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What is the area of rectangular region R? 1. Each diagonal [#permalink]
 17 Feb 2008, 17:17
Question Stats:
76% (01:22) correct
23% (02:10) wrong based on 0 sessions
What is the area of rectangular region R? 1. Each diagonal of R has length 5. 2. the perimeter of R is 14.
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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
17 Feb 2008, 20:58
You can use the Pythagorean theorum, but it won't help you without more information.
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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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
19 Feb 2008, 01:18
C should be the correct answer.
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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
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. Answer: A
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Re: DS: Rectangular [#permalink]
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.
Answer: A According to you; Area=3*4=12 What if: one side =1; other side =4\sqrt{6} and hypotenuse =5Area = 1*4\sqrt{6}=4\sqrt{6}OR one side =\sqrt{12.5}; other side =\sqrt{12.5} and hypotenuse =5Area = \sqrt{12.5}*\sqrt{12.5}=12.5There are infinite such possibilities because we are NOT GIVEN THAT SIDES ARE INTEGERS. OA: "C" is correct.
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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
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: 51^2+(2\sqrt{6})^2=5^2And a square is a specialized rectangle in GMAT.
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Re: DS: Rectangular [#permalink]
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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Re: DS: Rectangular [#permalink]
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: 51^2+(2\sqrt{6})^2=5^2And 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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Re: DS: Rectangular [#permalink]
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: 51^2+(2\sqrt{6})^2=5^2And 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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Re: DS: Rectangular [#permalink]
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: 51^2+(2\sqrt{6})^2=5^2And 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 4.9Like this, we have infinite possibilities because there are infinite real numbers between 0 and 5, exclusive.
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Re: DS: Rectangular [#permalink]
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.
Answer is C.
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Re: DS: Rectangular
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23 Jun 2011, 22:35
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