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# What is the area of rectangular region R? 1. Each diagonal

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What is the area of rectangular region R? 1. Each diagonal [#permalink]

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17 Feb 2008, 17:17
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What is the area of rectangular region R?
1. Each diagonal of R has length 5.
2. the perimeter of R is 14.
[Reveal] Spoiler: OA
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17 Feb 2008, 18:59
el1981 wrote:
What is the area of rectangular region R?
1. Each diagonal of R has length 5.
2. the perimeter of R is 14.

1) $$L^2+B^2 = 25$$ -- Not sufficient

2) $$2 (L+B) = 14$$ => $$L+B = 7$$ -- Not sufficient

But with both 1 & 2 $$L^2+B^2 = 25$$ => $$(L+B)^2 - 2(L*B) = 25$$ => $$L*B = \frac{(49 - 25)}{2}$$
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17 Feb 2008, 19:59
since we know its a rectangle, and we know the length of the diagonal is 5, cant we say that the other two sides are 3 and 4 ?

if so, then the answer is A
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17 Feb 2008, 20:09
pmenon wrote:
since we know its a rectangle, and we know the length of the diagonal is 5, cant we say that the other two sides are 3 and 4 ?

if so, then the answer is A

That's assuming the sides and diagonal form a perfect square. We can't assume that.
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17 Feb 2008, 20:15
oh, okay. i had thought that with rectangles, their corners were always 90 degrees. I guess we cant assume that because of shapes like rhombuses, etc ?

edit: wait a sec, if the corners arent 90 degrees, then why can we use pythagorean theorem and say that ^2+b^2 = 5 ?
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17 Feb 2008, 20:58

3-4-5 triangle works for an area of 12 (3*4)

but so does:

sqrt(20)-sqrt(5)-5 because 20+5 = 25 and that would give you an area of 10

so yes, it has a 90 degree angle, but you need to know more about the side lengths to get an area.
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18 Feb 2008, 14:50
neelesh wrote:
el1981 wrote:
What is the area of rectangular region R?
1. Each diagonal of R has length 5.
2. the perimeter of R is 14.

1) $$L^2+B^2 = 25$$ -- Not sufficient

2) $$2 (L+B) = 14$$ => $$L+B = 7$$ -- Not sufficient

But with both 1 & 2 $$L^2+B^2 = 25$$ => $$(L+B)^2 - 2(L*B) = 25$$ => $$L*B = \frac{(49 - 25)}{2}$$

neelesh, could you please clarify why 1&2 sufficient. Thanks.
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18 Feb 2008, 19:22
el1981 wrote:
neelesh wrote:
el1981 wrote:
What is the area of rectangular region R?
1. Each diagonal of R has length 5.
2. the perimeter of R is 14.

1) $$L^2+B^2 = 25$$ -- Not sufficient

2) $$2 (L+B) = 14$$ => $$L+B = 7$$ -- Not sufficient

But with both 1 & 2 $$L^2+B^2 = 25$$ => $$(L+B)^2 - 2(L*B) = 25$$ => $$L*B = \frac{(49 - 25)}{2}$$

neelesh, could you please clarify why 1&2 sufficient. Thanks.

Because I am taking the Statement 2 and using it in statement 1

$$L^2+B^2 = 25$$
=> $$L^2 + B^2 + 2(L*B) - 2(L*B) = 25$$
=> $$(L+B)^2 - 2(L*B) = 25$$
=> $$(7)^2 - 2(L*B) = 25$$ /* Using statement-2 */

Hence C.
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19 Feb 2008, 01:17
pmenon wrote:
oh, okay. i had thought that with rectangles, their corners were always 90 degrees. I guess we cant assume that because of shapes like rhombuses, etc ?

edit: wait a sec, if the corners arent 90 degrees, then why can we use pythagorean theorem and say that ^2+b^2 = 5 ?

The angles of a rectangle are ALWAYS 90 degrees.

But if the hypo is 5, its not necessary that the other two sides will always be 3 and 4.
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19 Feb 2008, 01:18
C should be the correct answer.
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20 May 2011, 12:09
1) L^2 + W^2 = 25 Not sufficient

2) L + W = 7 Not sufficient

But with both 1 & 2
(L+W)^2 = 7^2
=> L^2 +W^2 = 49
=> L^2 +W^2 + 2LW = 49
=> 25+ LW = 49
=> 2LW = 49 – 25
=> LW = 12

Ans. C
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20 May 2011, 17:12
Rectangle is a parallelogram of which all angles are 90 degree, and opposite sides are equal.

So, two sides of the rectangle are 3 and 4. Therefore, we can derive the area of rectangle from option 1 making it sufficient enough.

But option 2 alone is not sufficient.

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20 May 2011, 17:47
bibhas wrote:
Rectangle is a parallelogram of which all angles are 90 degree, and opposite sides are equal.

So, two sides of the rectangle are 3 and 4. Therefore, we can derive the area of rectangle from option 1 making it sufficient enough.

But option 2 alone is not sufficient.

According to you; Area=3*4=12

What if:
one side$$=1$$; other side$$=4\sqrt{6}$$ and hypotenuse$$=5$$
$$Area = 1*4\sqrt{6}=4\sqrt{6}$$

OR

one side$$=\sqrt{12.5}$$; other side$$=\sqrt{12.5}$$ and hypotenuse$$=5$$
$$Area = \sqrt{12.5}*\sqrt{12.5}=12.5$$

There are infinite such possibilities because we are NOT GIVEN THAT SIDES ARE INTEGERS.

OA: "C" is correct.
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20 May 2011, 18:07
Your option cannot form a triangle and second option makes the parallelogram a Square (questions says Rectangular region). So neither of them is sufficient enough to justify your answer.
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20 May 2011, 18:33
bibhas wrote:
Your option cannot form a triangle and second option makes the parallelogram a Square (questions says Rectangular region). So neither of them is sufficient enough to justify your answer.

I meant to say; any combination of two positive numbers whose squares add up to 25 i.e. 5^2 will form a triangle with hypotenuse 5.

one side: $$2\sqrt{6}$$ and other side: $$1$$; hypotenuse: $$5$$

$$1^2+(2\sqrt{6})^2=5^2$$

And a square is a specialized rectangle in GMAT.
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20 May 2011, 21:38
a+b gives,

l^2 -7l + 12 = 0 l = length

(l-3)(l-4)=0
hence lw = 12 in either cases, as l+w = 7.

C
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23 Jun 2011, 11:52
fluke wrote:
bibhas wrote:
Your option cannot form a triangle and second option makes the parallelogram a Square (questions says Rectangular region). So neither of them is sufficient enough to justify your answer.

I meant to say; any combination of two positive numbers whose squares add up to 25 i.e. 5^2 will form a triangle with hypotenuse 5.

one side: $$2\sqrt{6}$$ and other side: $$1$$; hypotenuse: $$5$$

$$1^2+(2\sqrt{6})^2=5^2$$

And a square is a specialized rectangle in GMAT.

Hi ,
Your statement "any combination of two positive numbers whose squares add up to 25 i.e. 5^2 will form a triangle with hypotenuse 5. " is not true if its a right angle traingle with diagonal as 5. as per Pythagoras theorem other 2 sides should be 3 and 4 so A is sufficient alone
Pl clarify incase I am missing anything
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23 Jun 2011, 12:01
sameershintrein wrote:
fluke wrote:
bibhas wrote:
Your option cannot form a triangle and second option makes the parallelogram a Square (questions says Rectangular region). So neither of them is sufficient enough to justify your answer.

I meant to say; any combination of two positive numbers whose squares add up to 25 i.e. 5^2 will form a triangle with hypotenuse 5.

one side: $$2\sqrt{6}$$ and other side: $$1$$; hypotenuse: $$5$$

$$1^2+(2\sqrt{6})^2=5^2$$

And a square is a specialized rectangle in GMAT.

Hi ,
Your statement "any combination of two positive numbers whose squares add up to 25 i.e. 5^2 will form a triangle with hypotenuse 5. " is not true if its a right angle traingle with diagonal as 5. as per Pythagoras theorem other 2 sides should be 3 and 4 so A is sufficient alone
Pl clarify incase I am missing anything

Pythagoras theorem says ..
Hyp^2 = sum of the squares of other 2 sides..
it never said the all the sides are phythagoras triplets like 3,4,5 and 9,12,15
so if the hyp = 5, yes its easier to assume that other 2 sides follow the triplet format and are 3 and 4
but nothing stops us from assuming that they can be 1 and 2 sqrt 6.
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23 Jun 2011, 12:14
sameershintrein wrote:
fluke wrote:
bibhas wrote:
Your option cannot form a triangle and second option makes the parallelogram a Square (questions says Rectangular region). So neither of them is sufficient enough to justify your answer.

I meant to say; any combination of two positive numbers whose squares add up to 25 i.e. 5^2 will form a triangle with hypotenuse 5.

one side: $$2\sqrt{6}$$ and other side: $$1$$; hypotenuse: $$5$$

$$1^2+(2\sqrt{6})^2=5^2$$

And a square is a specialized rectangle in GMAT.

Hi ,
Your statement "any combination of two positive numbers whose squares add up to 25 i.e. 5^2 will form a triangle with hypotenuse 5. " is not true if its a right angle traingle with diagonal as 5. as per Pythagoras theorem other 2 sides should be 3 and 4 so A is sufficient alone
Pl clarify incase I am missing anything

3,4,5 is just one of the infinite possibilities.

Why don't you draw it and see it yourself.

Draw a horizontal line-segment(AB) of 1 unit . Draw a perpendicular ray directly upward from point A. Now, using divider pointing at point B, and setting the divider to 5 units, make a small arc so that it cuts the ray at some point, say C. Join BC. You now have a right triangle with hypotenuse 5, one side 1 unit, and another side $$\sqrt{5^2-1} = \sqrt{24}= 2 \sqrt{6} \approx 4.9$$

Like this, we have infinite possibilities because there are infinite real numbers between 0 and 5, exclusive.
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23 Jun 2011, 22:35
1. Not sufficient

l^2+w^2 = 25

l ,w can have different values . for different values we will have different areas.

2. Not sufficient

we know the l+w =7,

still we can chose different combinations of l , w and different values yield different area.

together,

consider (l+w)^2 = l^2+w^w+2lw
we can find lw with the values we have from 1 and 2.

Re: DS: Rectangular   [#permalink] 23 Jun 2011, 22:35

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