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Geometry

Expert replies
Source: — Problem Solving |

by N:Dure » Sun Dec 19, 2010 8:55 am
He wants to prove that ABCD is a square with sides AB BC CD DA all equal.

Info. 1 tells us that the diagonals intersect and they have 90 degrees together, which doesn't really help.

2 tells us that the addition of the 2 opposite sides are equal. if we assume that AB+CD=BC+AD= 6 AB & CD, BC & AD could have different combinations like (4,2) (3,3) (2,4) (5,1) ..etc so we can't prove that they're all equal.

Thus there isn't sufficient info to prove that this is a square.
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by shovan85 » Sun Dec 19, 2010 9:42 am
Koala wrote:If ABCD is quadrilateral, is AB=BC=CD=DA?

1) AC is perpendicular to BD
2) AB+CD=BC+AD

OA is E
1. AC is perpendicular to BD

Think about the geometrical shapes (as quadrilateral) that has perpendicularly intersecting diagonals.

Being a quadrilateral a square, or a Rhombus, or a Kite can have perpendicular. Though square and rhombus has all the sides equal, Kite does not have equal sides.

Thus insufficient.

2. AB+CD=BC+AD

Take simple numbers to verify.

3+4 = 2+5 (=7) so does this mean that 3 = 4 = 2 = 5? Never.

Thus insufficient.

IMO E

PS: If you have any kind of concerns about geometrical figure have a look at the below image for an idea. (Honestly I did not know what is a Kite (in terms of geometrical shape ;)))
Attachments
quadrilateral_class_112.gif
If the problem is Easy Respect it, if the problem is tough Attack it
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by Rahul@gurome » Sun Dec 19, 2010 10:02 am
Koala wrote:If ABCD is quadrilateral, is AB=BC=CD=DA?

1) AC is perpendicular to BD
2) AB+CD=BC+AD
Statement 1: AC is perpendicular to BD.
As there is no restrictions on the lengths of the sides, they can be same or different.

Not sufficient


Statement 2: AB + CD = BC + AD
There can be various possible combinations of lengths satisfying this condition.

Not sufficient

1 & 2 Together: Choose AB = AD and CD = AD, then statement 2 satisfies and we can make AC perpendicular to BD (as shown in the figure below) but still lengths of all the sides are not same.
Image
Same case for AB = BC and CD = AD.

Not sufficient.

The correct answer is E.
N:Dure wrote:He wants to prove that ABCD is a square with sides AB BC CD DA all equal.
No. The question is asking for whether ABCD is a rhombus or not.
Square is a special type of rhombus where all the angles are right angle.
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