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CAT Exam Model Paper 5 with solutions for free online practice
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© examsnet.com
Question : 9
Total: 60
ABCD is a parallelogram. E is a point on AB such that AE : BE = 2 : 3. A line EF is drawn parallel to AD and it meets CD at F. G is a point on BC such that GB : GC = 1 :4 . What is the ratio of the area of ΔDEC to the area of AEFG?
3 :5
10:3
25:12
None of these
Validate
Solution:
Area of ΔDEC = Area of ΔDBC........ (Δ's between same parallel line and same base)
=
1
2
×
Area of parallelogram ABCD
Area of AEFG =
1
2
×
Area of parallelogram EBCF
=
1
2
×
3
5
×
Area of parallelogram ABCD
So,
Δ
DEC
Δ
EFG
=
1
2
×
area
(
ABCD
)
1
2
×
3
5
×
area
(
ABCD
)
=
5
3
© examsnet.com
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