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karlfurt
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Fig
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I got B, too. Not sure if this is too simplistic but....
from the stem we have 2L + 2w = 300 where L is length and W is width

1) Tells us that LW/4 = an integer. It's not enough. So, INSUF. (Is this right?)

2) We get 2*30 + 2W = 300. W = 120. and L = 30. Area is 3600.
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Yes, Fig is right, A is insufficient, I wasn't thinking clearly.
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Damager
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Fig
(B) for me :)

From 1:
Let set s : the side of 1 square

As a square is a special rectangle but so still a rectangle, we have 2 cases:
> The 4 cubes in a raw (the lenght = 4s and the width = s):
o perimeter = 2*(s + 4*s) = 300
o s = 150/5 = 30
o aera = 4s*s = 4s^2 = 4*30^2 = 3600

or

> The 2 cubes in a raw and 2 raws forming a scare (the lenght = 2s and the width = 2s):
o perimeter = 4*2*s = 300
o s = 150/4 = 75/2
o aera = (2s)^2 = (75)^2

INSUFF

From 2:
Let set L = lenght of the rectangle

2*(L+30) = 300
<=> L = 120

Thus,

Aera = L*30 = 120 * 30 = 3600

SUFF




SQARE SHOULD BE TREATED LIKE A SQARE IF U SAY SQAURE IS RECTANGLE AND CONCLUDE SQAURE DOES NOT HAVE EQUAL SIDES THEN ITS NOT A SQARE

DEFINITION OF SQARE SUPERSIDES THE DEFINITION OF RECTANGLE

(1)( L+L)4=3600 we have only one value of L hence only one area

(2) already explained by u
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Fig
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Damager
Fig
(B) for me :)

From 1:
Let set s : the side of 1 square

As a square is a special rectangle but so still a rectangle, we have 2 cases:
> The 4 cubes in a raw (the lenght = 4s and the width = s):
o perimeter = 2*(s + 4*s) = 300
o s = 150/5 = 30
o aera = 4s*s = 4s^2 = 4*30^2 = 3600

or

> The 2 cubes in a raw and 2 raws forming a scare (the lenght = 2s and the width = 2s):
o perimeter = 4*2*s = 300
o s = 150/4 = 75/2
o aera = (2s)^2 = (75)^2

INSUFF

From 2:
Let set L = lenght of the rectangle

2*(L+30) = 300
<=> L = 120

Thus,

Aera = L*30 = 120 * 30 = 3600

SUFF



SQARE SHOULD BE TREATED LIKE A SQARE IF U SAY SQAURE IS RECTANGLE AND CONCLUDE SQAURE DOES NOT HAVE EQUAL SIDES THEN ITS NOT A SQARE

DEFINITION OF SQARE SUPERSIDES THE DEFINITION OF RECTANGLE

(1)( L+L)4=3600 we have only one value of L hence only one area

(2) already explained by u


I believe that the defintions to qualify a figure are based on fundamentals that make independant the way to call this figure.

In addition, they are never excluding one another. Every time, we go in deeper level or sub group of a bigger group.

Level 0 : The figure is a polygon
Level 1 : The figure is a quadrilateral : a polygon of 4 vertices and 4 sides
Level 2 : The figure is a parallelogram: a quadrilateral with parallel opposite sides
Level 3 : The figue is a rectangle : a parallelogram with 1 right angle
Level 4 : The figue is a scare : a rectangle with 4 sides of the same lenght.
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OA is B!
I didn't know that a square is a rectangle...
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Fig
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karlfurt
OA is B!
I didn't know that a square is a rectangle...


:)....

Actually, it could be also considered a rhombus :)... Something that could be useful ;) :)
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Fig,

1. states that the rectangle can be divided into 4 equal squares. Does that not require the sides to be equal?

Can you please explain?

Thanks
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Fig
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gk3.14
Fig,

1. states that the rectangle can be divided into 4 equal squares. Does that not require the sides to be equal?

Can you please explain?

Thanks


Yes :)... I suggest u to have a look on the figure just below. In this picture, I have drawn the 2 possible cases. The sides of the squares are equal within a same shape. :)

Comparing the 2 ways to build a possibility, we notice that the squares used are different in order to obtain a same perimeter for both ways :). On my draw, well, I used the same squares but the perimeters are different (U could count the number of square side that create each perimeter : 10 and 8) :)
Attachments

Fig_2 ways to cut in 4 equal squares.GIF
Fig_2 ways to cut in 4 equal squares.GIF [ 2.46 KiB | Viewed 1868 times ]

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That was excellent..
Thanks Fig..
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Fig
gk3.14
Fig,

1. states that the rectangle can be divided into 4 equal squares. Does that not require the sides to be equal?

Can you please explain?

Thanks

Yes :)... I suggest u to have a look on the figure just below. In this picture, I have drawn the 2 possible cases. The sides of the squares are equal within a same shape. :)

Comparing the 2 ways to build a possibility, we notice that the squares used are different in order to obtain a same perimeter for both ways :). On my draw, well, I used the same squares but the perimeters are different (U could count the number of square side that create each perimeter : 10 and 8) :)


Excellent one!!! gr8 explanation!!
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Great explanation by Fig
I was also about to do the mistake ..
Square is a special type of rectangle
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Fig
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Thanks guys :)
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Scary - I would've picked B if this was the test and not given it a second thought.

What is source of question?



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