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anonymous
 one year ago
geometry problem
anonymous
 one year ago
geometry problem

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In triangle ABC, D, E and F are points on BC, AC and AB respectively such that BD: DC = 1:1, CE: EA = 1:2 and AF: FB = 1:3. If the area of triangle ABC = 48 cm2, find the area of triangle DEF.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433796310810:dw

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1I'm not sure that your diagram is correct You have used 'x' for BD, DC correctly, but I don't think you can use the 'same x' for AF and FB The diagram is correct but the nomenclature is wrong I think.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah think so but its in ratio it shoudnt matter much

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433755799791:dw

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 There is no indication of HOW you arrived at those values Care to give an explanation instead of an unverified answer?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Sure, thought id let others try too instead of spoiling the fun for them :)

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1@aryandecoolest It does matter  if look at BDF your diagram implies BF is 3 times longer than BD, but that is not a correct interpretation of the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0idk.. it's the exact question i have posted here... @ganeshie8 i didn't get... did we prove triangles similar or something

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1Haven't had time to look in detail starting with area fro Side angle side a = absinC/2 and we have the same angle so we can compare ratios Looking at triangle BCA and triangle DCE DC/BC=1/2 CE/CA = 1/3 So the area of DCE = 1/6 ABC = 8 @ganeshie8 Not compatible with your answer

perl
 one year ago
Best ResponseYou've already chosen the best response.0Was that supposed to be x , not z ?

perl
 one year ago
Best ResponseYou've already chosen the best response.0@aryandecoolest Did the problem come with a picture, or did you draw that based on the directions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah it came with the picture....the same i have posted

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1for bdf area = 1/2 * 3/4 = 3/8 ABC = 18 agreed for AFE Area = 1/4* 2/3 = 1/8 ABC = 8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is there any theorem like area is equal to square of the proportional sides of 2 similar triangles?

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1@aryandecoolest use the area for 'side angle side' Area = absinC/2 apply that to the large triangle and the 3 corner triangles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sine law ? but no angle given it's the ratios

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1it's not sine law it's area from side angle side the ratios cancel the angles out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah!!! fine... fine i get @MrNood

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I think that this problem is equivalent to this one: "find three natural numbers, n_1, n_2, and n_3, such that the first one is disible by 2, the second one is divisible by 3, and the third one is divisible by 4": \[\begin{gathered} {n_1} = 2k \hfill \\ {n_2} = 3k \hfill \\ {n_3} = 4k \hfill \\ \end{gathered} \] nameli n_1, n_2 and n_3 are the sides of our triangle

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2furthermore, the subsequent condition holds: \[\begin{gathered} 2p = 9k \hfill \\ \frac{{9k}}{2} \times \frac{{5k}}{2} \times \frac{{3k}}{2} \times \frac{k}{2} = {48^2} \hfill \\ \end{gathered} \] where 2p is the perimeter of our triangle and the second formula comes from the application of Eron's formula

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the solution is: \[\begin{gathered} {n_1} = 2k = \frac{8}{{\sqrt[4]{{15}}}} \hfill \\ {n_2} = 3k = \frac{{12}}{{\sqrt[4]{{15}}}} \hfill \\ {n_3} = 4k = \frac{{16}}{{\sqrt[4]{{15}}}} \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please give me a sec to understand....

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now we have: \[48 = \frac{{{n_3}{n_1}}}{2}\sin \beta \] dw:1433758100593:dw from which: \[\sin \beta = \frac{3}{4}\sqrt {15} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2finally we have: area of triangle FDB= \[ = \frac{{3{n_3}}}{4} \times \frac{{{n_1}}}{2} \times \sin \beta \times \frac{1}{2} = 18\] dw:1433758258337:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2similarly for other triangles DCE, and AFE

perl
 one year ago
Best ResponseYou've already chosen the best response.0if you assume x= y , then you can use herons formula http://www.wolframalpha.com/input/?i=sqrt%28%284x%2B3x%2B2x%29%2F2+*%28%284x%2B3x%2B2x%29%2F2+4x%29*%28%284x%2B3x%2B2x%29%2F23x%29*%28%284x%2B3x%2B2x%29%2F22x%29%29%3D48 I thought x and y were not necessarily equal

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2even if those numbers are irrational numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thanks a ton @Michele_Laino @ganeshie8 @perl . I am still understanding... i'll contact u if i get a prob. Thanks a lot all of you. Hope i had more medals to give.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! @aryandecoolest

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1there is NO indication that x and z (or y) are equal BD is NOT given as BF/3

perl
 one year ago
Best ResponseYou've already chosen the best response.0this problem is not well posed

perl
 one year ago
Best ResponseYou've already chosen the best response.0The directions should say that x = y :)

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1it is well posed  if you read the WORDS the problem is that somehow 'x' has been used to define the ratios of BC AND BA BF = 3z AF = z fits the words dw:1433759367588:dw

perl
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino using heron's I got k = 8/ 15^(1/4) therefore n1 = 2k = 16/ 15^(1/4) n2 = 3k = 24/ 15^(1/4) n3= 4k= 32/ 15^(1/4)

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1@perl there is no given relationship between the sides this is ANY triangle

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433759547800:dw

perl
 one year ago
Best ResponseYou've already chosen the best response.0@MrNood he stated that the diagram was given, z = x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh mistake, should be 8 and 14

perl
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @perl @aryandecoolest Did the problem come with a picture, or did you draw that based on the directions? \(\color{blue}{\text{End of Quote}}\) \(\color{blue}{\text{Originally Posted by}}\) @aryandecoolest yeah it came with the picture....the same i have posted \(\color{blue}{\text{End of Quote}}\)

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1It does not state that in the words

perl
 one year ago
Best ResponseYou've already chosen the best response.0Can you explain how you solved it without assuming x=y=z .

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1yes I can the area of a triangle is given by Area = ab sinC/2

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1so look at angle c for instance ABC area = (AC * CD sin C)/2 DCE area = (DC * CE * sin C)/2 so the ratio is the ratio of the product of the edges in this case theproduct is 1/6 so the area is 1/6 of ABC = 8 do the same for the two ther small triangles

perl
 one year ago
Best ResponseYou've already chosen the best response.0ok that makes sense. i just see a small typo

perl
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @MrNood so look at angle c for instance ABC area = (AC * `CD` sin C)/2 that should be `CB` or `BC` DCE area = (DC * CE * sin C)/2 so the ratio is the ratio of the product of the edges in this case theproduct is 1/6 so the area is 1/6 of ABC = 8 do the same for the two ther small triangles \(\color{blue}{\text{End of Quote}}\)

perl
 one year ago
Best ResponseYou've already chosen the best response.0I'll just type out the argument and you can see if i have an error. CD = 1/2 BC CE = 1/3 AC ABC area = (AC * BC sin C)/2 = 48 DCE area = (CD * CE * sin C)/2 = ( 1/2 BC * 1/3 AC ) * sin C / 2 = 1/2 * 1/3 ( BC * AC) * sin C /2 = 1/6 * 48 = 8

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1yes  that is correct  my apologies

perl
 one year ago
Best ResponseYou've already chosen the best response.0we can make a similar argument for angle B

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1as I said  do the samer for the other small triangles dw:1433761604217:dw

perl
 one year ago
Best ResponseYou've already chosen the best response.0then subtract from 48 to get the middle area

perl
 one year ago
Best ResponseYou've already chosen the best response.0interestingly if you do happen to assume x=y=z, you still get the same answer

perl
 one year ago
Best ResponseYou've already chosen the best response.0but you shouldn't since it is not stated in the directions

perl
 one year ago
Best ResponseYou've already chosen the best response.0and if you assume x=y=z then the problem can be solved using Heron's formula. do you have any questions @aryandecoolest

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, it's getting clear. thanks a lot

MrNood
 one year ago
Best ResponseYou've already chosen the best response.1@perl The solution I gave above is a general solution for all triangles It is not therefore 'interesting' that it works for the 'special case' x=y=z'

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Yes! you are right! @perl
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