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Rhombus / Parallelogram ?

Expert replies
by mmslf75 » Thu Dec 10, 2009 1:25 am
Source MGMAT CAT 2

What is the area of the quadrilateral with vertices A, B, C, and D?

(1) The perimeter of ABCD is equal to 16.

(2) Quadrilateral ABCD is a rhombus.


OA E


I could sense that in explanation provided by MGMAT guys on test is that
When we combine both st1 and st2 , we have a rhombus with sides 4 each
Now the Area of rhombus can be ( 4 * 4 ) = 16
OR
The area of Rhombus can be ( Base * Height = 4 * 2 root3 )

So, are they considering Rhombus to be a Parallelogram ?? qestion 1
FACT : I knew that Area of Rhombus = .5 * diagonal1 * diagonal 2.. This is something new I learnt..


Is every Square a Rectangle ?
Is everry Rhombus a Parallelogram?


What Am I missing here ??[/spoiler]
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Source: — Data Sufficiency |

by punitkaur » Thu Dec 10, 2009 11:04 am
Hi mmslf75,

how did you calculate the height of the rhombus given the side?
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by Stuart@KaplanGMAT » Thu Dec 10, 2009 11:23 am
mmslf75 wrote:Source MGMAT CAT 2

What is the area of the quadrilateral with vertices A, B, C, and D?

(1) The perimeter of ABCD is equal to 16.

(2) Quadrilateral ABCD is a rhombus.


OA E


I could sense that in explanation provided by MGMAT guys on test is that
When we combine both st1 and st2 , we have a rhombus with sides 4 each
Now the Area of rhombus can be ( 4 * 4 ) = 16
OR
The area of Rhombus can be ( Base * Height = 4 * 2 root3 )

So, are they considering Rhombus to be a Parallelogram ?? qestion 1
FACT : I knew that Area of Rhombus = .5 * diagonal1 * diagonal 2.. This is something new I learnt..


Is every Square a Rectangle ?
Is everry Rhombus a Parallelogram?


What Am I missing here ??[/spoiler]
The OA is correct.

Special shapes also belong to general categories.

Every square is a rhombus, rectangle, parallelogram and quadrilateral.
Every rectangle is a parallelogram and quadrilateral.
Every rhombus is a parallelogram and quadrilateral.
Every parallelogram is a quadrilateral.

Basically, if a shape fits the definition of a more general shape, it also belongs to that class. Since a parallelogram is simply a quadrilateral with 2 pairs of parallel sides, and since squares, rectangles and rhombii all have 2 pairs of parallel sides, they're all parallelograms.

The formula for the area of a parallelogram is simply:

area = base * height

Combining the statements, we know that we have a parallelogram with sides of 4 (by definition a rhombus has 4 equal sides, so we calculate the length of a side by taking perimiter/4); however, we have no clue how "slanted" the rhombus is, so there's no way to calculate the height. For example, the shape could be a square with an area of 16 or could have any area smaller than 16, depending on the angles.
Image

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by punitkaur » Thu Dec 10, 2009 11:36 am
Yup.. thats what I was wondering, where did the height come from.
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by mmslf75 » Fri Dec 11, 2009 9:37 am
punitkaur wrote:Hi mmslf75,

how did you calculate the height of the rhombus given the side?
In MGMAT explanation it is assumed ABCD to be a ll gram...with angle 60 degrees...therefore height 2 root 3
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by Stuart@KaplanGMAT » Fri Dec 11, 2009 10:42 am
mmslf75 wrote:
punitkaur wrote:Hi mmslf75,

how did you calculate the height of the rhombus given the side?
In MGMAT explanation it is assumed ABCD to be a ll gram...with angle 60 degrees...therefore height 2 root 3
Is it assumed that the angle is 60 degrees, or is that just given as one possibility, in order to show that the statements are insufficient?

As I stated in my post above, a rhombus is a sub-class of parallelogram.
Image

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Join the discussion

by mmslf75 » Fri Dec 11, 2009 10:47 pm
Stuart Kovinsky wrote:
mmslf75 wrote:
punitkaur wrote:Hi mmslf75,

how did you calculate the height of the rhombus given the side?
In MGMAT explanation it is assumed ABCD to be a ll gram...with angle 60 degrees...therefore height 2 root 3
Is it assumed that the angle is 60 degrees, or is that just given as one possibility, in order to show that the statements are insufficient?

As I stated in my post above, a rhombus is a sub-class of parallelogram.




Assumed 60 degrees..

I actually went by my own theory that since we are asked to find area of rhombus, I can say area = .5 d1 d2 and since d1 and d2 are not possible..to find out....I marked E

but u explained with the other concept..Do we actually need to go that far ?
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by mmslf75 » Wed Dec 16, 2009 9:44 am
mmslf75 wrote:
Stuart Kovinsky wrote:
mmslf75 wrote:
punitkaur wrote:Hi mmslf75,

how did you calculate the height of the rhombus given the side?
In MGMAT explanation it is assumed ABCD to be a ll gram...with angle 60 degrees...therefore height 2 root 3
Is it assumed that the angle is 60 degrees, or is that just given as one possibility, in order to show that the statements are insufficient?

As I stated in my post above, a rhombus is a sub-class of parallelogram Assumed 60 degrees..

I actually went by my own theory that since we are asked to find area of rhombus, I can say area = .5 d1 d2 and since d1 and d2 are not possible..to find out....I marked E

but u explained with the other concept..Do we actually need to go that far ?


think I understand now..

IN GENERAL AREA OF A PARALLELO is BASE * HEIGHT ( and PARALLELO being the SUPERSET of ALL QUADS ) we took
the area of rhombus to be the same


Wow !! I got it

thanks
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by agarwalmanoj2000 » Tue Jul 12, 2011 12:56 pm
Should answer be C? Please advise.

- Area of Rhombus = (D1*D2)/2
- 2 diagonals of Rhombus are equal and perpendicular bisector.

Based on statement 1 & 2, each side of the Rhombus is 4. As the diagonals are perpendicular bisector, they bisect each other making 90 degree angle. Diagonals makes 4 isosceles right angle triangle with hypotenuse equal to 4, so the other sides of each triangle will be 4/Sq.root 2 and hence diagonals will equal 2*(4/Sq.root 2) = 8/Sq.root 2.

Therefore area of the Rhombus = [(8/Sq.root 2) * (8/Sq.root 2)] *2 = 64.
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by newgmattest » Tue Jul 12, 2011 5:19 pm
Please remember that diagonals of Rhombus are *NOT* equal in length and they just bisect the other at 90 degree.
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by amit2k9 » Tue Jul 12, 2011 11:18 pm
a+b

base = 4 but height = ?
E it is.
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by gmattesttaker2 » Fri Oct 04, 2013 9:51 pm
Hello,

I was trying to solve as follows:

1 - Insuff. Since we don't know what kind of quadrilateral it is.

2 - Insuff. No information about rhombus ABCD. Hence, can't calculate area.

1 and 2:

Perimeter of rhombus ABCD = 16. Hence, each side is 4 (since all sides are equal).

Area of rhombus = 1/2 * (d1 + d2) , where d1 and d2 are diagonals of the rhombus

However, I was not sure after this point. Can you please tell what information is needed to find the diagonals of a rhombus? Thanks a lot.

Best Regards,
Sri
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by theCodeToGMAT » Fri Oct 04, 2013 10:08 pm
gmattesttaker2 wrote:Hello,

I was trying to solve as follows:

1 - Insuff. Since we don't know what kind of quadrilateral it is.

2 - Insuff. No information about rhombus ABCD. Hence, can't calculate area.

1 and 2:

Perimeter of rhombus ABCD = 16. Hence, each side is 4 (since all sides are equal).

Area of rhombus = 1/2 * (d1 + d2) , where d1 and d2 are diagonals of the rhombus

However, I was not sure after this point. Can you please tell what information is needed to find the diagonals of a rhombus? Thanks a lot.

Best Regards,
Sri
At this point, we cannot calculate the measure of Diagnol.. had this question mentioned that, side angles are 90*.. then we could have move forward..

So [spoiler]{E}[/spoiler]
R A H U L
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by GMATGuruNY » Sat Oct 05, 2013 3:36 am
mmslf75 wrote:Source MGMAT CAT 2

What is the area of the quadrilateral with vertices A, B, C, and D?

(1) The perimeter of ABCD is equal to 16.

(2) Quadrilateral ABCD is a rhombus.
When the statements are combined, ABCD could look like either of the following:

Case 1:
Image

Case 2:
Image

Thus, different areas are possible.

The correct answer is E.
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by Anaira Mitch » Sat Dec 31, 2016 1:56 am
Just curious to know if in statement 2 it would have been "(2) Quadrilateral ABCD is a Square". I believe answer should have been C.
Anyone can confirm on this?
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