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16 Sep 2014, 00:54
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In quadrilateral \(ABCD\), is \(AC\) longer than \(BD\)?
(1) \(\angle ABC \gt \angle BCD\).
(2) \(AB = BC = CD = DA\).
(1) \(\angle ABC \gt \angle BCD\).
(2) \(AB = BC = CD = DA\).
16 Sep 2014, 00:54
Kudos
Bookmarks
Official Solution:
In quadrilateral \(ABCD\), is \(AC\) longer than \(BD\)?
Note that the question essentially asks whether, in quadrilateral \(ABCD\), diagonal \(AC\) is longer than diagonal \(BD\).
(1) \(\angle ABC \gt \angle BCD\).
Consider two kite-shaped figures, which give two different answers:
(2) \(AB = BC = CD = DA\).
The above statement implies that the quadrilateral is a rhombus. However, we can have \(AC > BD\), \(AC < BD\), or, in the case of a square, even \(AC = BD\). Not sufficient.
(1)+(2) Since from (2) the quadrilateral is a rhombus, then the fact that \(\angle ABC \gt \angle BCD\) implies that the side opposite \(\angle ABC\), which is \(AC\), is longer than the side opposite \(\angle BCD\), which is \(BD\). Sufficient.
Answer: C
In quadrilateral \(ABCD\), is \(AC\) longer than \(BD\)?
Note that the question essentially asks whether, in quadrilateral \(ABCD\), diagonal \(AC\) is longer than diagonal \(BD\).
(1) \(\angle ABC \gt \angle BCD\).
Consider two kite-shaped figures, which give two different answers:
(2) \(AB = BC = CD = DA\).
The above statement implies that the quadrilateral is a rhombus. However, we can have \(AC > BD\), \(AC < BD\), or, in the case of a square, even \(AC = BD\). Not sufficient.
(1)+(2) Since from (2) the quadrilateral is a rhombus, then the fact that \(\angle ABC \gt \angle BCD\) implies that the side opposite \(\angle ABC\), which is \(AC\), is longer than the side opposite \(\angle BCD\), which is \(BD\). Sufficient.
Answer: C
