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26 Jan 2023, 02:30
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D
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Difficulty:
65%
(hard)
Question Stats:
27% (01:14) correct
73%
(01:20)
wrong
based on 26
sessions
History
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Not Attempted Yet
Is quadrilateral ABCD a square?
(1) ABCD is a rectangle whose diagonals intersect at right angles.
(2) AC, a diagonal of ABCD, is also the hypotenuse of isosceles right triangles ABC and ADC.
Hard geometry problem.
(1) ABCD is a rectangle whose diagonals intersect at right angles.
(2) AC, a diagonal of ABCD, is also the hypotenuse of isosceles right triangles ABC and ADC.
Hard geometry problem.
Updated on: 26 Jan 2023, 13:20
Originally posted by Maria240895chile on 26 Jan 2023, 07:31.
Last edited by Maria240895chile on 26 Jan 2023, 13:20, edited 2 times in total.
Last edited by Maria240895chile on 26 Jan 2023, 13:20, edited 2 times in total.
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(I) its diagonals intersect in right angles, and we know ABCD is a rectangle, then is a square. SUFFICIENT
(II) if the diagonal is the hypothenuse of both isosceles triangles, then is a square
IMO D
(II) if the diagonal is the hypothenuse of both isosceles triangles, then is a square
IMO D
26 Jan 2023, 13:16
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Bunuel
Statement 1
(1) ABCD is a rectangle whose diagonals intersect at right angles.
Given, ABCD is a rectangle. If the diagonals intersect at right angles, we can conclude that the sides are equal. Hence the figure is a square. The statement is sufficient.
Statement 2
(2) AC, a diagonal of ABCD, is also the hypotenuse of isosceles right triangles ABC and ADC.
\(\triangle ABC\)
- \(AB = BC\)
- \(\angle ABC = 90\)
\(\triangle ADC\)
- \(AD = DC\)
- \(\angle ADC = 90\)
As AC is the common sides between the two triangles, we can conclude that all the sides are equal.
Hence this statement is also sufficient.
Option D
