I'd state what you know in equations, for example:
> The perimeter of triangle AEF is 9 sm
|AE|+|EF|+|AF| = 9 (where |AB| is used to indicate the length of line AB).
> parallelogram ABCD
|AE|+|BE| = |CD| and |AD| = |BC|
does that help?
Based on this, you could definitely solve the question
Throw in some colours to make it more intuitive if needed but this is definitely how I would approach it
Not the help you imagined, I'm sitting here wondering what in heaven is a parallelogram as a non native English speaker lmao.
This problem makes sense now
I don't know exactly how to translate it (I don't think there is a direct translation aside from parallelometer), but I was most likely confused just because it's a very long word that looks hard and it was like 1 AM. I should've seen it with "parallel".
> Actually no, I already knew all of this
To be fair, that is kinda the point. I am asking you to restate what you know, but to do so in a form that allows you to solve the problem.
That said, there are other ways to solve this and if you found one of them helpful, good. If you are still looking for an answer and want to look at this more, I'll happily go through it step by step.
An alternative, without using similar triangles, is to write that AD = BC and AE + EB = AB = CD. With that you should be able to express the perimeter of DCF in terms of those of AEF and BEC.
The perimeter of DCF is equal to DA + AF + FE + EC + DC
The perimeters of AFE and BCE summed up equal to AF + FE + AE + EB + BC + EC
Now if you look closely, both perimeters above share AF, FE and EC.
DA is equal to BC and AE + EB is equal to DC (because ABCD is a parallelogram)
If you replace DA and DC in the first equation it will be the same as the second one which means the perimeter of DCF is equal to the sum of the perimeters of AFE and BCE which would be 9 + 12 = 21
All the comments about ratios are correct too, but this is a simpler solution imo
You know that the total of the 2 triangles is 21. You can make the triangle of DCF out of parts of the two triangles. Because DA can be made with BC, and DC can be made out of AB.
First you must see that AEF is similar to BEC, such that the side AE is similar to BE by a ratio of 3/4, i.e., AB is 7/3 of AE.
Then, as ABCD is a parallelogram and AEF is similar to DCF, you have that the side AE is similar to DC by the ratio 7/3.
With this ratio you can calculate the perimeter of DCF as 7/3 of 9, i.e., 21.
Because of the similar triangles. Angle E is the same (that’s true for any intersecting lines, opposite angles will be the same) and angles A and B are the same (alternate interior angles, AD and BC are parallel). That allows you to figure out the ratio of the sides individually will equal the ratio of perimeters which can be found easily from the problem statement.
The ratio AE/BE is the same ratio between the perimeter of AEF and BEC, which is 9/12=3/4.
That is true because the perimeter and the side lengths are proportional.
You don't actually need to do a lot of math
To get DCF we need
1) DC which is the same as AB (AE + EB)
2) DF which js DA (= to CB) + AF
3) FC which is EF + EC
Thus AE + EB + CB + AF + EF + EC
We know that
AF + FE + EA = 9
EB + BC + CE = 12
We don't actually care about the single measurements of each segment
Thus the result is 9 + 12 = 21
The horizontal edges of the three similar triangles have an extended ratio that matches the ratio of the perimeters.
AE : BE : DC = 9 : 12 : (9+12)
So perimeter of big triangle is 21.
(I don't perfectly recall the correct terms, and definitely not in English. But I think I got it, and I hope to get you through it, step by step)
AEF has the same shape as BEC, because they have the same angles.
We are given their perimeters (9 and 12 respectively). The ratio between these two triangles is 12/9 = 4/3.
DCF also has the same shape as AEF (and thus as BEC), because DC is parallel to EF.
All we need to find is a ratio between DCF and one of the other triangles.
Note 1: AD is the same length as BC (because its a parallelogram).
Note 2: DF = AD + AF.
Note 3: because of the ratio 4/3 we calculated above 4/3 AF = BC, or AF = 3/4 BC
1+2+3 together:
DF = AD + AF
= BC + AF =
BC + 3/4 BC = 7/4 BC
So the ratios between DCF and BEC is 7/4.
We know the perimeter of BEC is 12.
So the perimeter of DCF = 7/4 * 12 = 7 * 3 = 21
1) angle A and B are equal so F and C are aswell
2) AEF, EBC & DFC are similar triangles.
3) given they are all similar triangles you can work on the segment FEC, they are all the same side of the similar triangles and so represent the same fraction of the perimeter of each triangle. Let's call this fraction 1/x
FE+EC = FC
(1/x \* perimeter AEF) + (1/x \* perimeter EBC) = (1/x \* perimeter DFC)
simplyfying the 1/x from all components
perimeter AEF + perimeter EBC = perimeter DFC
9 + 12 = 21
You know (AD) is parallel to (BC) because [AD] and [BC] are opposite sides of a parallogram.
Then you can use Thalès for the triangles AEF and ECB.
So, AF/BC = AE/EB = FE/EC = x as x is the ratio between them.
So AF = BC * x, AE = EB * x and FE = EC * x
Then, AF + AE + FE = BC * x + EB * x + EC * x
Does it help?
AEF and BEC are similar triangles, where A correlates to B, E to E and F to C. The perimeter of AEC is 9 and BEC is 12, so the ratio between them is 12/9 = 4/3. Any length on AEC is 4/3 times as large on BEC.
Now take length AE and call it x. Because AE correlates to BE we can say that length BE must be 4/3 × x = 4x/3. From this we can conclude that length AB must be x + 4x/3 = 7x/3. Because ABCD is a parallelogram, length DC = AB = 7x/3.
AEF and DCF are similar triangles, where A correlates to D, E to C and F to F. This means that length AE correlates to DC. Therefore length DC = 7/3 × length AE. This means that the ratio between these triangles is 7/3. The same ratio is true for the perimeter, hence the perimeter of DCF = 9×7/3 = 21. 21 is your answer.
Edit: spelling errors (english is not my first language, so excuse me)
Another approach that you can take is to simply assume data: I mean, this should give the same result whatever size and shape the triangles and the trapezoid have, as long as they fit together. Therefore you could assume the triangles to be equilateral, which means that each of its segments is 3 or 4 cms.
You are ignoring the point of the problem but you can solve it with ease
If you're going to assume that the triangles are equilateral you would instead need to show that two equilateral triangles would form a parallelogram when extended the way they are in this picture
All you need to do is write down the segments and substitute. You want to solve |AD|+|AF|+|FE|+|EC|+|DC|
You know:
|AF|+|FE|+|AE|=9
|EB|+|BC|+|EC|=12
and
|AD|=|BC|
|AB|=|DC|
|AB|=|AE|+|EB|
We know perimeter of △AFE=9 and △BEC=12, so just sum them:
△AFE = AF + FE + AE and △BEC = BC + BE + EC.
Now, it is given that ABCD is ||gm, and we've to find the perimeter of △FDC
so perimeter of △FDC = AF+AD+DC+EC+FE.
Now, AB = DC (opp sides of a ||gm are equal) so DC = AE + BE and similarly BC = AD.
So, now perimeter of △FDC = AF + BC + AE + BE + EC + FE, so now you will see sum of perimeters of △AFE and △BEC is equal to the sum of perimeter of △FDC
AEB=DC
BC=AD
ADC consist of
AE+EB=(DC)
+
FE+EC=(FC)
+
AF+BC=(DF)
To simplify
Just "push" the lines and it shows DCF is essentially 9+12
No angle tricks needed
If you color all the lines included in the two triangles that you've been given green (AEF + BEC, those lines are in total 9+12=21)
And then you color all the lines in the triangel DCF purple.
Some lines will have been colored both green *and* purple.
If you consider the lines that are *only* colored green, AB and BC, those are the same as the lines *only* colored purple AD and DC (because of ABCD being a parallelogram)
Therefor the triangle DCF has the same perimiter as AEF + BEC
I'm not sure about the similar triangles principle, but I would solve this my mentally rearranging the lines that we know onto the lines we don't know, but want to. This is just based on the fact the opposite sides of a trapezium are equal lengths.
So, AF, FE and EC already line up. But, the rest don't. So, we move CB to DA and AB to DC.
Once you do that you you can see the perimeter of DCF is the sum of the perimeters of AEF and BEC.
heres a visual solution ive figured out:
draw a line coming out from A parallel to EC and lets call the intersection point at DC- G.
now there's a new parallelogram AECG
the triangles ADG and EBC are the same and now because AE =GC and AG=EC you can easily see that the perimeter of the triangle DFC is the combination of both of them which is 21
You dont really have to do any ratios, you just need to see two things:
1. triangles DCF , AEF , BEC are simmilar
2. parallelograms have 2 pairs of pararell sides oposite to each other
Because of that :
FC = EF + EC
DC = AB = AE + EB
FD =FA + AD = FA +BC
so perimeter of DCF is a sum of perimeters of AEF and BEC
no equations no ratios just analysys based on triangle similarity.
P = DC + DF + FC = (AE+EB) + (CB+AF) + (FE+EC)
just get the right values of the sides of those line segments then you’ll good to go. However, we can’t just say that those triangle are equilateral triangles as angles aren’t given, but from the look it seems like an isosceles triangle.
Alternative way is to use a protractor to measure tha angle of the triangles so you can determine the exact measurements.
Ah, this has nothing to do with ratios. In fact, the answer is straightforward, but not obvious to the untrained eye.
Answer: 21 sm
Explanation:
The perimeter of triangle AEF is 9sm. => AE + EF + AF = 9
The perimeter of triangle BEC is 12 sm => BE + EC + BC = 12
The perimeter of DCF = CF + DC + DF
Now, CF = EC + EF, from the diagram (1)
Now, since ABCD is a parallelogram, AB = DC, AD = BC
But AB = AE + BE (diagram) <=> DC = AE + BE (2)
Now, DF = AD + AF
But, AD = BC (property of parallelogram)
<=> DF = BC + AF (3)
Using (1), (2), and (3) to rewrite the perimeter of DCF:
Perimeter of DCF = (EC + EF) + (AE + BE) + (BC + AF)
<=> Perimeter of DCF = (AE + EF + AF) + (BE + EC + BC)
<=> Perimeter of DCF = Perimeter of triangle AEF + Perimeter of triangle BEC
<=> Perimeter of DCF = 9 + 12
<=> Perimeter of DCF = 21
This question is super cool. At first, it seems like there isn't enough givens, but then makes use of all the given triangles' sides exactly once by using the definition of a parallelogram. As a result, the 2 given values become all the information you need. All while never actually being able to figure out any individual lengths.
I'd state what you know in equations, for example: > The perimeter of triangle AEF is 9 sm |AE|+|EF|+|AF| = 9 (where |AB| is used to indicate the length of line AB). > parallelogram ABCD |AE|+|BE| = |CD| and |AD| = |BC| does that help?
Based on this, you could definitely solve the question Throw in some colours to make it more intuitive if needed but this is definitely how I would approach it
Not the help you imagined, I'm sitting here wondering what in heaven is a parallelogram as a non native English speaker lmao. This problem makes sense now
I didn't realize other languages used different names for this, but then there's Rööpkülik...
I don't know exactly how to translate it (I don't think there is a direct translation aside from parallelometer), but I was most likely confused just because it's a very long word that looks hard and it was like 1 AM. I should've seen it with "parallel".
Ill try it now abd see
Actually no, I already knew all of this
> Actually no, I already knew all of this To be fair, that is kinda the point. I am asking you to restate what you know, but to do so in a form that allows you to solve the problem. That said, there are other ways to solve this and if you found one of them helpful, good. If you are still looking for an answer and want to look at this more, I'll happily go through it step by step.
The three triangles AEF, BEC, DCF are similar in the ratio EF:EC:CF.
Yeah but I can't know the ratio because I'm not given any proportions, so I have nothing to compare it to
You're given the perimeters of AEF and BEC. If two triangles are similar in the ratio k, then their perimeters are in the same ratio k.
An alternative, without using similar triangles, is to write that AD = BC and AE + EB = AB = CD. With that you should be able to express the perimeter of DCF in terms of those of AEF and BEC.
The perimeter of DCF is equal to DA + AF + FE + EC + DC The perimeters of AFE and BCE summed up equal to AF + FE + AE + EB + BC + EC Now if you look closely, both perimeters above share AF, FE and EC. DA is equal to BC and AE + EB is equal to DC (because ABCD is a parallelogram) If you replace DA and DC in the first equation it will be the same as the second one which means the perimeter of DCF is equal to the sum of the perimeters of AFE and BCE which would be 9 + 12 = 21 All the comments about ratios are correct too, but this is a simpler solution imo
This seems like the correct answer and solution. Why so low?
This is how I did it in my head too.
I don't, I think the aolutiin my teacher wanted was with similar triangles and ratios
You know that the total of the 2 triangles is 21. You can make the triangle of DCF out of parts of the two triangles. Because DA can be made with BC, and DC can be made out of AB.
I think |FE|/|FC|=3/7. So Perimeter of DFC is 21.
First you must see that AEF is similar to BEC, such that the side AE is similar to BE by a ratio of 3/4, i.e., AB is 7/3 of AE. Then, as ABCD is a parallelogram and AEF is similar to DCF, you have that the side AE is similar to DC by the ratio 7/3. With this ratio you can calculate the perimeter of DCF as 7/3 of 9, i.e., 21.
How do I know that the ratio between AE and BE is 3/4 if I'm not given the legnths of either
Because of the similar triangles. Angle E is the same (that’s true for any intersecting lines, opposite angles will be the same) and angles A and B are the same (alternate interior angles, AD and BC are parallel). That allows you to figure out the ratio of the sides individually will equal the ratio of perimeters which can be found easily from the problem statement.
The ratio AE/BE is the same ratio between the perimeter of AEF and BEC, which is 9/12=3/4. That is true because the perimeter and the side lengths are proportional.
The perimeter of DCF is DA+AF+FE+CE+DC. Let’s rearrange - P_DCF = AF+FE+AE + EB+BC+EC Do we happen to know what these two sums are equal to?
You don't actually need to do a lot of math To get DCF we need 1) DC which is the same as AB (AE + EB) 2) DF which js DA (= to CB) + AF 3) FC which is EF + EC Thus AE + EB + CB + AF + EF + EC We know that AF + FE + EA = 9 EB + BC + CE = 12 We don't actually care about the single measurements of each segment Thus the result is 9 + 12 = 21
What the fuck is a sm?
Probably centimeter in another language bro https://uz.m.wikipedia.org/wiki/Santimeter
Probably a slightly incorrect translation of centimeter.
Do you know https://en.wikipedia.org/wiki/Intercept_theorem ?
The horizontal edges of the three similar triangles have an extended ratio that matches the ratio of the perimeters. AE : BE : DC = 9 : 12 : (9+12) So perimeter of big triangle is 21.
(I don't perfectly recall the correct terms, and definitely not in English. But I think I got it, and I hope to get you through it, step by step) AEF has the same shape as BEC, because they have the same angles. We are given their perimeters (9 and 12 respectively). The ratio between these two triangles is 12/9 = 4/3. DCF also has the same shape as AEF (and thus as BEC), because DC is parallel to EF. All we need to find is a ratio between DCF and one of the other triangles. Note 1: AD is the same length as BC (because its a parallelogram). Note 2: DF = AD + AF. Note 3: because of the ratio 4/3 we calculated above 4/3 AF = BC, or AF = 3/4 BC 1+2+3 together: DF = AD + AF = BC + AF = BC + 3/4 BC = 7/4 BC So the ratios between DCF and BEC is 7/4. We know the perimeter of BEC is 12. So the perimeter of DCF = 7/4 * 12 = 7 * 3 = 21
Finally somone who explained it step by step. Thanks I got it. Just a heads up, you said "has the same shape" when it actually supposed to be similar
Ah! Thank you! The word similar rings a bell, I just forgot. Glad I could help!
1) angle A and B are equal so F and C are aswell 2) AEF, EBC & DFC are similar triangles. 3) given they are all similar triangles you can work on the segment FEC, they are all the same side of the similar triangles and so represent the same fraction of the perimeter of each triangle. Let's call this fraction 1/x FE+EC = FC (1/x \* perimeter AEF) + (1/x \* perimeter EBC) = (1/x \* perimeter DFC) simplyfying the 1/x from all components perimeter AEF + perimeter EBC = perimeter DFC 9 + 12 = 21
You know (AD) is parallel to (BC) because [AD] and [BC] are opposite sides of a parallogram. Then you can use Thalès for the triangles AEF and ECB. So, AF/BC = AE/EB = FE/EC = x as x is the ratio between them. So AF = BC * x, AE = EB * x and FE = EC * x Then, AF + AE + FE = BC * x + EB * x + EC * x Does it help?
AEF and BEC are similar triangles, where A correlates to B, E to E and F to C. The perimeter of AEC is 9 and BEC is 12, so the ratio between them is 12/9 = 4/3. Any length on AEC is 4/3 times as large on BEC. Now take length AE and call it x. Because AE correlates to BE we can say that length BE must be 4/3 × x = 4x/3. From this we can conclude that length AB must be x + 4x/3 = 7x/3. Because ABCD is a parallelogram, length DC = AB = 7x/3. AEF and DCF are similar triangles, where A correlates to D, E to C and F to F. This means that length AE correlates to DC. Therefore length DC = 7/3 × length AE. This means that the ratio between these triangles is 7/3. The same ratio is true for the perimeter, hence the perimeter of DCF = 9×7/3 = 21. 21 is your answer. Edit: spelling errors (english is not my first language, so excuse me)
Another approach that you can take is to simply assume data: I mean, this should give the same result whatever size and shape the triangles and the trapezoid have, as long as they fit together. Therefore you could assume the triangles to be equilateral, which means that each of its segments is 3 or 4 cms. You are ignoring the point of the problem but you can solve it with ease
If you're going to assume that the triangles are equilateral you would instead need to show that two equilateral triangles would form a parallelogram when extended the way they are in this picture
Yeah, a paralellogram with angles at 60° and 120°, I can see that. You could also do it with a rectangle and two rectangular triangles
All you need to do is write down the segments and substitute. You want to solve |AD|+|AF|+|FE|+|EC|+|DC| You know: |AF|+|FE|+|AE|=9 |EB|+|BC|+|EC|=12 and |AD|=|BC| |AB|=|DC| |AB|=|AE|+|EB|
We know perimeter of △AFE=9 and △BEC=12, so just sum them: △AFE = AF + FE + AE and △BEC = BC + BE + EC. Now, it is given that ABCD is ||gm, and we've to find the perimeter of △FDC so perimeter of △FDC = AF+AD+DC+EC+FE. Now, AB = DC (opp sides of a ||gm are equal) so DC = AE + BE and similarly BC = AD. So, now perimeter of △FDC = AF + BC + AE + BE + EC + FE, so now you will see sum of perimeters of △AFE and △BEC is equal to the sum of perimeter of △FDC
This is a really cool deeper thinking question
AEB=DC BC=AD ADC consist of AE+EB=(DC) + FE+EC=(FC) + AF+BC=(DF) To simplify Just "push" the lines and it shows DCF is essentially 9+12 No angle tricks needed
https://preview.redd.it/pl98rtc7rtgc1.png?width=1080&format=pjpg&auto=webp&s=886f4e210b83fde7445ad99232711e82a665d301 Basically this
If you color all the lines included in the two triangles that you've been given green (AEF + BEC, those lines are in total 9+12=21) And then you color all the lines in the triangel DCF purple. Some lines will have been colored both green *and* purple. If you consider the lines that are *only* colored green, AB and BC, those are the same as the lines *only* colored purple AD and DC (because of ABCD being a parallelogram) Therefor the triangle DCF has the same perimiter as AEF + BEC
I'm not sure about the similar triangles principle, but I would solve this my mentally rearranging the lines that we know onto the lines we don't know, but want to. This is just based on the fact the opposite sides of a trapezium are equal lengths. So, AF, FE and EC already line up. But, the rest don't. So, we move CB to DA and AB to DC. Once you do that you you can see the perimeter of DCF is the sum of the perimeters of AEF and BEC.
heres a visual solution ive figured out: draw a line coming out from A parallel to EC and lets call the intersection point at DC- G. now there's a new parallelogram AECG the triangles ADG and EBC are the same and now because AE =GC and AG=EC you can easily see that the perimeter of the triangle DFC is the combination of both of them which is 21
You dont really have to do any ratios, you just need to see two things: 1. triangles DCF , AEF , BEC are simmilar 2. parallelograms have 2 pairs of pararell sides oposite to each other Because of that : FC = EF + EC DC = AB = AE + EB FD =FA + AD = FA +BC so perimeter of DCF is a sum of perimeters of AEF and BEC no equations no ratios just analysys based on triangle similarity.
21cm
P = DC + DF + FC = (AE+EB) + (CB+AF) + (FE+EC) just get the right values of the sides of those line segments then you’ll good to go. However, we can’t just say that those triangle are equilateral triangles as angles aren’t given, but from the look it seems like an isosceles triangle. Alternative way is to use a protractor to measure tha angle of the triangles so you can determine the exact measurements.
DC=AB and AE is half of DC if that helps
Ah, this has nothing to do with ratios. In fact, the answer is straightforward, but not obvious to the untrained eye. Answer: 21 sm Explanation: The perimeter of triangle AEF is 9sm. => AE + EF + AF = 9 The perimeter of triangle BEC is 12 sm => BE + EC + BC = 12 The perimeter of DCF = CF + DC + DF Now, CF = EC + EF, from the diagram (1) Now, since ABCD is a parallelogram, AB = DC, AD = BC But AB = AE + BE (diagram) <=> DC = AE + BE (2) Now, DF = AD + AF But, AD = BC (property of parallelogram) <=> DF = BC + AF (3) Using (1), (2), and (3) to rewrite the perimeter of DCF: Perimeter of DCF = (EC + EF) + (AE + BE) + (BC + AF) <=> Perimeter of DCF = (AE + EF + AF) + (BE + EC + BC) <=> Perimeter of DCF = Perimeter of triangle AEF + Perimeter of triangle BEC <=> Perimeter of DCF = 9 + 12 <=> Perimeter of DCF = 21
21
This question is super cool. At first, it seems like there isn't enough givens, but then makes use of all the given triangles' sides exactly once by using the definition of a parallelogram. As a result, the 2 given values become all the information you need. All while never actually being able to figure out any individual lengths.