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09 Jan 2023, 11:15
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C
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Difficulty:
65%
(hard)
Question Stats:
53% (02:20) correct
47%
(02:16)
wrong
based on 34
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In the given figure ABCD is a rectangle and AEG and FEB are two triangles inscribed in the rectangle ABCD such that the area of triangle AEG is half of the area of triangle FEB. What is the area of rectangle ABCD?
(1) EFCB is a square of side length 5 centimetres.
(2) A line passing through point G and parallel to AB divides the rectangle ABCD into 2 equal halves
Attachment:
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
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yashikaaggarwal
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09 Jan 2023, 14:13
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(1) EFCB is a square of side length 5 centimetres.
It only justifies that EFCB is a square. ADEF can be rectangle.
(2) A line passing through point G and parallel to AB divides the rectangle ABCD into 2 equal halves.
From this we know that rectangle is divided in 2 equal parts, but we are not sure about the value of any side of rectangle.
Combining both, A and B
It gives the value of side of rectangle as 10.
Area is solvable.
Answer is C
Posted from my mobile device
It only justifies that EFCB is a square. ADEF can be rectangle.
(2) A line passing through point G and parallel to AB divides the rectangle ABCD into 2 equal halves.
From this we know that rectangle is divided in 2 equal parts, but we are not sure about the value of any side of rectangle.
Combining both, A and B
It gives the value of side of rectangle as 10.
Area is solvable.
Answer is C
Posted from my mobile device
09 Jan 2023, 23:50
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Bunuel
Solution:
Pre Analysis:
- ABCD is a rectangle
- We are also given Area of triangle AEG \(=\frac{1}{2}\times\) Area of triangle FEB
- We are asked the area of rectangle ABCD
- TRAPS!!
- We cannot assume E and G are midpoints of AB and FE respectively
- We cannot assume FE is perpendicular to AB
Statement 1: EFCB is a square of side length 5 centimetres
- From this statement, we can infer
- Area of FEBC = 25
- Area of triangle FEB \(=\frac{1}{2}\times 25=\frac{25}{2}\)
- Area of triangle AEG \(=\frac{1}{2}\times\) Area of triangle FEB \(=\frac{1}{2}\times \frac{25}{2}=\frac{25}{4}\)
- FE is perpendicular to AB
- \(\frac{1}{2}\times GE\times AE=\frac{25}{4}\) or \(AE\times GE=\frac{25}{2}\)
- We still don't know the length of AE or FD thus we cannot get the area of AEFD or AGFD
- Thus, statement 1 alone is not sufficient and we can eliminate options A and D
Statement 2: A line passing through point G and parallel to AB divides the rectangle ABCD into 2 equal halves
- From this statement, we can infer that G is the midpoint of FE which is obviously not sufficient to get the answer
- Thus, statement 2 alone is also not sufficient
Combining:
- From statement 1, we get:
- FE is perpendicular to AB
- \(FE=5\)
- \(AE\times GE=\frac{25}{2}\)
- From statement 2, we get that G is the midpoint of FE
- Upon combining, we can get the value of AE which is also the value of DF
- Ultimately, we can get the area of AEFD, and then we can add it to area of FEBC to get our answer
Hence the right answer is Option C
