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

[BTGmoderatorDC]

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

[swerve]

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