Parallelogram ABCD: m(<ABC)=120*, BC>CD, P_{ABCD}=26cm, r=sqrt{3}cm. Find...

[politexnik]

Good afternoon, please help, I’ll also say that you need to solve it normally without inserted numbers that will fit.

 

[khansaheb]

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Good afternoon, please help, I’ll also say that you need to solve it normally without inserted numbers that will fit.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem

 

[stapel]

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When you reply with your thoughts and efforts so far, please also provide the following information:

i) B is a vertex, not an angle. Where you say "B = 120^o", do you actually mean "the measure of the angle ABC is 120^o?

ii) When you say "BC > CD", do you actually mean "|BC| > |CD|" (that is, that the length of the segment BC is more than is the length of the segment CD"?

iii) What does "P_ABCD = 26cm" mean? Is this maybe the length of the perimeter of the parallelogram?

iv) What does "S_ABCD" mean?

 

[The Highlander]

Good afternoon, please help, I’ll also say that you need to solve it normally without inserted numbers that will fit.

Looking through your "history" in the forum, I note that you always post your problem(s) unaccompanied by any initial attempts to solve them (or identification of where exactly your difficulty lies) or any thoughts you may have on an approach to solving them, you never post any final solutions you have reached on the problems after help has been given (despite this having been requested of you more than once) and you are not good at responding to requests made of you in general! ?

We need you to help us to help you by responding to requests made of you (to provide further information and show us your work)!

I had the same thoughts as @stapel (above) about the imprecise use of terminology (and lack of clear definition of terms used) but the very first question that came to my mind was: Is this a parallelogram?
(and what is meant by: "you need to solve it normally without inserted numbers that will fit.")

Assuming it is a parallelogram (and the other issues mentioned in Post #3) then...
∠D = ∠B = 120° and ∴ ∠C = ∠A = 60°
P_ABCD = The Perimeter of ABCD (= 26 cm) and, if S_ABCD is the Semiperimeter of ABCD, then S_ABCD = 13 cm. Yes?
(Is it possible that S_ABCD should have been S_BCD, ie: the Semiperimeter of ΔBCD?)
Please confirm all of the foregoing (or explain if otherwise).

Given those assumptions, it occurred to me to begin by adding a few "bits" to your diagram (see below) as this would allow the calculation of |CI| (the distance of the incentre from vertex A) but I have no intention of exploring your problem(s) any further unless and until we get some response(s) from you (as per the forum guidelines). ?

​

 

[Dr.Peterson]

P_ABCD = The Perimeter of ABCD (= 26 cm) and, if S_ABCD is the Semiperimeter of ABCD, then S_ABCD = 13 cm. Yes?
(Is it possible that S_ABCD should have been S_BCD, ie: the Semiperimeter of ΔBCD?)

I have seen S used to mean area (either Surface Area, or some word in another language), so that is another possibility.

 

[Dr.Peterson]

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Good afternoon, please help, I’ll also say that you need to solve it normally without inserted numbers that will fit.

I can tell you that I have solved it, and the first step was to find CF on @The Highlander's marked up drawing, which you should find very helpful.

If you would tell us anything about what you have learned or what you have tried, it would be very helpful to us. Do you know anything about incircles, perimeters, and areas, for example? How about lengths of tangents from a point? And what does "solve it normally" mean to you?

But I see you have been given a solution on another site, so the polite thing to do (presumably unrelated to your name) would be to tell us that you don't need any more help (and to stop posting here, since it clearly is not help that you want).

 

[The Highlander]

I can tell you that I have solved it, and the first step was to find CF on @The Highlander's marked up drawing, which you should find very helpful.

But I see you have been given a solution on another site, so the polite thing to do (presumably unrelated to your name) would be to tell us that you don't need any more help (and to stop posting here, since it clearly is not help that you want).

Agreed! (Absolutement!)

If s/he's not going to play by the rules then maybe s/he should just stay in the locker room (or go to a different stadium altogether!). ?
[OK, OK, I know I'm guilty of the odd offence myself but mine never warrant more than a yellow card ?? (to mix my metaphors ?)]

I know I said I wasn't going to pursue this problem any further (until we heard from the OP) but, of course, I couldn't resist going back to it when I had a spot of free time (the telly's rubbish in the summer! ?)

It took me a wee while to realize that it was the other side of the R-A triangle (FC instead of CI) that I should have solved but once I got onto the side "combinations" I soon made some headway. ?
There was another (lengthy) pause after I got BD but I eventually figured ot how to solve for the remainder. ?

The answers I arrived at were:-

S_ABCD = 13 cm.
BD   =   7 cm.
P_BCD  =  20 cm.
BC   =   8 cm.
CD   =   5 cm.

Are those the same answers that the OP was given (?) elsewhere? (I presume you came up with the same ones that s/he was gifted too?)

(I just want to be sure I did it correctly myself, lol.)

If those are the right answers, then I'll post the full solution (in glorious technicolour ?) in the thread (for the benefit of any visitors) unless you want to post your own solution?

 

[The Highlander]

As the statutory 4 days have now passed (and it's clear that we will get nothing further on this topic from the OP), I will post my solution to this problem for the benefit of any future viewers (who may not be able to solve it themselves) and for the sake of 'completeness' (so the thread has a 'resolution' in it).
Furthermore, @Dr.Peterson has pointed out that the OP obtained a solution to the problem on another site (or sites) so I think it's only fair that a solution should also be present here (for interested observers).

I have marked up the original drawing to illustrate that there are several pairs of line segments of equal length (because the length of the two tangents from an external point to the points of tangency on a circle is the same) so I have used different colours to show these pairs of line segments.

​

For the sake of simplicity (and clarity) I will refer to segment lengths as, for example, FC rather than |FC| (and, occasionally, just as "lines" rather then segments or segment lengths).

I have also assumed that the notations P_ABCD and S_ABCD mean the Perimeter of the figure ABCD and Semiperimeter of the figure ABCD respectively.

FC just happens to be the only line whose length we can calculate at the outset.
Because r is at right angles to the side CD we therefore know that...

\(\displaystyle \sf tan~30°=\frac{r}{F\color{#E25041}{C}}\implies F\color{#E25041}{C}\color{black}=\frac{r}{tan~30°}\implies F\color{#E25041}{C}\color{black}=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=\sqrt{3}\times\sqrt{3}=\color{#E25041}{3}\) cm.​

Now that we know the length of FC, that also gives us the length of EC as 3 cm too and we can now proceed to calculating the lengths of the remaining segments...

We can start by equating line lengths.

We are given that P_ABCD = 26 cm ∴ S_ABCD = 13 cm, so we can say that...

(BC+CD) = 13 cm (I will drop the cm units henceforth).

Thus (BC+CD) = (BE+EC+CF+FD) = 13
(I am using FC to represent only the magnitude (length) of line segments so CF = FC and so on.)

Now BD = (BG+GD) but BG = BE and GD = FD so BD = (BE+FD) too.

So we have...

(BC+CD) = (BE+EC+CF+FD) = 13

but we already know that (EC+CF) = 6

\(\displaystyle \implies\) (BE+FD) + 6 = 13 \(\displaystyle \implies\) (BE+FD) = 7.

Therefore, BD [= (BG+GD) = (BE+FD)] =7 (cm)

That's as far as manipulating the segment lengths can take us but, since we now know the length of the Perimeter of the triangle BCD as being 20 cm (S_ABCD + 7) we can now take a different tack to solve for the remaining sides (BC & CD).

Apparently what I want to post is too big to put into a single post (even though MSWord™ tells me that it way less than 10,000 characters! ?) so I will continue in a subsequent post...

 

[The Highlander]

Now consider the Area of the triangle.
(Note that in my revised diagram I have also labelled the sides of the triangle BCD in line with the convention that each side is given the lower case letter of the vertex opposite it.)

​

We have the following (well-known?) formula for the Area of a triangle...

Area = ½bd Sin C​

Where b & d are (the lengths of) two sides of the triangle and C is (the measure of) the included angle (between them).

But there is another (less publicized?) formula for the Area of a triangle related to its Incircle (the circle inscribed in the triangle such that each side of the triangle is tangential to the circle) as is shown in the diagram.

That formula is...

Area = sr​

Where r is the radius of the Incircle (given as √3) and s is the Semiperimeter of the triangle which we now know is 10 (half of its (20 cm) Perimeter, ie: 20 ÷ 2 = 10)

So we may now calculate the Area of the triangle BCD as...

A = sr = 10√3​

And now that we know its Area we can equate that answer to our other equation and get...

½bd Sin C = 10√3

\(\displaystyle \implies bd=\frac{2\times10\sqrt{3}}{Sin~60°}\)

\(\displaystyle \implies bd=\frac{20\sqrt{3}}{\frac{\sqrt{3}}{2}}\)    

 \(\displaystyle \implies bd=20\sqrt{3}\times\frac{2}{\sqrt{3}}\)

 \(\displaystyle \implies bd=20\times 2=40\)​

So we now know that...

b × d = 40 and b + d = 13
(Those familiar with factorizing quadratics will, no doubt, be jumping
straight ahead at this point but let's go through it methodically anyway. ?)​

If b + d = 13 then d = (13 - b)

Substituting that into our other equation we can say...

     b(13 - b) = 40
\(\displaystyle \implies 13b-b^2=40\)
\(\displaystyle \implies 0=40-13b+b^2 \text{ or } b^2-13b+40=0\)
      (and now factorizing that)
\(\displaystyle \implies (b-8)(b-5)=0\)

and so b = 8 or 5.

Now,
if b = 8 then d = 5
and                      (because b + d = 13)
if b = 5 then d = 8.

But we were given that d > b (BC > CD) so d must be 8 (and ∴ b = 5)

Now we know the lengths of all three sides of the ΔBCD that allows us to also determine all the remaining line segment lengths as...
BG = BE = 5 (8 - 3) cm, GD = FD = 2 (5 - 3) cm and (we already knew that) EC = FC = 3 cm.

And we can now summarize the required answers as...

S_ABCD =  13 cm.
BD   =  7 cm.
(P_BCD  =  20 cm.)
BC    =  8 cm.
CD   =  5 cm.

Hope that helps. ?​

 

[Dr.Peterson]

So we may now calculate the Area of the triangle BCD as...

A = sr = 10√3​

If you recall my suggestion (post #5) that S_ABCD might mean area, as I've seen in some questions in the past, you can see why they might have asked for that! It's just twice what you've calculated, which is central to the process of finding the other quantities (though BD came first). So that may have been a little hint.

(b−8)(b−5)=40

and so b = 8 or 5.

You meant (b - 8)(b - 5) = 0, right?

Furthermore, @Dr.Peterson has pointed out that the OP obtained a solution to the problem on another site (or sites) so I think it's only fair that a solution should also be present here (for interested observers).

Actually, on the other site it was only revealed that "BD = 7, and the rest is easy". But that was taken as a satisfactory answer, so perhaps the rest was easy to them. (I'd say finding BD is the easy part.)

 

[The Highlander]

If you recall my suggestion (post #5) that S_ABCD might mean area, as I've seen in some questions in the past, you can see why they might have asked for that! It's just twice what you've calculated, which is central to the process of finding the other quantities (though BD came first). So that may have been a little hint.

Indeed but I just used S to mean Semiperimeter because it suited my purposes. ?
If they did mean Area then, as you say, it's simply twice the area of the triangle. ie: 20√3 cm²

You meant (b - 8)(b - 5) = 0, right?

Absolutely! I will request that slip-up be amended. (It was a lengthy treatise to spot every mistake in it! ?)

Actually, on the other site it was only revealed that "BD = 7, and the rest is easy". But that was taken as a satisfactory answer, so perhaps the rest was easy to them. (I'd say finding BD is the easy part.)

I would agree completely.

That is also why I have 'aimed' my solution at 14/15 year-olds upwards. I wouldn't expect High School students to know the "sr" formula for area of a triangle (it certainly isn't taught in HS here) whereas factorizing quadratics and the other formula for the area of a triangle are (at least to the majority of our students).

(I also note, having had a quick look at the site you mention, that there is a mistake in their explanation (of the "hard part")! Their line: "BC+CFCD=BE+EC+DF+FC=BE+DF+6=13" is wrong! Maybe the author there wasn't so sure themself how to complete it and just suggested it might be "easy" once BD was known? ?)

Good to know I'm not the only one guilty of the odd proof reading error. ?

(@stapel: Would you be kind enough to fix the mistake the good Doctor has pointed out in the line below this one in Post #9, please? It should, of course say 0 not 40. Thank you.)

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