Complimentary methods
Given
A
D
=
2
D
F
,
C
E
=
2
E
D
,
C
F
=
2
F
E
, and the area of triangle
D
F
E
is
1
cm
2
, what is the area of the triangle
A
B
C
?
The answer is 19.
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2 solutions
Same. Your representation is nice.
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thank you very much
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You are welcome. How did you draw so big diagram ? I use paint brush and my big diagram becomes small in brilliant!! Thank you.
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@Niranjan Khanderia – u can use geo gebra
How you found these areas, can you explain more plz
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Δ s DEF and ADE have same height, but base AD =2 base DF ∴ ADE is 2. ADE and AEC same height but base twice ∴ A E C = 2 ∗ 2 = 4 . S i m i l a r l y o t h e r 2 a n d 4 . EDB=3, EBC has same height but twice base, so it is 3*2=6 I hope this satisfies your question.
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Sorry but you got it wrong at ADE and AEC same height but base twice, because we have no information for that. BE=2DE not CE=2DE, the same mistake repeats at EDB and EBC.
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@Thai-Anh Nguyen – i've re-corrected it
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@Ahmed Moh AbuBakr – the information CF=2FE is still not relevant, maybe BF=2FE?
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@Thai-Anh Nguyen – Yes. BF=2FE, would be relevant. Solution is based on this.
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@Niranjan Khanderia – There is no correction up this moment.
@Thai-Anh Nguyen – sorry ,, but not me who made a mistake because i am sure i put it correctly ,, i think the stuff who edited it made that
@Thai-Anh Nguyen – but it now correct
area (BDF)=2, area(BDA)=4 so area(BFA)=6 same method area(CDA)=6 and area(BEC)=6 so area(ABC)=3*6+1=19 sq cm