RaviChandra wrote:
In triangle ABC, D and E are points on AC and AB such DE || BC and length of DE is one-third of BC. If the area of triangle ABC is 216 square units, find the area of the shaded triangle.
a. 12
b. 18
c. 24
d. 16
Attachment:
2.PNG
Note that AED is similar to ABC with each side as 1/3rd of the original
=> Area(AED) = 1/9 * Area(ABC) =24
Area(ABD) = Area(ACE) = 1/3 * Area(ABC) = 72
[same height, 1/3rd base]
Area(ABD) = Area(ACE)
=> Area(ABD) - Area(AEDF) = Area(ACE) - Area(AEDF)
=> Area(BEF) = Area(CDF) = x (say)
Finally note that DEF is similar to BFC with one side in the ratio of 1:3
Thus area(DEF) = 1/9 * area(BFC) = y (say)
Area(EDC) = Area(EDF)+Area(DFC) = x + y
ALSO
Area(EDC) = Area(AEC) - Area(AED) =72 - 24 = 48
So x+y = 48
Area(BEDC) = Area(BEF) + Area(CDF) + Area(EDF) + Area(BFC) = x+x+y+9y
ALSO
Area(BEDC) = Area(ABC)-Area(AED)=216-24=192
So x+5y=96
Subtracting the 2 equations
4y=48
y=12
Hence, Area(DEF)=12
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