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In the figure shown above, point E is the intersection point of the diagonals AC and BD of

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by BTGmoderatorDC » Wed Sep 29, 2021 6:11 pm

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Screenshot 2020-04-01 at 8.55.41 PM.png
In the figure shown above, point E is the intersection point of the diagonals AC and BD of rectangle ABCD and the length of AB = 7.25 cm. Is AB greater than BC?
1. EC = 5 cm
2. The perimeter of AEB is greater than the perimeter of BEC


OA D

Source: e-GMAT
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Source: — Data Sufficiency |
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Re: In the figure shown above, point E is the intersection point of the diagonals AC and BD of

by swerve » Thu Sep 30, 2021 8:00 am
BTGmoderatorDC wrote: ↑
Wed Sep 29, 2021 6:11 pm
 Screenshot 2020-04-01 at 8.55.41 PM.png

In the figure shown above, point E is the intersection point of the diagonals AC and BD of rectangle ABCD and the length of AB = 7.25 cm. Is AB greater than BC?
1. EC = 5 cm
2. The perimeter of AEB is greater than the perimeter of BEC


OA D

Source: e-GMAT
Diagonals of rectangle are equal and bisect each other.

Statement 1-

\(EC^2=\left(\dfrac{AB}{2}\right)^2+\left(\dfrac{BC}{2}\right)^2\)

We know EC and AB; hence, we can find BC. Sufficient \(\Large{\color{green}\checkmark}\)

Statement 2-

AE +BE+AB > BE +EC+BC

AE=EC { Diagonals of rectangle are equal and bisect each other.}

AB > BC. Sufficient \(\Large{\color{green}\checkmark}\)

Therefore, D
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