geometry problem

- anonymous

geometry problem

- jamiebookeater

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- 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.

- anonymous

|dw:1433796310810:dw|

- 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.

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## More answers

- anonymous

Yeah think so but its in ratio it shoudnt matter much

- ganeshie8

|dw:1433755799791:dw|

- MrNood

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

- ganeshie8

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

- 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

- anonymous

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

- 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

- perl

Was that supposed to be x , not z ?

- perl

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

- anonymous

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

- MrNood

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

- anonymous

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

- perl

ok one moment :)

- MrNood

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

- anonymous

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

- MrNood

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

- anonymous

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

- 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

- 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

- Michele_Laino

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

- anonymous

please give me a sec to understand....

- Michele_Laino

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

- 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|

- Michele_Laino

similarly for other triangles DCE, and AFE

- 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

- Michele_Laino

even if those numbers are irrational numbers

- 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.

- Michele_Laino

ok! @aryandecoolest

- MrNood

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

- perl

this problem is not well posed

- MrNood

|dw:1433759138500:dw|

- perl

The directions should say that x = y :)

- 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|

- 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)

- MrNood

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

- ganeshie8

|dw:1433759547800:dw|

- MrNood

|dw:1433759530762:dw|

- perl

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

- ganeshie8

Ahh mistake, should be 8 and 14

- 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
\(\color{blue}{\text{End of Quote}}\)

- MrNood

It does not state that in the words

- perl

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

- MrNood

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

- 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

- MrNood

|dw:1433760746644:dw|

- MrNood

|dw:1433760901106:dw|

- perl

ok that makes sense.
i just see a small typo

- 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
\(\color{blue}{\text{End of Quote}}\)

- 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

- MrNood

yes - that is correct - my apologies

- perl

we can make a similar argument for angle B

- MrNood

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

- perl

then subtract from 48 to get the middle area

- perl

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

- perl

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

- perl

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

- anonymous

no, it's getting clear. thanks a lot

- 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'

- Michele_Laino

Yes! you are right! @perl

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