trapezoids similarity
$begingroup$
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:
geometry
asked Dec 21 '18 at 9:16
IndulaIndula
506
$endgroup$
|
show 5 more comments
$begingroup$
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:
geometry
asked Dec 21 '18 at 9:16
IndulaIndula
506
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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:
geometry
asked Dec 21 '18 at 9:16
IndulaIndula
506
$endgroup$
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:
geometry
geometry
asked Dec 21 '18 at 9:16
IndulaIndula
506
asked Dec 21 '18 at 9:16
IndulaIndula
506
edited Jan 24 at 15:41
Indula
asked Dec 21 '18 at 9:16
IndulaIndula
506
asked Dec 21 '18 at 9:16
IndulaIndula
506
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
|
show 5 more comments
$begingroup$
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
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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:
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.
answered Dec 21 '18 at 12:31
Jaap ScherphuisJaap Scherphuis
4,167717
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.
answered Dec 21 '18 at 13:52
Sameer BahetiSameer Baheti
5718
$endgroup$
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%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
$begingroup$
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:
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.
answered Dec 21 '18 at 12:31
Jaap ScherphuisJaap Scherphuis
4,167717
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
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.
answered Dec 21 '18 at 12:31
Jaap ScherphuisJaap Scherphuis
4,167717
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
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.
answered Dec 21 '18 at 12:31
Jaap ScherphuisJaap Scherphuis
4,167717
$endgroup$
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:
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.
answered Dec 21 '18 at 12:31
Jaap ScherphuisJaap Scherphuis
4,167717
edited Dec 21 '18 at 15:22
answered Dec 21 '18 at 12:31
Jaap ScherphuisJaap Scherphuis
4,167717
answered Dec 21 '18 at 12:31
Jaap ScherphuisJaap Scherphuis
4,167717
answered Dec 21 '18 at 12:31
Jaap ScherphuisJaap Scherphuis
4,167717
4,167717
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.
answered Dec 21 '18 at 13:52
Sameer BahetiSameer Baheti
5718
$endgroup$
add a comment |
$begingroup$
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, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.
answered Dec 21 '18 at 13:52
Sameer BahetiSameer Baheti
5718
$endgroup$
add a comment |
$begingroup$
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, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.
answered Dec 21 '18 at 13:52
Sameer BahetiSameer Baheti
5718
$endgroup$
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, $e=6-sqrt{7}+frac{5sqrt7}{3}=6+frac{2sqrt7}{3}$.
answered Dec 21 '18 at 13:52
Sameer BahetiSameer Baheti
5718
answered Dec 21 '18 at 13:52
Sameer BahetiSameer Baheti
5718
answered Dec 21 '18 at 13:52
Sameer BahetiSameer Baheti
5718
answered Dec 21 '18 at 13:52
Sameer BahetiSameer Baheti
5718
5718
add a comment |
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%2f3048322%2ftrapezoids-similarity%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
$begingroup$
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