In the figure shown above, point E is the intersection point of the diagonals AC and BD of
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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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Diagonals of rectangle are equal and bisect each other.BTGmoderatorDC wrote: ↑Wed Sep 29, 2021 6:11 pmScreenshot 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
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