In the figure, the areas of parallelograms EBFD adn AECF are 3 and 2, respectively. What is the area of rectangle ABCD?
1. 3
2. 4
3. 5
4. 4(root3)
5. 7
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fskilnik@GMATH
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Hi there!
The areas of parallelograms EBFD and AECF may be seen as DF*AD and FC *AD, therefore considering AD = k (any positive real constant), we get DF = 3/k and FC = 2/k.
Please note that DC = 5/k therefore DC*AD = 5/k * k = 5, and we are done.
Regards,
Fabio.
The areas of parallelograms EBFD and AECF may be seen as DF*AD and FC *AD, therefore considering AD = k (any positive real constant), we get DF = 3/k and FC = 2/k.
Please note that DC = 5/k therefore DC*AD = 5/k * k = 5, and we are done.
Regards,
Fabio.
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GMATGuruNY
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Plug in BC = 1.bajwa2307 wrote:In the figure, the areas of parallelograms EBFD adn AECF are 3 and 2, respectively. What is the area of rectangle ABCD?
1. 3
2. 4
3. 5
4. 4(root3)
5. 7
Area of EBFD = bh = 3.
Since b = DF and h = BC = 1, we get:
DF * 1 = 3
DF = 3.
Area of AECF = bh = 2
Since b = FC and h = BC = 1, we get:
FC * 1 = 2
FC = 2.
Area of ABCD:
b = DF + FC = 3+2 = 5.
h = BC = 1.
Area = bh = 5*1 = 5.
The correct answer is C.
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As a tutor, I don't simply teach you how I would approach problems.
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