Geometry: Similar Shapes

[paulb203]

Homework Statement

See screenshot in main body

Relevant Equations

I don't know if there is a generic equation
Maybe something like AB(SF)=AC ?

Why are there two possible values of x?

Here's what I've done so far.

AE(SF)=AD
12(SF)=15
SF=15/12
=5/4

AB(SF)=AC
8(5/4)=AC
AC=10

AC-AB=x
x=10-8
x=2

A hint please, as to my next step

 

[SammyS]

Homework Statement: See screenshot in main body
Relevant Equations: I don't know if there is a generic equation
Maybe something like AB(SF)=AC ?

View attachment 363169
Why are there two possible values of x?

Here's what I've done so far.

AE(SF)=AD
12(SF)=15
SF=15/12
=5/4

AB(SF)=AC
8(5/4)=AC
AC=10

AC-AB=x
x=10-8
x=2

A hint please, as to my next step

Are you assuming anything about how side CD is related to side EB ?

 

[paulb203]

Are you assuming anything about how side CD is related to side EB ?

View attachment 363174

Thanks.

I wasn't until now, as I didn't think it was relevant, but you've got me thinking about it.

My guess is the relationship is always the same; the side of the small triangle * SF = side of the large triangle?

 

[fresh_42]

Thanks.

I wasn't until now, as I didn't think it was relevant, but you've got me thinking about it.

My guess is the relationship is always the same; the side of the small triangle * SF = side of the large triangle?

These are similar with even three equal angles, but do not have parallel lines.

 

[SammyS]

Thanks.

I wasn't until now, as I didn't think it was relevant, but you've got me thinking about it.

My guess is the relationship is always the same; the side of the small triangle * SF = side of the large triangle?

What is "SF" ?
I suppose may be Scale Factor ?

The given figure may be fooling you into assuming more than what is stated.

 

[DaveC426913]

My guess is the relationship is always the same; the side of the small triangle * SF = side of the large triangle?

No. Not always the same. Therein lies your assumption.

 

[TensorCalculus]

View attachment 363211
These are similar with even three equal angles, but do not have parallel lines.

similar to this one:

ABC and AED are similar

maybe you have assumed something about the diagram that you should not have... hopefully this diagram helps you figure it out?

EDIT: sorry, that D looks a bit like an O, incase there's any confusion

 

[SammyS]

similar to this one:

View attachment 363218
ABC and AED are similar

maybe you have assumed something about the diagram that you should not have... hopefully this diagram helps you figure it out?

EDIT: sorry, that D looks a bit like an O, incase there's any confusion

What does this figure have to do with the given problem?

 

[TensorCalculus]

What does this have to do with the given problem?

I'll DM you, so I don't give the answer away to the OP.

 

[DaveC426913]

What does this have to do with the given problem?

It's a push in the right direction for the OP.

 

[Mark44]

Homework Statement: See screenshot in main body
Relevant Equations: I don't know if there is a generic equation
Maybe something like AB(SF)=AC ?

AE(SF)=AD

What is "SF" ?

Yes, what is "SF"?

Two triangles are said to be similar if their corresponding angles are equal. In similar triangles, the corresponding sides are in the same proportion. With regard to your drawing in post #1, this means that AB: AC :: AE : AD
Another way to say the same is $\frac{AB}{AC} = \frac{AE}{AD}$
Or, using ratios and the dimensions in your drawing,
$\frac 8 {8 + x} = \frac {12}{15}$

 

[renormalize]

Or, using ratios and the dimensions in your drawing,
$\frac 8 {8 + x} = \frac {12}{15}$

Your equation has the unique solution $x=2$. So why does the original problem in post #1 state:

?

 

[DaveC426913]

You are making the same assumption as the OP (that's not the only expression for x)

No. Renormalize's post is rhetorical. He is drawing attention to the fact that it's Mark44 who is the one making the same assumption as the OP.

 

[Mark44]

After taking a third look at the drawing, I realized that the problem is purposely tricky, with sides CD and BE only appearing to be parallel.

 

[TensorCalculus]

No. Renormalize's post is rhetorical. He is drawing attention to the fact that it's Mark44 who is the one making the same assumption as the OP.

Oh I didn't understand that it was rhetorical... my bad

 

[Steve4Physics]

@paulb203, if you are still struggling, you have probably hit a bit of a mental block. Here’s an extra problem to try which should help. (Hope it’s not giving too much away.)

For each of the following diagrams:
a) explain why each pair of triangles are similar
b) express x in terms of a, b and c.

 

[paulb203]

Thanks, guys.
Loads for me to think about; I'll get back to this shortly.
For now; yes, by (SF) I mean Scale Factor. Sorry about that.

 

[paulb203]

Back to basics.

Are the two triangles they are referring to as follows;

ABE, the small triangle, bottom left?
And, ACD, the large triangle, which is the whole picture, and incorporates ABE?
Is AED, the base of the large triangle, 15cm (12+3)?
Is AE, the base of the small triangle 12/15ths of the base AED? And is the Scale Factor (SF) therefore 5/4

(12*(5/4)=15)?

Is ABE, the left side of the large triangle, 8+xcm?
Is AB, the left side of the small triangle, 8/(8+x)ths of ABC. I thought at this point I could infer the SF from the my first calculations but now I’m not so sure. I was thinking 8*(5/4)=10; is that incorrect?
I’ve been nudged towards looking at CD in relation to BE. I assumed those were parallel, as I thought similar shapes were proportional versions of each other (?).

Hmm..?

I might have to revisit the similar shape units of Maths Genie GCSE Revision...

 

[paulb203]

View attachment 363211
These are similar with even three equal angles, but do not have parallel lines.

Thanks.
Can similar shapes be equal in side lengths and angles; basically the same , except for how they are positioned?

 

[paulb203]

After taking a third look at the drawing, I realized that the problem is purposely tricky, with sides CD and BE only appearing to be parallel.

Thanks.
I thought they had to be parallel in order for the triangles to be similar (?).

 

[TensorCalculus]

I’ve been nudged towards looking at CD in relation to BE. I assumed those were parallel, as I thought similar shapes were proportional versions of each other (?).

Do CD and BE have to be parallel? Look at what I posted previously, #7, where both triangles are similar. What do you notice about sides DE and BC?

Thanks.
Can similar shapes be equal in side lengths and angles; basically the same , except for how they are positioned?

If the side lengths and the angles are all the same, then you have a special case of similar shapes: called congruent shapes. Similarity means that the ratio of the side lengths are the same, and the angles are the same (e.g. a triangle with side lengths 3,4,5 could be similar to a triangle of side lengths 6,8,10 since the ratio of the lengths of the sides are the same)
Note that there is no mention of orientation: shapes can be similar and oriented in completely different directions.

 

[paulb203]

@paulb203, if you are still struggling, you have probably hit a bit of a mental block. Here’s an extra problem to try which should help. (Hope it’s not giving too much away.)

For each of the following diagrams:
a) explain why each pair of triangles are similar
b) express x in terms of a, b and c.
View attachment 363238

Thanks, Steve.
I'll give this a go after I've revisited the very basics; I fear I've got myself in a right muddle :)

 

[Mark44]

Can similar shapes be equal in side lengths and angles; basically the same , except for how they are positioned?

Yes, and as already noted, if both the corresponding sides and angles are equal, the two figures are congruent.

 

[fresh_42]

Thanks.
Can similar shapes be equal in side lengths and angles; basically the same , except for how they are positioned?

Similar only means that the angles are the same, not the size. If, e.g., two angles are equal, then we can have the situation as the figure suggests, with parallel lines, or we can swap between the two equal angles and get a completely different picture. All we have is, that S:S:S and A:A:A are equal, where S are the side lengths and A the three angles. You have only considered S:S but you need to take all three.

What we know from similarity is that
$$
\overline{AB}\, : \,\overline{BE}\, : \,\overline{EA} \,=\, \overline{AC}\, : \,\overline{CD}\, : \,\overline{DA}.
$$
This translates to
$$
8\, : \,\overline{BE}\, : \,12\,=\,(8+x)\, : \,\overline{CD}\, : \,15,
$$
possibly renaming the vertices of one of the triangles, the problem statement isn't very precise in that regard.

 

[DaveC426913]

I thought similar shapes were proportional versions of each other (?).

They are.

I assumed those were parallel,

Not necessarily...

How could this possibly be?

Here is a pair of similar triangles:

 

[pbuk]

I'll give this a go after I've revisited the very basics; I fear I've got myself in a right muddle :)

You haven't got yourself in a muddle, it is a trick question and you are in a muddle because you haven't spotted the trick.

I'm not sure revisiting the basics will help, the only thing that will help is spotting the trick. The first thing to notice is that the line segment DE that is 3 cm long is drawn almost the same length as AD which is 12 cm: the diagram has been deliberately drawn to mislead you.

In particular, the diagram tricks you into thinking that $\angle ABE = \angle ACD \text{ and } \angle BEA = \angle CDA$. What other possibility is there?

 

[Mark44]

The first thing to notice is that the line segment DE that is 3 cm long is drawn almost the same length as AD which is 12 cm: the diagram has been deliberately drawn to mislead you.

I dislike problems like this where the figure is maliciously drawn way out of scale.

 

[DaveC426913]

I dislike problems like this where the figure is maliciously drawn way out of scale.

Malicious is an unusual choice of words. It's deliberate, but it's for a purpose.

 

[Mark44]

I stand by malicious.

 

[Gavran]

The equation which describes both options should be written in the next form $$ \frac{8+x}{12+3}=\left(\frac{8}{12}\right)^{\pm1} $$.

 

[Steve4Physics]

I think the best approach for a student at this level is to draw two diagrams corresponding to the two possible solutions.

One diagram has BE and CD in parallel; the other doesn't . The OP needs to draw both diagrams (roughly to scale). If they can't draw them, they can't understand and answer the question properly. (That's why I gave a big hint in Post #16.)

Minor edits.

 

[DaveC426913]

I stand by malicious.

OK. I disagree, and I'll explain why.

Malice, at its core, involves intent to harm.

Drawing a diagram that is not representative of the final solution is an important tool to teach the student to be aware of exactly the kind of pitfall we are all witnessing in this thread. Geometry is not just about the student getting the answer by rote; it is also about the student understanding what is being asked.

If the opening question had been drawn as you expect, the student would have learned nothing about real-world problems that are not always obvious. They would go out into the world seeing two lines that look to their naive eye to be parallel, but in fact are not. They would have not learned rigour in geometry. And geometry is all about rigour.

The student must learn to see when such things as parallelism are not true - and notably, that similar triangles need not be un-mirrored images of each other. They're not going to learn that if every question's mathematically rigorous answer can be short-circuited by enabling eyeball estimation.

It's not malicious - not harm - it is, instead, a critical part of the lesson.

The OP's struggle is a perfect example. Because of this particular question, they will never make this mistake again.

 

[Mark44]

OK. I disagree, and I'll explain why.

Malice, at its core, involves intent to harm.

Or, in the case of the drawing in this thread, the harm arises by the effort of the problem poser to show how clever he is, and conversely, how stupid the poor student who is the target.

Drawing a diagram that is not representative of the final solution is an important tool to teach the student to be aware of exactly the kind of pitfall we are all witnessing in this thread. Geometry is not just about the student getting the answer by rote; it is also about the student understanding what is being asked.

Baloney. It would be one thing if the inaccuracy of the drawing were inadvertent. But as pointed out by @pbuk in post #26, the drawing is wildly inaccurate, which is not an accident.

 

[renormalize]

Baloney. It would be one thing if the inaccuracy of the drawing were inadvertent. But as pointed out by @pbuk in post #26, the drawing is wildly inaccurate, which is not an accident.

Would you find the problem more palatable if its author had added an explicit warning such as:
"Note: illustration is not necessarily drawn to scale."?

 

[DaveC426913]

Or, in the case of the drawing in this thread, the harm arises by the effort of the problem poser to show how clever he is, and conversely, how stupid the poor student who is the target.

You are second-guessing the intent of the poser. Why look for malice when there is a perfectly valid educational basis for the set up of the problem?

How does a student learn rigour of they can just eyeball the answer? How would they exercise their analytical muscle if the work is done for them?

In fact: why did we bother letting this thread get 34 posts in, if not to encourage the OP to think it through himself. Isn't that equally "malicious"? Aren't we all guilty of that which what you accuse the problem poser?

Can we accuse you of showing you how clever you are, and how stupid the OP is - when you didn't tell them the twist in the problem right off the bat? (Or did you, too, have your reasons for not spoon-feeding the OP?)

Baloney. It would be one thing if the inaccuracy of the drawing were inadvertent. But as pointed out by @pbuk in post #26, the drawing is wildly inaccurate, which is not an accident.

Nobody ever suggested it was an accident. It is a deliberate, tried-and-true technique to teach the student a critical geometry analysis skill.

Learning is more than rote number-crunching; learning starts with problem-solving: what is being asked. In geometry, more than many other subjects, that is a matter of deducing and what is given, and - notably - what is not given.

 

[Mark44]

Would you find the problem more palatable if its author had added an explicit warning such as:
"Note: illustration is not necessarily drawn to scale."?

Yes, I would not object so strenuously.

You are second-guessing the intent of the poser. Why look for malice when there is a perfectly valid educational basis for the set up of the problem?

How about sadism?

Nobody ever suggested it was an accident. It is a deliberate, tried-and-true technique to teach the student a critical geometry analysis skill.

I agree that it probably was not accidental, but posing a problem that is deliberately misleading is NOT a tried-and-true pedagogical technique.

Learning is more than rote number-crunching; learning starts with problem-solving: what is being asked. In geometry, more than many other subjects, that is a matter of deducing and what is given, and - notably - what is not given.

I maintain that what was given was deliberately misleading. As I said before, the intent appears to be to convince the student how clever the teacher is, and less so about problem-solving techniques.

@DaveC426913 I taught mathematics at the college level for about 25 years. I don't recall that any of my peers or myself posing any problems as willfully misleading as this one. What's your experience?

 

[fresh_42]

I think it is important to teach not making hidden assumptions, and being led to false conclusions only because pictures are easier to read than text. This is not only important in mathematics, but also important for any kind of investigation. Maybe a small hint like "e.g." at the picture label would have been helpful, but that is a matter of taste, and of how easy an exercise is intended to be.

 

[TensorCalculus]

Honestly, it is good that the OP gets exposed to this sort of question in homework. Whether questions like this are a good thing or not (though I think I find myself leaning more towards @DaveC426913 and @fresh_42 's side - even if the question is written to try and show how smart the question writer is, it still forces the person doing the question to not rely on the diagram and make assumptions, which is a valuable lesson) they will almost certainly come up in formal exams (GCSE, A-Level, etc) so practicing them, and struggling on them, means that they have a better chance of not making similar assumptions on the actual exam, where whether they get the question right or wrong does actually matter.

That being said, technically the diagram is correct for the first possible value of x (x=2), just not the second, less obvious one (which has probably been done on purpose). To the OP (I hope I am not giving too much away with this): we have already established the maybe BE and CD are not parallel. What happens when we get rid of this assumption? Can you draw a diagram where this assumption is not the case? If BE and CD aren't parallel, then $\angle ABE \not = \angle ACD$ and the same can be said for $\angle ABE$ and $\angle ADC$
but for triangles to be similar, then they must have 3 equal angles. How can BE and CD be parallel, yet the triangles still be similar? There is one other possibility...

 

[DaveC426913]

I taught mathematics at the college level for about 25 years.

I was afraid of that ...

 

[hutchphd]

Malicious is an unusual choice of words. It's deliberate, but it's for a purpose.

And that purpose seems to me to be malicious.
Is this a restatement of the parental admonition to "do as I say, not as I do": "do as I say not as I draw?"
I am not seeing this as interesting........am I still not "getting it"?

 

[pbuk]

[questions like this] will almost certainly come up in formal exams (GCSE, A-Level, etc)

Edit: I have just realised that this is Q.22 from Edexcel Maths Paper 1 November 2017. I'm off to read the examiner's report.

They almost certainly will not. GCSE and A-Level exams have been developed over many years to test the ability to apply knowledge, not the ability to spot trick questions. Any (UK) exam board that let a question like this slip through the net of pre-testing would be heavily criticised within the profession.

 

[pbuk]

Nobody ever suggested it was an accident. It is a deliberate, tried-and-true technique to teach the student a critical geometry analysis skill.

No, far from being a 'tried and true technique', undermining a pupil's confidence by setting trick questions is bad practice.

From the OP's posts in this thread we can see that this poor question has taught nothing, and has destroyed any confidence they had in their knowledge of similar triangles:

I might have to revisit the similar shape units of Maths Genie GCSE Revision...

I'll give this a go after I've revisited the very basics; I fear I've got myself in a right muddle :)

Hang on in there @paulb203, the whole world is not out to get you! I'd advise you to use a different source for learning reinforcement questions in future though (where did you get this shocker from?)

 

[pbuk]

Edit: I have just realised that this is Q.22 from Edexcel Maths Paper 1 November 2017. I'm off to read the examiner's report.

They almost certainly will not. GCSE and A-Level exams have been developed over many years to test the ability to apply knowledge, not the ability to spot trick questions. Any (UK) exam board that let a question like this slip through the net of pre-testing would be heavily criticised within the profession.

So here is the examiners' report for this question:

"It was good to see a number of students having some understanding of similar shapes and awarding 2 marks for a value of x =2 correctly found wasn’t uncommon. Almost no students were then able to deduce the second assumption that could be made and as such no further marks were awarded."​

Doesn't look to me like anything to be proud of.

I cannot find anything similar (please excuse the pun) in more recent papers.

 

[DaveC426913]

No, far from being a 'tried and true technique', undermining a pupil's confidence by setting trick questions is bad practice.

It's not a "trick question". Students are taught to not trust diagrams.

An angle that looks about 90 degrees - is not 90 degrees unless it is labeled so.
Two lines that look sortta parallel - are not parallel unless they are labeled so.

This is how one teaches geometry.

Students are taught that similar triangles can be mirrors of each other.

The OP didn't get it because he has not yet learned the lesson that this particular problem is designed to challenge him with.

From the OP's posts in this thread we can see that this poor question has taught nothing, and has destroyed any confidence they had in their knowledge of similar triangles:

We should spoon feed them only answers that don't challenge them?

Would you be comfortable driving on a bridge made by a student who was fed only the easy answers?

Would you drive on a bridge built by a student that did not know similar triangles can be mirrors of each other?

 

[DaveC426913]

So here is the examiners' report for this question:

"It was good to see a number of students having some understanding of similar shapes and awarding 2 marks for a value of x =2 correctly found wasn’t uncommon. Almost no students were then able to deduce the second assumption that could be made and as such no further marks were awarded."​

Doesn't look to me like anything to be proud of.

All the more reason why this kind of question needs to be taught.

Or:

"We found that understanding stress fractures in airplane structures was too hard, and many students were unable to deduce the correct answer. In light of that, we have decided to simply not give them hard problems. We will just hand out a cheat sheet of all the formulae they'll need for their aviation engineering degrees and give every single one of them a degree."

 

[DaveC426913]

posing a problem that is deliberately misleading is NOT a tried-and-true pedagogical technique.

How exactly is the problem "deliberately misleading"?

Are you asserting that

- that in a given geometric diagram, two lines that look pretty parallel can be assumed to be parallel?

- that, in this diagram, I can assume AD and BC are parallel simply because they look parallel?

- that this diagram is "maliciously misleading"?

 

[mathwonk]

I am also a retired college math teacher (over 40 years teaching). So I am on the side of Mark44 in this matter. I do not disagree with anything anyone else has said. I just know from decades of experience that it is hard enough to teach even the most basic facts, that one does not ever have time to waste tricking 98% of people with questions like this.

When ever you pose a test question, you should ask yourself what it is testing. Then ask yourself whether it furthers your announced classroom goals to test that property. And ask yourself also if it is consistent with your grading policy whether getting it right should count towards a grade.

However I also know that it is useless to discuss teaching philosophy in the abstract, as in this thread, as no one ever agrees. People only grasp what one intends by his teaching philosophy when they see him carry it out in class. But I could not leave Mark out on a limb here, given my actual opinion, as well as my respect for those who disagree.

yes one does want to encourage caution in students who ignore unstated assumptions. It would be easy to do that here in a more helpful way I think. One could simply ask if there is a second interpretation of the question that yields a second answer, for extra credit, and remind students that it is necessary to specify the actual correspondence of vertices when invoking a similarity.

The goal is for most of the students to actually learn something, not for the question to be phrased so as to fool almost all of them.
I believe negative reinforcement is not actually the most highly recommended method of effective learning.

Apologies if I have misinterpreted the many cogent arguments that have been suggested. I admit that, even after 40 years, I myself never mastered the art of teaching.

peace.

 

[Mark44]

How exactly is the problem "deliberately misleading"?

You're kidding, right?
Here's from post #26, as to why the figure in the problem is deliberately misleading.

The first thing to notice is that the line segment DE that is 3 cm long is drawn almost the same length as AD which is 12 cm: the diagram has been deliberately drawn to mislead you.

It's not a "trick question".

See @pbuk's opinion on this (copied above) as well as @mathwonk's (copied below).

The goal is for most of the students to actually learn something, not for the question to be phrased so as to fool almost all of them.
I believe negative reinforcement is not actually the most highly recommended method of effective learning.

Students are taught to not trust diagrams.

???
Are they also taught not to trust definitions and theorems? This seems ridiculous.

 

[A.T.]

... the harm arises by the effort of the problem poser to show how clever he is,...

Ironically, it can also suggest that the problem poser wasn't clever enough to come up with a problem that is interesting by itself, and instead had to resort to stuffing a boring problem with misleading queues.

As for the pedagogical value, it makes perfect sense and does no harm to give students this as exercise that is not graded, to teach them to watch out for misleading pictures. And that includes not fooling yourself with your own exemplary diagrams that you draw when solving a problem.

But I would not put that into a graded test.

See also:

 

[fresh_42]

The exercise mentions explicitly two possible values of x. What more hint do you need to think about your hidden assumptions? It is literally an invitation to reflect on the situation and re-read the problem statement.

A criminal investigator won't be given such hints in his daily work. Where should he learn not to trust the obvious? The real world is complex, and we far too often fall for the simplest explanation, disregarding its complexity. Calling this question malicious, despite it actually stating that there is more than one solution, is ridiculous. It calls for a reflection of the problem statement, and that seems to be the goal to be learnt, not the interception theorems.