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# The lengh of a parallelogram's one side is 6 and another

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The lengh of a parallelogram's one side is 6 and another [#permalink]

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20 Dec 2005, 20:48
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The lengh of a parallelogram's one side is 6 and another side is 7. What is(are) possibly the area of the parallelogram?

I) 21 II) 42 III) 84

1) I only
2) II only
3) III only
4) I and II only
5) I and III only
6) II and III only
7) I, II, and III
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20 Dec 2005, 20:59
21 - 1 alone.
Area of a parallellogram is base x height
So area cannot be greater than 42 as height cannot be greater than either 6 or 7.
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JAI HIND!

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20 Dec 2005, 21:11
JAI HIND wrote:
21 - 1 alone.
Area of a parallellogram is base x height
So area cannot be greater than 42 as height cannot be greater than either 6 or 7.

I would go for 4) I and II... note that a rectangle is a special type of parallelogram. so 7*6 = 42 is a possible area.
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20 Dec 2005, 21:31
yb wrote:
JAI HIND wrote:
21 - 1 alone.
Area of a parallellogram is base x height
So area cannot be greater than 42 as height cannot be greater than either 6 or 7.

I would go for 4) I and II... note that a rectangle is a special type of parallelogram. so 7*6 = 42 is a possible area.

good catch!
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21 Dec 2005, 17:26
4. I & II

Parallelogram can also be rectangle
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Re: PS - Area of a parallelogram [#permalink]

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22 Dec 2005, 03:55
gamjatang wrote:
The lengh of a parallelogram's one side is 6 and another side is 7. What is(are) possibly the area of the parallelogram?

I) 21 II) 42 III) 84

1) I only
2) II only
3) III only
4) I and II only
5) I and III only
6) II and III only
7) I, II, and III

Area=AB*AC*Sin(^BAC)
42=6*7*Sin(90)
21=6*7*Sin(30)

So if angle will be 30 grad it will be 21

so 4)
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22 Dec 2005, 10:00
I am sure that II is definitely true.
But not sure about the I.

Is it that
Area of a parallelogram = Product of adjacent sides * sin (included angle)
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22 Dec 2005, 20:39
A. (1) only.

A(parallelogram)=height * side

For the same measures of length and width, area of rectangle is always larger than the area of parallelogram ( since the length of hypotenuse is always larger than the length of the leg).

In this case, if the figure was a rectangle, area would be equal to 6*7=42

Area (parallelogram)<42.
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22 Dec 2005, 20:43
Thanks, guys.

I didn't know the rectangle is a special type of paralelogram.
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08 Jan 2006, 16:48
JAI HIND wrote:
21 - 1 alone.
Area of a parallellogram is base x height
So area cannot be greater than 42 as height cannot be greater than either 6 or 7.

Folks I just dont understand how can the area of the llgm be 21.

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08 Jan 2006, 18:07
hkm_gmat wrote:
JAI HIND wrote:
21 - 1 alone.
Area of a parallellogram is base x height
So area cannot be greater than 42 as height cannot be greater than either 6 or 7.

Folks I just dont understand how can the area of the llgm be 21.

hkm_gmat, at first I was also confused.

I think that the question asks what are the possible values "among the floowing".

Assuming 7 as the base, if we calculate the largest possible area, it would come out to 7*6 = 42.

Infact the area of the parallelogram would be such that:
0 < Area of ll of gram <= 42

So (III) 84 is out. The other two values 21 & 42 are possible.
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09 Jan 2006, 00:40
Wouldnt it be exact 42... when the 2 sides of a parallaleogram are given then how can the area have a minimum and a max value??
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09 Jan 2006, 00:50
andy_gr8 wrote:
Wouldnt it be exact 42... when the 2 sides of a parallaleogram are given then how can the area have a minimum and a max value??

Andy, the angle between the sides of the parallelogram is not given. Depending on the angle, you would get different heights and hence different areas.
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09 Jan 2006, 00:50
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