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a parallelogram has 2 sides: 14, 18 one of the diagonals is

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a parallelogram has 2 sides: 14, 18 one of the diagonals is [#permalink] New post 24 Sep 2005, 06:03
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a parallelogram has 2 sides: 14, 18 one of the diagonals is 16, what is the length of the othe diagonal?
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 [#permalink] New post 24 Sep 2005, 14:45
Is there some data missing, like an angle? I'm not sure if there is sufficient information, unless, there is a property of a parallelogram that I missed.

How do we solve this or whats the OE?
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 [#permalink] New post 24 Sep 2005, 15:56
Length of the second diagnol is 44.

Area(parallelogram)=18X14=252/2=176 (Area of each triangle)

Base/2Xheight=Area(triangles)

16/2XH=176

H=22

22X2=44
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 [#permalink] New post 24 Sep 2005, 16:04
GMATT73 wrote:
Length of the second diagnol is 44.

Area(parallelogram)=18X14=252/2=176 (Area of each triangle)

Base/2Xheight=Area(triangles)

16/2XH=176

H=22

22X2=44


Area of a parallelogram = base * height. How did you calculate height?
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 [#permalink] New post 25 Sep 2005, 05:44
anymore guys. the question reminds you an important concept/formulae.
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 [#permalink] New post 25 Sep 2005, 17:38
i give a shot : 2sqreroot33 :lol:
but actually it's really because I want to participate :)
Come on, what's the trick ?
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 [#permalink] New post 25 Sep 2005, 20:30
It's a very good question.
lets a and b - sides
d1 and d2 diagonals
The key formular is:
d1^2+d2^2=2*(a^2+b^2).

from that another digonal is 28 ( if I'm not mistaken in calculations).
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 [#permalink] New post 26 Sep 2005, 07:33
Please demonstrate Natalya.
I don't think there is sufficient info.
Himalaya, do u have an answer?
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 [#permalink] New post 26 Sep 2005, 07:37
Quote:
d1^2+d2^2=2*(a^2+b^2).

I'm not sure if that formula applies to a parallelogram. Looks more like what would apply to a kite (adjacent sides equal and diagonals perpendicular).

Couldn't find this anywhere through google. Does anyone have a link?
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 [#permalink] New post 27 Sep 2005, 03:41
If anyone doesn't believe the answer is 28,
you can work it out by drawing all the right angles in and applying Pythagoras' theorem several times and substituting.

So draw a rectangle which is 18+x wide, y high. In the top left take out a triangle x wide, with hypotenuse of 14. Then 14^2 = y^2 + x^2

and so on
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 [#permalink] New post 27 Sep 2005, 05:34
How do the diagonals come in the picture?
Richard.. can you elaborate on that? I tried like you said but somehow the diagonals dont get included on any of the pythagorean equations... What am I missing?

Thanks.
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 [#permalink] New post 27 Sep 2005, 06:42
Difficult to describe.

Draw a big rectangle on a sheet of paper.
Its height is y.
Its length is 18+x.
On the top mark x from the left.
On the bottom mark x from the right.
Draw in the parallelogram inscribed within this rectangle.
(By drawing line from bottom left corner to point on top x from left.)
Mark a rectangle on the left, which is x wide, y high.
The diagonal is 14. Side of parallelogram.

So x^2 + y^2 = 14^2 = 196 (E1)

Let z = 18 - x (E2), z is the middle width portion of the rectangle.

Then you can split the bottom line, into three parts,
x, z, x.
Now we know the length of the shorter diagonal 16.
But this is the hypotenuse of a triangle.
The sides are y and z.

So y^2 + z^2 = 16^2 = 256 (E3)

(E3) - (E1)
z^2 - x^2 = 60
From (E2) z = 18 - x
So z^2 = 18^2 - 36x + x^2

(324 - 36x + x^2) - x^2 = 60
(324 - 60) = 36x
x = 264 / 36 = 22 / 3

Hence (E2)
z = 18 - 22/3 = 32/3

And (E1)
y^2 = 196 - x^2 = 196 - (22/3)^2
y^2 = (1764 - 484) / 9 = 1280/9

Width of rectangle,
w = 18 + x = (54+22)/3 = 76/3

Diagonal d
d^2 = w^2 + y^2
d^2 = (76^2)/9 + 1280/9
d^2 = (5776+1280)/9 = 7056/9
d = 84/3 = 28

But there would not be time to do this in the real GMAT.
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 [#permalink] New post 27 Sep 2005, 06:49
Richard,
That did clear my confusion. Thanks for the detailed explanation.
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 [#permalink] New post 27 Sep 2005, 14:06
vikramm wrote:
Quote:
d1^2+d2^2=2*(a^2+b^2).

I'm not sure if that formula applies to a parallelogram. Looks more like what would apply to a kite (adjacent sides equal and diagonals perpendicular).

Couldn't find this anywhere through google. Does anyone have a link?


I too was wondering where from this new formula came. A little trignometry helped derive formula though. Anyone interested in derivation can look into attached doc.

For others, get this formula in your kitty.
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  [#permalink] 27 Sep 2005, 14:06
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