 at 15:19












  • $begingroup$
    Thanks a lot! That was really helpful!
    $endgroup$
    – Indula
    Dec 21 '18 at 15:25



















0












$begingroup$

Since the trapezium is right, base length of smaller $triangle$ by Pythagoras theorem is $sqrt{4^2-3^2}=sqrt{7}$.



So remnant of the base of smaller trapezium is $6-sqrt{7}$.



Clearly $f=frac83$ by similarity of triangles.



Now finding the base of bigger $triangle$ (say $E$) by similarity of triangles:
$$frac{E}{sqrt{7}}=frac{4+frac83}{4}$$
or $E=frac{5sqrt7}{3}$



So that, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.






share|cite|improve this answer









$endgroup$













    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048322%2ftrapezoids-similarity%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









    1












    $begingroup$

    You wrote that you drew a line parallel to the left side from the top-right corner down to the base. That line will also be split into lengths of exactly $3$ and $2$, just like the left side. This is because you have parallelograms on the left. This mean that the ratios on the triangles on the right is $3:(2+3)$, and this gives you that the right hand side of the trapezoid is $4*5/3=20/3$, making $f=20/3 - 4 = 8/3$.



    To find $e$ you will have to make some assumption. From the right hand end of the horizontal line of length $6$, you can draw a line down to the base, parallel to the left side of the trapezoid. To the left of that line you have a parallelogram, so that line is of length $2$, and the part of the base to the left of it is of length $6$. So to find $e$, you need the part of the base to the right of it. That unknown part is the base of a triangle with sides $2$ and $f=8/3$. Unfortunately, it is impossible to find that third length unless we know one of the angles of that triangle.



    To illustrate this, I have drawn the diagram in two ways:



    enter image description here



    You can clearly see the length of the base $e$ differs, even though the given lengths and the length of $f$ all remain unchanged.



    Presumably the left side is supposed to be perpendicular to the base. In that case you can use Pythagoras to find out the exact length.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks so much! There was a typo in the very first post I made which indicated f=1.6666... which should have been f=8/3=2.666... Thanks for verifying my method for finding the length f. I was thinking about assuming that the trapezium is right-angled (it looks like that after all) and using the Pythagorean theorem to find the length e but was hesitant about explicitly adopting that assumption seeing as they haven't indicated that explicitly.
      $endgroup$
      – Indula
      Dec 21 '18 at 15:15










    • $begingroup$
      But on another note, if it hadn't been right-angled, and instead had been something akin to what you've sketched in your first diagram, is it possible to find the length, even if we knew the degrees of the angle?
      $endgroup$
      – Indula
      Dec 21 '18 at 15:17






    • 1




      $begingroup$
      Yes, in that case you will have to use the law of cosines, which can be considered an extension of the Pythagorean theorem that works for an arbitrary angle, not just a right angle.
      $endgroup$
      – Jaap Scherphuis
      Dec 21 '18 at 15:19












    • $begingroup$
      Thanks a lot! That was really helpful!
      $endgroup$
      – Indula
      Dec 21 '18 at 15:25
















    1












    $begingroup$

    You wrote that you drew a line parallel to the left side from the top-right corner down to the base. That line will also be split into lengths of exactly $3$ and $2$, just like the left side. This is because you have parallelograms on the left. This mean that the ratios on the triangles on the right is $3:(2+3)$, and this gives you that the right hand side of the trapezoid is $4*5/3=20/3$, making $f=20/3 - 4 = 8/3$.



    To find $e$ you will have to make some assumption. From the right hand end of the horizontal line of length $6$, you can draw a line down to the base, parallel to the left side of the trapezoid. To the left of that line you have a parallelogram, so that line is of length $2$, and the part of the base to the left of it is of length $6$. So to find $e$, you need the part of the base to the right of it. That unknown part is the base of a triangle with sides $2$ and $f=8/3$. Unfortunately, it is impossible to find that third length unless we know one of the angles of that triangle.



    To illustrate this, I have drawn the diagram in two ways:



    enter image description here



    You can clearly see the length of the base $e$ differs, even though the given lengths and the length of $f$ all remain unchanged.



    Presumably the left side is supposed to be perpendicular to the base. In that case you can use Pythagoras to find out the exact length.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks so much! There was a typo in the very first post I made which indicated f=1.6666... which should have been f=8/3=2.666... Thanks for verifying my method for finding the length f. I was thinking about assuming that the trapezium is right-angled (it looks like that after all) and using the Pythagorean theorem to find the length e but was hesitant about explicitly adopting that assumption seeing as they haven't indicated that explicitly.
      $endgroup$
      – Indula
      Dec 21 '18 at 15:15










    • $begingroup$
      But on another note, if it hadn't been right-angled, and instead had been something akin to what you've sketched in your first diagram, is it possible to find the length, even if we knew the degrees of the angle?
      $endgroup$
      – Indula
      Dec 21 '18 at 15:17






    • 1




      $begingroup$
      Yes, in that case you will have to use the law of cosines, which can be considered an extension of the Pythagorean theorem that works for an arbitrary angle, not just a right angle.
      $endgroup$
      – Jaap Scherphuis
      Dec 21 '18 at 15:19












    • $begingroup$
      Thanks a lot! That was really helpful!
      $endgroup$
      – Indula
      Dec 21 '18 at 15:25














    1












    1








    1





    $begingroup$

    You wrote that you drew a line parallel to the left side from the top-right corner down to the base. That line will also be split into lengths of exactly $3$ and $2$, just like the left side. This is because you have parallelograms on the left. This mean that the ratios on the triangles on the right is $3:(2+3)$, and this gives you that the right hand side of the trapezoid is $4*5/3=20/3$, making $f=20/3 - 4 = 8/3$.



    To find $e$ you will have to make some assumption. From the right hand end of the horizontal line of length $6$, you can draw a line down to the base, parallel to the left side of the trapezoid. To the left of that line you have a parallelogram, so that line is of length $2$, and the part of the base to the left of it is of length $6$. So to find $e$, you need the part of the base to the right of it. That unknown part is the base of a triangle with sides $2$ and $f=8/3$. Unfortunately, it is impossible to find that third length unless we know one of the angles of that triangle.



    To illustrate this, I have drawn the diagram in two ways:



    enter image description here



    You can clearly see the length of the base $e$ differs, even though the given lengths and the length of $f$ all remain unchanged.



    Presumably the left side is supposed to be perpendicular to the base. In that case you can use Pythagoras to find out the exact length.






    share|cite|improve this answer











    $endgroup$



    You wrote that you drew a line parallel to the left side from the top-right corner down to the base. That line will also be split into lengths of exactly $3$ and $2$, just like the left side. This is because you have parallelograms on the left. This mean that the ratios on the triangles on the right is $3:(2+3)$, and this gives you that the right hand side of the trapezoid is $4*5/3=20/3$, making $f=20/3 - 4 = 8/3$.



    To find $e$ you will have to make some assumption. From the right hand end of the horizontal line of length $6$, you can draw a line down to the base, parallel to the left side of the trapezoid. To the left of that line you have a parallelogram, so that line is of length $2$, and the part of the base to the left of it is of length $6$. So to find $e$, you need the part of the base to the right of it. That unknown part is the base of a triangle with sides $2$ and $f=8/3$. Unfortunately, it is impossible to find that third length unless we know one of the angles of that triangle.



    To illustrate this, I have drawn the diagram in two ways:



    enter image description here



    You can clearly see the length of the base $e$ differs, even though the given lengths and the length of $f$ all remain unchanged.



    Presumably the left side is supposed to be perpendicular to the base. In that case you can use Pythagoras to find out the exact length.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 21 '18 at 15:22

























    answered Dec 21 '18 at 12:31









    Jaap ScherphuisJaap Scherphuis

    4,129717




    4,129717












    • $begingroup$
      Thanks so much! There was a typo in the very first post I made which indicated f=1.6666... which should have been f=8/3=2.666... Thanks for verifying my method for finding the length f. I was thinking about assuming that the trapezium is right-angled (it looks like that after all) and using the Pythagorean theorem to find the length e but was hesitant about explicitly adopting that assumption seeing as they haven't indicated that explicitly.
      $endgroup$
      – Indula
      Dec 21 '18 at 15:15










    • $begingroup$
      But on another note, if it hadn't been right-angled, and instead had been something akin to what you've sketched in your first diagram, is it possible to find the length, even if we knew the degrees of the angle?
      $endgroup$
      – Indula
      Dec 21 '18 at 15:17






    • 1




      $begingroup$
      Yes, in that case you will have to use the law of cosines, which can be considered an extension of the Pythagorean theorem that works for an arbitrary angle, not just a right angle.
      $endgroup$
      – Jaap Scherphuis
      Dec 21 '18 at 15:19












    • $begingroup$
      Thanks a lot! That was really helpful!
      $endgroup$
      – Indula
      Dec 21 '18 at 15:25


















    • $begingroup$
      Thanks so much! There was a typo in the very first post I made which indicated f=1.6666... which should have been f=8/3=2.666... Thanks for verifying my method for finding the length f. I was thinking about assuming that the trapezium is right-angled (it looks like that after all) and using the Pythagorean theorem to find the length e but was hesitant about explicitly adopting that assumption seeing as they haven't indicated that explicitly.
      $endgroup$
      – Indula
      Dec 21 '18 at 15:15










    • $begingroup$
      But on another note, if it hadn't been right-angled, and instead had been something akin to what you've sketched in your first diagram, is it possible to find the length, even if we knew the degrees of the angle?
      $endgroup$
      – Indula
      Dec 21 '18 at 15:17






    • 1




      $begingroup$
      Yes, in that case you will have to use the law of cosines, which can be considered an extension of the Pythagorean theorem that works for an arbitrary angle, not just a right angle.
      $endgroup$
      – Jaap Scherphuis
      Dec 21 '18 at 15:19












    • $begingroup$
      Thanks a lot! That was really helpful!
      $endgroup$
      – Indula
      Dec 21 '18 at 15:25
















    $begingroup$
    Thanks so much! There was a typo in the very first post I made which indicated f=1.6666... which should have been f=8/3=2.666... Thanks for verifying my method for finding the length f. I was thinking about assuming that the trapezium is right-angled (it looks like that after all) and using the Pythagorean theorem to find the length e but was hesitant about explicitly adopting that assumption seeing as they haven't indicated that explicitly.
    $endgroup$
    – Indula
    Dec 21 '18 at 15:15




    $begingroup$
    Thanks so much! There was a typo in the very first post I made which indicated f=1.6666... which should have been f=8/3=2.666... Thanks for verifying my method for finding the length f. I was thinking about assuming that the trapezium is right-angled (it looks like that after all) and using the Pythagorean theorem to find the length e but was hesitant about explicitly adopting that assumption seeing as they haven't indicated that explicitly.
    $endgroup$
    – Indula
    Dec 21 '18 at 15:15












    $begingroup$
    But on another note, if it hadn't been right-angled, and instead had been something akin to what you've sketched in your first diagram, is it possible to find the length, even if we knew the degrees of the angle?
    $endgroup$
    – Indula
    Dec 21 '18 at 15:17




    $begingroup$
    But on another note, if it hadn't been right-angled, and instead had been something akin to what you've sketched in your first diagram, is it possible to find the length, even if we knew the degrees of the angle?
    $endgroup$
    – Indula
    Dec 21 '18 at 15:17




    1




    1




    $begingroup$
    Yes, in that case you will have to use the law of cosines, which can be considered an extension of the Pythagorean theorem that works for an arbitrary angle, not just a right angle.
    $endgroup$
    – Jaap Scherphuis
    Dec 21 '18 at 15:19






    $begingroup$
    Yes, in that case you will have to use the law of cosines, which can be considered an extension of the Pythagorean theorem that works for an arbitrary angle, not just a right angle.
    $endgroup$
    – Jaap Scherphuis
    Dec 21 '18 at 15:19














    $begingroup$
    Thanks a lot! That was really helpful!
    $endgroup$
    – Indula
    Dec 21 '18 at 15:25




    $begingroup$
    Thanks a lot! That was really helpful!
    $endgroup$
    – Indula
    Dec 21 '18 at 15:25











    0












    $begingroup$

    Since the trapezium is right, base length of smaller $triangle$ by Pythagoras theorem is $sqrt{4^2-3^2}=sqrt{7}$.



    So remnant of the base of smaller trapezium is $6-sqrt{7}$.



    Clearly $f=frac83$ by similarity of triangles.



    Now finding the base of bigger $triangle$ (say $E$) by similarity of triangles:
    $$frac{E}{sqrt{7}}=frac{4+frac83}{4}$$
    or $E=frac{5sqrt7}{3}$



    So that, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since the trapezium is right, base length of smaller $triangle$ by Pythagoras theorem is $sqrt{4^2-3^2}=sqrt{7}$.



      So remnant of the base of smaller trapezium is $6-sqrt{7}$.



      Clearly $f=frac83$ by similarity of triangles.



      Now finding the base of bigger $triangle$ (say $E$) by similarity of triangles:
      $$frac{E}{sqrt{7}}=frac{4+frac83}{4}$$
      or $E=frac{5sqrt7}{3}$



      So that, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since the trapezium is right, base length of smaller $triangle$ by Pythagoras theorem is $sqrt{4^2-3^2}=sqrt{7}$.



        So remnant of the base of smaller trapezium is $6-sqrt{7}$.



        Clearly $f=frac83$ by similarity of triangles.



        Now finding the base of bigger $triangle$ (say $E$) by similarity of triangles:
        $$frac{E}{sqrt{7}}=frac{4+frac83}{4}$$
        or $E=frac{5sqrt7}{3}$



        So that, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.






        share|cite|improve this answer









        $endgroup$



        Since the trapezium is right, base length of smaller $triangle$ by Pythagoras theorem is $sqrt{4^2-3^2}=sqrt{7}$.



        So remnant of the base of smaller trapezium is $6-sqrt{7}$.



        Clearly $f=frac83$ by similarity of triangles.



        Now finding the base of bigger $triangle$ (say $E$) by similarity of triangles:
        $$frac{E}{sqrt{7}}=frac{4+frac83}{4}$$
        or $E=frac{5sqrt7}{3}$



        So that, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 13:52









        Sameer BahetiSameer Baheti

        5168




        5168






























            draft saved

            draft discarded




















































            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048322%2ftrapezoids-similarity%23new-answer', 'question_page');
            }
            );

            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







            -

            Popular posts from this blog

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more