## anonymous one year ago geometry problem

1. anonymous

In 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.

2. anonymous

|dw:1433796310810:dw|

3. MrNood

I'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.

4. anonymous

Yeah think so but its in ratio it shoudnt matter much

5. ganeshie8

|dw:1433755799791:dw|

6. MrNood

@ganeshie8 There is no indication of HOW you arrived at those values Care to give an explanation instead of an unverified answer?

7. ganeshie8

Sure, thought id let others try too instead of spoiling the fun for them :)

8. MrNood

@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

9. anonymous

idk.. it's the exact question i have posted here... @ganeshie8 i didn't get... did we prove triangles similar or something

10. MrNood

Haven'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

11. perl

Was that supposed to be x , not z ?

12. perl

@aryandecoolest Did the problem come with a picture, or did you draw that based on the directions?

13. anonymous

yeah it came with the picture....the same i have posted

14. MrNood

for bdf area = 1/2 * 3/4 = 3/8 ABC = 18 agreed for AFE Area = 1/4* 2/3 = 1/8 ABC = 8

15. anonymous

is there any theorem like area is equal to square of the proportional sides of 2 similar triangles?

16. perl

ok one moment :)

17. MrNood

@aryandecoolest use the area for 'side angle side' Area = absinC/2 apply that to the large triangle and the 3 corner triangles

18. anonymous

sine law ? but no angle given it's the ratios

19. MrNood

it's not sine law it's area from side angle side the ratios cancel the angles out

20. anonymous

Ah!!! fine... fine i get @MrNood

21. Michele_Laino

I 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

22. Michele_Laino

furthermore, 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

23. Michele_Laino

the solution is: $\begin{gathered} {n_1} = 2k = \frac{8}{{\sqrt[4]{{15}}}} \hfill \\ {n_2} = 3k = \frac{{12}}{{\sqrt[4]{{15}}}} \hfill \\ {n_3} = 4k = \frac{{16}}{{\sqrt[4]{{15}}}} \hfill \\ \end{gathered}$

24. anonymous

please give me a sec to understand....

25. Michele_Laino

now we have: $48 = \frac{{{n_3}{n_1}}}{2}\sin \beta$ |dw:1433758100593:dw| from which: $\sin \beta = \frac{3}{4}\sqrt {15}$

26. Michele_Laino

finally 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|

27. Michele_Laino

similarly for other triangles DCE, and AFE

28. perl

if 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%2F2-3x%29*%28%284x%2B3x%2B2x%29%2F2-2x%29%29%3D48 I thought x and y were not necessarily equal

29. Michele_Laino

even if those numbers are irrational numbers

30. anonymous

okay 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.

31. Michele_Laino

ok! @aryandecoolest

32. MrNood

there is NO indication that x and z (or y) are equal BD is NOT given as BF/3

33. perl

this problem is not well posed

34. MrNood

|dw:1433759138500:dw|

35. perl

The directions should say that x = y :)

36. MrNood

it 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|

37. perl

@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)

38. MrNood

@perl there is no given relationship between the sides this is ANY triangle

39. ganeshie8

|dw:1433759547800:dw|

40. MrNood

|dw:1433759530762:dw|

41. perl

@MrNood he stated that the diagram was given, z = x

42. ganeshie8

Ahh mistake, should be 8 and 14

43. perl

$$\color{blue}{\text{Originally Posted by}}$$ @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 $$\color{blue}{\text{End of Quote}}$$

44. MrNood

It does not state that in the words

45. perl

Can you explain how you solved it without assuming x=y=z .

46. MrNood

yes I can the area of a triangle is given by Area = ab sinC/2

47. MrNood

so 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

48. MrNood

|dw:1433760746644:dw|

49. MrNood

|dw:1433760901106:dw|

50. perl

ok that makes sense. i just see a small typo

51. perl

$$\color{blue}{\text{Originally Posted by}}$$ @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 $$\color{blue}{\text{End of Quote}}$$

52. perl

I'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

53. MrNood

yes - that is correct - my apologies

54. perl

we can make a similar argument for angle B

55. MrNood

as I said - do the samer for the other small triangles |dw:1433761604217:dw|

56. perl

then subtract from 48 to get the middle area

57. perl

interestingly if you do happen to assume x=y=z, you still get the same answer

58. perl

but you shouldn't since it is not stated in the directions

59. perl

and if you assume x=y=z then the problem can be solved using Heron's formula. do you have any questions @aryandecoolest

60. anonymous

no, it's getting clear. thanks a lot

61. MrNood

@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'

62. Michele_Laino

Yes! you are right! @perl