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What is the area of parallelogram ABCD? 1. AB = BC = CD = DA

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What is the area of parallelogram ABCD? 1. AB = BC = CD = DA [#permalink] New post 29 Oct 2007, 07:34
What is the area of parallelogram ABCD?

1. AB = BC = CD = DA = 1
2. AC = BD = sqrt(2)
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 [#permalink] New post 29 Oct 2007, 07:40
1) SUFF
the parallelogram is a square
2) SUFF
using pythagorean theory we can see that the sides must be equal to 1, with hypotenuse of sqrt(2).

D.
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 [#permalink] New post 29 Oct 2007, 08:48
(C) for me :)

Stat1
We have a rhombus...

INSUFF.

Stat2
We have a rectangle...

INSUFF.

Both 1 and 2
We are sure to have a square.

SUFF.
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 [#permalink] New post 29 Oct 2007, 08:54
Got it --
1) So even though you know the sides, this is not sufficient to solve for base & height of the rhombus.
INSUFF

2) Rectangle, not rhombus here.

Both:
Rectangle with equal sides = square with sides 1
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Re: Parallelogram [#permalink] New post 29 Oct 2007, 08:54
bmwhype2 wrote:
What is the area of parallelogram ABCD?

1. AB = BC = CD = DA = 1
2. AC = BD = sqrt(2)


1 alone could yield either a square or a rhombus.

2 alone tells nothing except that the diagonal is root 2.

1 and 2 combined give us an idea that it's a square indeed.

So C.
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 [#permalink] New post 29 Oct 2007, 09:01
In 2), would it be possible to use pythagoream theorem here to solve for the sides?

A^2 + B^2 = C^2 where c = sqrt(2)
so A^2 + B^2 = 2, so the sides must be 1 & 1, no?
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 [#permalink] New post 29 Oct 2007, 09:08
yuefei wrote:
In 2), would it be possible to use pythagoream theorem here to solve for the sides?

A^2 + B^2 = C^2 where c = sqrt(2)
so A^2 + B^2 = 2, so the sides must be 1 & 1, no?


Again you are assuming that the sides are at right angles (viz the figure is a rectangle/square) for the pythagorean theorem to be applicable.
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 [#permalink] New post 29 Oct 2007, 09:40
Got it... the parallelogram with congruent diagnols could be a trapezoid, thus the angles are not 90. Thanks
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 [#permalink] New post 29 Oct 2007, 10:29
For 2, to me, parallelogram ABCD + equal sizes of diagonal = rectangle :)

As a parallelogram ABCD implies that diagonales crosses one another at their middle points, the equal sizes of diagonal assure us that we have a square... not matter the angle between diagonale :)

A trapezoid cannot be a parallelogram :)
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 [#permalink] New post 30 Oct 2007, 02:48
Fig wrote:
For 2, to me, parallelogram ABCD + equal sizes of diagonal = rectangle :)

As a parallelogram ABCD implies that diagonales crosses one another at their middle points, the equal sizes of diagonal assure us that we have a square... not matter the angle between diagonale :)

A trapezoid cannot be a parallelogram :)


Are the diagonals in a rhombus equal?
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 [#permalink] New post 30 Oct 2007, 03:15
bmwhype2 wrote:
Fig wrote:
For 2, to me, parallelogram ABCD + equal sizes of diagonal = rectangle :)

As a parallelogram ABCD implies that diagonales crosses one another at their middle points, the equal sizes of diagonal assure us that we have a square... not matter the angle between diagonale :)

A trapezoid cannot be a parallelogram :)


Are the diagonals in a rhombus equal?


Not necessary...

Image

If so, it's also a square :)
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 [#permalink] New post 30 Oct 2007, 04:49
Fig wrote:
bmwhype2 wrote:
Fig wrote:
For 2, to me, parallelogram ABCD + equal sizes of diagonal = rectangle :)

As a parallelogram ABCD implies that diagonales crosses one another at their middle points, the equal sizes of diagonal assure us that we have a square... not matter the angle between diagonale :)

A trapezoid cannot be a parallelogram :)


Are the diagonals in a rhombus equal?


Not necessary...

Image

If so, it's also a square :)


thanks


A square is a rhombus and a rectangle. In other words, if each angle of a rhombus is 90° then it's a square.


1. AB = BC = CD = DA = 1
all sides equal, can be a rhombus or square.
Area of a square = s^2
Area of a rhombus = s*h
diff formulas, INSUFF

2. AC = BD = sqrt(2)
diagonals are equivalent. must be a square.
Area of a square = s^2
sides not defined
INSUFF

taken together, we see it is a s:s:sqrt(2) right triangle
area= 1^2=1
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 [#permalink] New post 30 Oct 2007, 04:54
bmw -- you stated the following in 2)

"diagonals are equivalent. must be a square.
Area of a square = s^2
sides not defined"

Can't you solve for s?

Create a triangle with hypotenuse sqrt(2) and side s.
s^2 + s^2 = sqrt(2)
s^2 = sqrt(2)/2

Is this possible?
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 [#permalink] New post 30 Oct 2007, 04:56
bmwhype2 wrote:
Fig wrote:
bmwhype2 wrote:
Fig wrote:
For 2, to me, parallelogram ABCD + equal sizes of diagonal = rectangle :)

As a parallelogram ABCD implies that diagonales crosses one another at their middle points, the equal sizes of diagonal assure us that we have a square... not matter the angle between diagonale :)

A trapezoid cannot be a parallelogram :)


Are the diagonals in a rhombus equal?


Not necessary...

Image

If so, it's also a square :)


thanks


A square is a rhombus and a rectangle. In other words, if each angle of a rhombus is 90° then it's a square.


1. AB = BC = CD = DA = 1
all sides equal, can be a rhombus or square.
Area of a square = s^2
Area of a rhombus = s*h
diff formulas, INSUFF

2. AC = BD = sqrt(2)
diagonals are equivalent. must be a square.
Area of a square = s^2
sides not defined
INSUFF

taken together, we see it is a s:s:sqrt(2) right triangle
area= 1^2=1


Well, one right angle is enough ;)... As a rhombhus is a paralelogram, 1 right angle creates 1 right angle by a parallel construction then the 2 last ones by another parallel construction :)
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 [#permalink] New post 30 Oct 2007, 11:30
Here is a similar question

Points A, B, C and D form a quadrilateral. Is AC longer than BD?

1. angle ABC > angle BCD
2. AB = BC = CD = DA
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 [#permalink] New post 30 Oct 2007, 11:45
yuefei wrote:
bmw -- you stated the following in 2)

"diagonals are equivalent. must be a square.
Area of a square = s^2
sides not defined"

Can't you solve for s?

Create a triangle with hypotenuse sqrt(2) and side s.
s^2 + s^2 = sqrt(2)
s^2 = sqrt(2)/2

Is this possible?


you dont know that the sides are equal. it can yield a rectangle.
  [#permalink] 30 Oct 2007, 11:45
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