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BCDF and ABDE are squares.
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01 Feb 2023, 06:22
01 Feb 2023, 06:22
Expert Reply
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Question Stats:
45% (02:00) correct
54% (01:57) wrong based on 24 sessions
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BCDF and ABDE are squares.
Quantity A |
Quantity B |
Twice the area of the shaded region |
Three times the area of BCDF |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Show Answer
Re: BCDF and ABDE are squares.
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18 Feb 2023, 06:28
18 Feb 2023, 06:28
1
Expert Reply
OE
No lengths are given, therefore, you are free to pick some easy numbers to work with. Draw the figure and fill in all given information, which here is that the figures are squares. For the sides of the small square, 1 is an easy number, as it makes the area of square BCDF equal (1)(1), which is 1:
Next, identify relationships. The figure implies that the dashed line BD is both an edge of square ABDE and the diagonal of square BCDF, making it the hypotenuse of equal right triangles BCD and BDF. By Pythagorean theorem on right triangle BCD, you get the following:
\(c^2=a^2+b^2\)
\(BD^2=1^2+1^2\)
\(BD=\sqrt{2}\)
So the area is \(\sqrt{2}*\sqrt{2}=2\)
No lengths are given, therefore, you are free to pick some easy numbers to work with. Draw the figure and fill in all given information, which here is that the figures are squares. For the sides of the small square, 1 is an easy number, as it makes the area of square BCDF equal (1)(1), which is 1:
Next, identify relationships. The figure implies that the dashed line BD is both an edge of square ABDE and the diagonal of square BCDF, making it the hypotenuse of equal right triangles BCD and BDF. By Pythagorean theorem on right triangle BCD, you get the following:
\(c^2=a^2+b^2\)
\(BD^2=1^2+1^2\)
\(BD=\sqrt{2}\)
So the area is \(\sqrt{2}*\sqrt{2}=2\)
gmatclubot
Re: BCDF and ABDE are squares. [#permalink]
18 Feb 2023, 06:28
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