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# the area of a parallelogram is 100. what is the perimeter of

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the area of a parallelogram is 100. what is the perimeter of [#permalink]

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15 Dec 2007, 16:02
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the area of a parallelogram is 100. what is the perimeter of the parallelogram ?

1) the base of the parallelogram is 10
2) one of the angles of the parallelogram is 45 degree
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15 Dec 2007, 19:59
bmwhype2 wrote:
the area of a parallelogram is 100. what is the perimeter of the parallelogram ?

1) the base of the parallelogram is 10
2) one of the angles of the parallelogram is 45 degree

I got C.

p = 20 + 20 sqrt 2
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15 Dec 2007, 21:19
bmwhype2 wrote:
the area of a parallelogram is 100. what is the perimeter of the parallelogram ?

1) the base of the parallelogram is 10
2) one of the angles of the parallelogram is 45 degree

C

1. Not suff . We can only know the height ie 100/10 = 10

2. Not suff . T

Combining we can find length of other parallel side ie 10sqrt2.
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16 Dec 2007, 19:02
why cant we infer this from A?

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16 Dec 2007, 22:37
bmwhype2 wrote:
why cant we infer this from A?

do you know one (two) of the angle of parrellogram is 45 deg?
CEO
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16 Dec 2007, 22:46
GMAT TIGER wrote:
bmwhype2 wrote:
why cant we infer this from A?

do you know one (two) of the angle of parrellogram is 45 deg?

isnt the height always perpendicular to the base? thus forming an isosceles right triangle?
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17 Dec 2007, 01:15
Actually, I was surprised that you were considering C ...

S = a * h

I. Suff: a*h = 100, a = 10 --> h = 10 --> we can infer b from 1:1:sqrt(2) triangle
P = 2*(a + b)

Last edited by Whatever on 17 Dec 2007, 01:19, edited 3 times in total.
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17 Dec 2007, 01:15
bmwhype2 wrote:
GMAT TIGER wrote:
bmwhype2 wrote:
why cant we infer this from A?

do you know one (two) of the angle of parrellogram is 45 deg?

isnt the height always perpendicular to the base? thus forming an isosceles right triangle?

you can draw a parallellogram with two opposit angles 1-1 (or 10-10, 100-100) and rest 179-179 (170-170, 80-80). do you still think that, it is an issoceles triangle if we draw a perpendicular on the base?

only opposit sides are equal not the adjacent sides. we know this facto nly after going through st 1 and 2.
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17 Dec 2007, 01:25
bmwhype2 wrote:
GMAT TIGER wrote:
bmwhype2 wrote:
why cant we infer this from A?

do you know one (two) of the angle of parrellogram is 45 deg?

isnt the height always perpendicular to the base? thus forming an isosceles right triangle?

you are right. the answer is D.

(1) find height and then hypotenuse SUFF

(2) 45 degrees gives an isosles triangle. since we know that area formula is heigh*base and it equals 100. in isosles triangle two sides are equal and hence one of them is sqrt 100. SUFF

good question!
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17 Dec 2007, 01:40
bmwhype2 wrote:
the area of a parallelogram is 100. what is the perimeter of the parallelogram ?

1) the base of the parallelogram is 10
2) one of the angles of the parallelogram is 45 degree

CaspAreaGuy wrote:
you are right. the answer is D.

(1) find height and then hypotenuse SUFF

(2) 45 degrees gives an isosles triangle. since we know that area formula is heigh*base and it equals 100. in isosles triangle two sides are equal and hence one of them is sqrt 100. SUFF

good question!

CaspAreaGuy, but what if the base of parallelogram = 100 and height = 1?
Then, the triangle has legs 1:1:sqrt(2) and hence the P = 100*2 + sqrt(2) * 2 = 200 + 2*sqrt(2)?
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17 Dec 2007, 01:51
Whatever wrote:
bmwhype2 wrote:
the area of a parallelogram is 100. what is the perimeter of the parallelogram ?

1) the base of the parallelogram is 10
2) one of the angles of the parallelogram is 45 degree

CaspAreaGuy wrote:
you are right. the answer is D.

(1) find height and then hypotenuse SUFF

(2) 45 degrees gives an isosles triangle. since we know that area formula is heigh*base and it equals 100. in isosles triangle two sides are equal and hence one of them is sqrt 100. SUFF

good question!

CaspAreaGuy, but what if the base of parallelogram = 100 and height = 1?
Then, the triangle has legs 1:1:sqrt(2) and hence the P = 100*2 + sqrt(2) * 2 = 200 + 2*sqrt(2)?

but if the base is 100 and height is 1, will it be an isosles triangle? didn't get your point, sorry.
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17 Dec 2007, 01:57
CaspAreaGuy wrote:
but if the base is 100 and height is 1, will it be an isosceles triangle? didn't get your point, sorry.

Consider base = 100 and height = 1
..---------------------------
/ | ............................. /
---------------------------

There is a triangle from the left formed with the height = 1, and with the angles 45-90 (hence height forms right angle)-45 --> triangle is isosceles
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17 Dec 2007, 02:08
I got your point. Thank you. Traingle can still be isosless even if the height =100
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17 Dec 2007, 05:10
You need both.
Square is a parallelogram with a perimeter of 40
So you need B information to give you the perimeter 20+20sqrt2
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17 Dec 2007, 06:51
GMAT TIGER wrote:
bmwhype2 wrote:
GMAT TIGER wrote:
bmwhype2 wrote:
why cant we infer this from A?

do you know one (two) of the angle of parrellogram is 45 deg?

isnt the height always perpendicular to the base? thus forming an isosceles right triangle?

you can draw a parallellogram with two opposit angles 1-1 (or 10-10, 100-100) and rest 179-179 (170-170, 80-80). do you still think that, it is an issoceles triangle if we draw a perpendicular on the base?

only opposit sides are equal not the adjacent sides. we know this facto nly after going through st 1 and 2.

alright i figured it out.

we can have that figure, or we can have a square. if we have that figure the sides are 10sqrt2 and 10. If we have a square, we have 10 and 10.
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17 Dec 2007, 08:07
No.
Square gives the minimum perimeter

so
1) 90 degrees square looks like this
----------

----------

2) 70 degrees looks like this
----------

----------

3) 45 degrees looks like this
----------

----------

4)
35 degrees looks like this
----------

----------

The perimeter could be infinite as the degrees approach 0
This is why you need both
45degrees gives you isosceles both on the base on the height and don't provide enough info in finding the perimeter
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17 Dec 2007, 08:09
Aaaaaa spaces are excluded
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19 Dec 2007, 19:34
Re: Parallelogram   [#permalink] 19 Dec 2007, 19:34
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