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Is quadrilateral ABCD

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by maihuna » Fri Dec 25, 2009 9:45 am
Is quadrilateral ABCD a rectangle?

1. Both angle ABC and angle BCD are right angle.
2. Two diagonals have the same length.
Last edited by maihuna on Sat Dec 26, 2009 7:04 am, edited 1 time in total.
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Source: — Data Sufficiency |
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by papgust » Fri Dec 25, 2009 7:04 pm
Statement II is not clear. Please post it correctly
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by harsh.gupta.175 » Fri Dec 25, 2009 7:26 pm
Hi,

Here is how i would answer it:

Statement I:

ABC and BCD are right angles. Hence the sides AB || DC. To be a rectangle the angles BAD and CDA should also be equal to 90 degree. All we know is that the sum of these two angles =180 since AB||DC. Hence there are more than one possible combination for these angles.

Not Suficient.

Statement II:

This is the tricky one and i am not sure about this part.

Any quadilateral in which the diagonals are equal will have the same height and base for both the diagonals. This implies that the opposite sides will be parallel and hence this quadilateral will be a paralleogram. Any parallegram with diagonals equal is a rectangle.

Hence sufficient.
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by rohan_vus » Fri Dec 25, 2009 7:42 pm
harsh.gupta.175 wrote:Hi,

Here is how i would answer it:

Statement I:

ABC and BCD are right angles. Hence the sides AB || DC. To be a rectangle the angles BAD and CDA should also be equal to 90 degree. All we know is that the sum of these two angles =180 since AB||DC. Hence there are more than one possible combination for these angles.

Not Suficient.

Statement II:

This is the tricky one and i am not sure about this part.

Any quadilateral in which the diagonals are equal will have the same height and base for both the diagonals. This implies that the opposite sides will be parallel and hence this quadilateral will be a paralleogram. Any parallegram with diagonals equal is a rectangle.

I disagee with this statement . First of all you dont if its a parallelogram . Its not mentioned anyhwere..Consider a trapezioid ..where slant sides are equal and and horzitonal sides are unequal . Here diagonals can be equal even if its not a parallelogram .

Hence sufficient.
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by linkinpark » Sat Dec 26, 2009 1:34 am
IMO A, since two opposite angles are given 90 each, rest of angles will be 90 hence its a rectangle.
from B it can be a rectanlge or a very small trapezoid that has two non parallerl sides of equal length.
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by papgust » Sat Dec 26, 2009 2:01 am
IMO A, since two opposite angles are given 90 each, rest of angles will be 90 hence its a rectangle.
from B it can be a rectanlge or a very small trapezoid that has two non parallerl sides of equal length.
That could not be the case because I says that base angles are 90 degrees each [Angles ABC and BCD]. Still the other 2 angles need not be 90 degrees.
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by maihuna » Sat Dec 26, 2009 7:09 am
papgust wrote:
IMO A, since two opposite angles are given 90 each, rest of angles will be 90 hence its a rectangle.
from B it can be a rectanlge or a very small trapezoid that has two non parallerl sides of equal length.
That could not be the case because I says that base angles are 90 degrees each [Angles ABC and BCD]. Still the other 2 angles need not be 90 degrees.
Can you draw two angle at right angle and other two not 90. Remember the right angles are on common side. If
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by papgust » Sat Dec 26, 2009 7:54 am
maihuna wrote: Can you draw two angle at right angle and other two not 90. Remember the right angles are on common side.
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by maihuna » Sat Dec 26, 2009 9:05 am
papgust wrote:
maihuna wrote: Can you draw two angle at right angle and other two not 90. Remember the right angles are on common side.
good one,
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by target790 » Sun Dec 27, 2009 7:39 am
OA please?

IMO -E


1>Not sufficient
2>Not sufficient
1> && 2> Not sufficient(can be rectangle or square)
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