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Quadrilateral

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Source: — Data Sufficiency |

by Stuart@KaplanGMAT » Fri Nov 27, 2009 1:46 pm
heshamelaziry wrote:Is quadrilateral RSTV a rectangle?

(1) The measure of ∠RST is 90 degrees
(2) The measure of ∠TVR is 90 degrees
Each statement alone is clearly insufficient. The question really comes down to combining them.

Based on the ordering of the letters, we know that RST and TVR are opposite angles. We certainly could draw a rectangle based on that information, but could we draw any other shape?

So, we really need to answer:

if the opposite angles in a quadrilateral are both 90 degrees, does the shape have to be a rectangle?

The answer turns out to be no. It's tough to demonstrate that without drawing a diagram, but picture two right angle triangles with the same hypotenuse but different legs. We can "glue" the triangles together to form a quadrilateral and, because the legs are different lengths, only the opposite angles will both be 90 degrees.

For example, if our triangles were:

5, 5root3, 10 (30/60/90 degree angles)

and

6, 8, 10 (not 30/60/90 degree angles)

We could glue them together on the 10 side to create a quadrilateral and only the two opposite angles would be 90 degrees (and the sides would be 5, 5root3, 6 and 8, clearly not a rectangle).

So, even after combination our shape may or may not be a rectangle: choose E.
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by wanttobeat » Tue Dec 01, 2009 12:06 am
Stuart Kovinsky wrote:
heshamelaziry wrote:Is quadrilateral RSTV a rectangle?

(1) The measure of ∠RST is 90 degrees
(2) The measure of ∠TVR is 90 degrees
Each statement alone is clearly insufficient. The question really comes down to combining them.

Based on the ordering of the letters, we know that RST and TVR are opposite angles. We certainly could draw a rectangle based on that information, but could we draw any other shape?

So, we really need to answer:

if the opposite angles in a quadrilateral are both 90 degrees, does the shape have to be a rectangle?

The answer turns out to be no. It's tough to demonstrate that without drawing a diagram, but picture two right angle triangles with the same hypotenuse but different legs. We can "glue" the triangles together to form a quadrilateral and, because the legs are different lengths, only the opposite angles will both be 90 degrees.

For example, if our triangles were:

5, 5root3, 10 (30/60/90 degree angles)

and

6, 8, 10 (not 30/60/90 degree angles)

We could glue them together on the 10 side to create a quadrilateral and only the two opposite angles would be 90 degrees (and the sides would be 5, 5root3, 6 and 8, clearly not a rectangle).

So, even after combination our shape may or may not be a rectangle: choose E.
the question says the shape is a quadrilateral. By glueing together the two triangles of various legs, we will not have a quadrilateral, will we?

I think the answer is E because a square also has two opposite sides, so its a quadrilateral but not a rectangle.
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by Stuart@KaplanGMAT » Tue Dec 01, 2009 1:09 am
wanttobeat wrote: the question says the shape is a quadrilateral. By glueing together the two triangles of various legs, we will not have a quadrilateral, will we?

I think the answer is E because a square also has two opposite sides, so its a quadrilateral but not a rectangle.
Two key points:

First, if we glue two triangles together at the hypotenuse, we have 4 "unglued" sides, indeed giving us a quadrilateral.

For example, a square is simply two identical 45/45/90 triangles glued at the hypotenuse (when you cut a square diagonally, you create those two triangles all over again).

Second, a square is a rectangle, since it has all the properties of a rectangle (two pairs of equal and opposite sides, four 90 degree angles).

"Special" shapes also belong to general groups.

For example, a quadrilateral is an enclosed 4 (straight) sided shape.

A rectangle is a special quadrilateral.

A square is a special rectangle which is also a special quadrilateral.
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by wanttobeat » Tue Dec 01, 2009 10:28 pm
thanks for the explanation...now its clear to me :)
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by viju9162 » Wed Dec 02, 2009 3:13 am
Thanks Stuart for the wonderful explaination.
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by ikaplan » Wed Jan 12, 2011 12:49 am
Stuart, can you please provide a drawing for the 'glued' triangles which share the hypotenuse?

Thank you!
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by ankurmit » Wed Jan 12, 2011 8:31 pm
Can anyone provide me drawing for this
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by Night reader » Thu Jan 13, 2011 2:41 am
ankurmit wrote:Can anyone provide me drawing for this
Hi Ankur, I am not artist so excuse my line drawing :)
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by ankurmit » Thu Jan 13, 2011 6:00 am
Thanks dude..

Its nice drawing :)
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by thebigkats » Sun Jan 16, 2011 2:42 pm
Quadrilaterals can come in variety of shapes including kites and complex quadrilaterals.

A kite is where adjacent sides are same length (so in a quadri ABCD - AB=AD and BC=CD). you can make a kite wherein angles where two different length sides meet is right angle. So in above example: ABC = CDA = 90

Similarly you can make a complex quadrilateral (kind of looks like two triangles touching each other at a common point). In this case also, it is possible to have two right angles while not having a rectangle.

So either of these two is NOT sufficient

cheers,
kats
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by meghal » Fri Sep 30, 2011 10:26 am
Nice explaination
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by prashant misra » Tue Nov 08, 2011 1:06 am
can anyone explain it with a diagram its not clear to me i am not able to understand this question.
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by bpdulog » Wed Nov 09, 2011 12:41 pm
So when do we consider a square and a rectangle as two separate items for exam purposes?
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by shankar.ashwin » Wed Nov 09, 2011 9:20 pm
All squares are rectangles too, but not the other way around.
There's a question i remember which tests this,
How many rectangles can be drawn in a 8*8 chessboard?

I clearly remember them counting possibilities of squares as well.
bpdulog wrote:So when do we consider a square and a rectangle as two separate items for exam purposes?
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