CAT Exam Model Paper 5 with solutions for free online practice

Question : 9 of 60

Marks: +1, -0

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 }

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