Why Are There Two Possible Values of x in Similar Shapes Geometry Problems?

[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 $$ \frac{8+x}{12+3}=\left(\frac{8}{12}\right)^{\pm1} $$.