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Re: NEED SOME Help on this DS question [#permalink]

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26 Nov 2010, 14:07

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ajit257 wrote:

What is the area of rectangular region R ? (1) Each diagonal of R has length 5 (2) The perimeter of R is 14

Please could someone explain this question ...thanks.

Let the sides of the rectangle be \(x\) and \(y\). Question: \(area=xy=?\)

(1) Each diagonal of R has length 5 --> as the diagonals in a rectangle are the hypotenuses for the sides then: \(x^2+y^2=5^2\), but we can not get the value of \(xy\) from this info. Not sufficient.

(2) The perimeter of R is 14 --> \(P=2(x+y)=14\) --> \(x+y=7\). Again we can not get the value of \(xy\) from this info. Not sufficient.

(1)+(2) We have \(x^2+y^2=25\) and \(x+y=7\). Square the second expression: \(x^2+2xy+y^2=49\), as \(x^2+y^2=5^2\) then \(25+2xy=49\) --> \(xy=12\). Sufficient.

Re: NEED SOME Help on this DS question [#permalink]

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27 Nov 2010, 10:22

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rockroars wrote:

I thought "Statement 1" alone is sufficient to solve this problem.

3,4,5 is the only Pythagorean triplet which supports 5 a diagonal of a right angled triangle.

Why can't the answer be A?

We are not told that the lengths of the sides are integers. So knowing that hypotenuse equals to 5 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 3:4:5. Or in other words: if \(x^2+y^2=5^2\) DOES NOT mean that \(x=3\) and \(y=4\). Certainly this is one of the possibilities but definitely not the only one. In fact \(x^2+y^2=5^2\) has infinitely many solutions for \(x\) and \(y\) and only one of them is \(x=3\) and \(y=4\).

For example: \(x=1\) and \(y=\sqrt{24}\) or \(x=2\) and \(y=\sqrt{21}\)...

So knowing that the diagonal of a rectangle (hypotenuse) equals to one of the Pythagorean triple hypotenuse value is not sufficient to calculate the sides of this rectangle.

Re: What is the area of rectangular region R? [#permalink]

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29 Feb 2012, 02:30

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From what I understand of rectangular diagonals or quadrilateral diagonals is that if they are the same length, then all sides should be of equal length. Also area of Rhombus = 1/2 * diagonal * diagonal? Correct me if I'm wrong here, just need clarification

Re: What is the area of rectangular region R? [#permalink]

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29 Feb 2012, 02:40

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calvin1984 wrote:

From what I understand of rectangular diagonals or quadrilateral diagonals is that if they are the same length, then all sides should be of equal length. Also area of Rhombus = 1/2 * diagonal * diagonal? Correct me if I'm wrong here, just need clarification

All rectangles have the diagonals of equal length, so (1) doesn't necessarily means that given rectangle is a rhombus.

Re: NEED SOME Help on this DS question [#permalink]

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21 Sep 2012, 09:15

Bunuel wrote:

ajit257 wrote:

What is the area of rectangular region R ? (1) Each diagonal of R has length 5 (2) The perimeter of R is 14

Please could someone explain this question ...thanks.

Let the sides of the rectangle be \(x\) and \(y\). Question: \(area=xy=?\)

(1) Each diagonal of R has length 5 --> as the diagonals in a rectangle are the hypotenuses for the sides then: \(x^2+y^2=5^2\), but we can not get the value of \(xy\) from this info. Not sufficient.

(2) The perimeter of R is 14 --> \(P=2(x+y)=14\) --> \(x+y=7\). Again we can not get the value of \(xy\) from this info. Not sufficient.

(1)+(2) We have \(x^2+y^2=25\) and \(x+y=7\). Square the second expression: \(x^2+2xy+y^2=49\), as \(x^2+y^2=5^2\) then \(25+2xy=49\) --> \(xy=12\). Sufficient.

Answer: C.

Area of a rectangular region = Product of two diagonals/2 We are given both are diagonals are equal to 5 So area would be = 25/2 = 12.5 Thus A is sufficient

Let me know why i am wrong.

Waiting for reply. _________________

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Re: NEED SOME Help on this DS question [#permalink]

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21 Sep 2012, 09:23

Expert's post

fameatop wrote:

Bunuel wrote:

ajit257 wrote:

What is the area of rectangular region R ? (1) Each diagonal of R has length 5 (2) The perimeter of R is 14

Please could someone explain this question ...thanks.

Let the sides of the rectangle be \(x\) and \(y\). Question: \(area=xy=?\)

(1) Each diagonal of R has length 5 --> as the diagonals in a rectangle are the hypotenuses for the sides then: \(x^2+y^2=5^2\), but we can not get the value of \(xy\) from this info. Not sufficient.

(2) The perimeter of R is 14 --> \(P=2(x+y)=14\) --> \(x+y=7\). Again we can not get the value of \(xy\) from this info. Not sufficient.

(1)+(2) We have \(x^2+y^2=25\) and \(x+y=7\). Square the second expression: \(x^2+2xy+y^2=49\), as \(x^2+y^2=5^2\) then \(25+2xy=49\) --> \(xy=12\). Sufficient.

Answer: C.

Area of a rectangular region = Product of two diagonals/2 We are given both are diagonals are equal to 5 So area would be = 25/2 = 12.5 Thus A is sufficient

Let me know why i am wrong.

Waiting for reply.

The red part is not correct. It's true about squares: \(area_{square}=\frac{diagonal^2}{2}\).

Re: What is the area of rectangular region R ? [#permalink]

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08 Dec 2012, 11:06

I screwed up on this one like an earlier poster. So hypothetically, if the question stem stated that the sides were integers, would A be sufficient alone?

I'm nervous on the DS. I got one wrong on the PS in the official guide and 6 wrong already on DS and I'm only on question 50 .

Re: What is the area of rectangular region R ? [#permalink]

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09 Dec 2012, 09:49

Expert's post

RonBagel wrote:

I screwed up on this one like an earlier poster. So hypothetically, if the question stem stated that the sides were integers, would A be sufficient alone?

Yes, if we were told that the lengths of the sides of the rectangle are integers, then the first statement would be sufficient: x^2+y^2=25 --> x=3 and y=4 or vise -versa --> xy=12. _________________

Re: NEED SOME Help on this DS question [#permalink]

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11 Apr 2014, 01:04

Bunuel wrote:

ajit257 wrote:

What is the area of rectangular region R ? (1) Each diagonal of R has length 5 (2) The perimeter of R is 14

Please could someone explain this question ...thanks.

Let the sides of the rectangle be \(x\) and \(y\). Question: \(area=xy=?\)

(1) Each diagonal of R has length 5 --> as the diagonals in a rectangle are the hypotenuses for the sides then: \(x^2+y^2=5^2\), but we can not get the value of \(xy\) from this info. Not sufficient.

(2) The perimeter of R is 14 --> \(P=2(x+y)=14\) --> \(x+y=7\). Again we can not get the value of \(xy\) from this info. Not sufficient.

(1)+(2) We have \(x^2+y^2=25\) and \(x+y=7\). Square the second expression: \(x^2+2xy+y^2=49\), as \(x^2+y^2=5^2\) then \(25+2xy=49\) --> \(xy=12\). Sufficient.

Answer: C.

Hi Bunuel,

Can't we apply the 1 sqrt3 2 theory to statement one? Since it's a rectangular then the angle created by the diagonal must be 90 and leaving the rest 30 and 60. So the sides must be 5/2 and 5/2(sqrt3).

Re: NEED SOME Help on this DS question [#permalink]

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11 Apr 2014, 02:35

Expert's post

aquax wrote:

Bunuel wrote:

ajit257 wrote:

What is the area of rectangular region R ? (1) Each diagonal of R has length 5 (2) The perimeter of R is 14

Please could someone explain this question ...thanks.

Let the sides of the rectangle be \(x\) and \(y\). Question: \(area=xy=?\)

(1) Each diagonal of R has length 5 --> as the diagonals in a rectangle are the hypotenuses for the sides then: \(x^2+y^2=5^2\), but we can not get the value of \(xy\) from this info. Not sufficient.

(2) The perimeter of R is 14 --> \(P=2(x+y)=14\) --> \(x+y=7\). Again we can not get the value of \(xy\) from this info. Not sufficient.

(1)+(2) We have \(x^2+y^2=25\) and \(x+y=7\). Square the second expression: \(x^2+2xy+y^2=49\), as \(x^2+y^2=5^2\) then \(25+2xy=49\) --> \(xy=12\). Sufficient.

Answer: C.

Hi Bunuel,

Can't we apply the 1 sqrt3 2 theory to statement one? Since it's a rectangular then the angle created by the diagonal must be 90 and leaving the rest 30 and 60. So the sides must be 5/2 and 5/2(sqrt3).

What's wrong with this explanation?

Thanks.

Let me ask you a question: why must the remaining angles be 30 and 60 degrees? Why cannot they be 25 or 65? Or 20 and 70? Basically any pair totaling 90? _________________

Re: NEED SOME Help on this DS question [#permalink]

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11 Apr 2014, 03:29

aquax wrote:

Bunuel wrote:

ajit257 wrote:

What is the area of rectangular region R ? (1) Each diagonal of R has length 5 (2) The perimeter of R is 14

Please could someone explain this question ...thanks.

Let the sides of the rectangle be \(x\) and \(y\). Question: \(area=xy=?\)

(1) Each diagonal of R has length 5 --> as the diagonals in a rectangle are the hypotenuses for the sides then: \(x^2+y^2=5^2\), but we can not get the value of \(xy\) from this info. Not sufficient.

(2) The perimeter of R is 14 --> \(P=2(x+y)=14\) --> \(x+y=7\). Again we can not get the value of \(xy\) from this info. Not sufficient.

(1)+(2) We have \(x^2+y^2=25\) and \(x+y=7\). Square the second expression: \(x^2+2xy+y^2=49\), as \(x^2+y^2=5^2\) then \(25+2xy=49\) --> \(xy=12\). Sufficient.

Answer: C.

Hi Bunuel,

Can't we apply the 1 sqrt3 2 theory to statement one? Since it's a rectangular then the angle created by the diagonal must be 90 and leaving the rest 30 and 60. So the sides must be 5/2 and 5/2(sqrt3).

What's wrong with this explanation?

Thanks.

Probably you are getting confused because of this example: "a rectangle is inscribed in a circle of radius r...."

The diagonal divides the rectangle in two right triangles, so sum of two angles need to be 90 but it can be 30: 60, 45:45...and so on...

Re: What is the area of rectangular region R ? [#permalink]

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22 Jul 2014, 08:33

Hi Bunuel,

To keep it straight, just because it says its a rectangle does not mean we have to have two 30-60-90 triangles, but if we put together two 30-60-90 triangles we get a rectangle? Correct? I picked "A" because I thought that since it said we had a rectangle, we had to have two of these triangles. From the discussion above, it looks like this is not a mandatory condition of a rectangle.

Re: What is the area of rectangular region R ? [#permalink]

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22 Jul 2014, 08:36

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jbdoyl3 wrote:

Hi Bunuel,

To keep it straight, just because it says its a rectangle does not mean we have to have two 30-60-90 triangles, but if we put together two 30-60-90 triangles we get a rectangle? Correct? I picked "A" because I thought that since it said we had a rectangle, we had to have two of these triangles. From the discussion above, it looks like this is not a mandatory condition of a rectangle.

Correct. But you can get a rectangle by putting together any two congruent right triangles, not necessarily 30-60-90 triangles. _________________

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