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# Figure ABCD is a rectangle with sides of length x centimete

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Figure ABCD is a rectangle with sides of length x centimete [#permalink]

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15 Jul 2007, 19:45
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Figure ABCD is a rectangle with sides of length x centimeters and width y centimeters, and a diagonal of length z centimeters. What is the measure, in centimeters, of the perimeter of ABCD ?

(1) x – y = 7
(2) z = 13
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Feb 2014, 05:28, edited 2 times in total.
Renamed the topic, edited the question, added the figure and the OA.
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Re: Figure ABCD is a rectangle with sides of length x [#permalink]

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03 Feb 2014, 04:12
dynocomet wrote:
Figure ABCD is a rectangle with sides of length x centimeters and width y centimeters, and a diagonal of length z centimeters. What is the measure, in centimeters, of the perimeter of ABCD? (The diagram shows line segment AD labeled as "x" and line segment DC labeled as "y")

(1) x - y = 7
(2) z = 13

I read that the answer is C, but cannot understand why. First of all, statement 1 seems like a contradiction -- how can the hypotenuse be SMALLER than either of the two sides?

Thanks.

The answer to the questions shows C. However why can we not state that from statement (2) we are dealing with a 5,12,13 triangle? Hence B would be sufficient to answer the question.
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Re: Figure ABCD is a rectangle with sides of length x [#permalink]

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03 Feb 2014, 05:37
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Mattd wrote:
dynocomet wrote:
Figure ABCD is a rectangle with sides of length x centimeters and width y centimeters, and a diagonal of length z centimeters. What is the measure, in centimeters, of the perimeter of ABCD? (The diagram shows line segment AD labeled as "x" and line segment DC labeled as "y")

(1) x - y = 7
(2) z = 13

I read that the answer is C, but cannot understand why. First of all, statement 1 seems like a contradiction -- how can the hypotenuse be SMALLER than either of the two sides?

Thanks.

The answer to the questions shows C. However why can we not state that from statement (2) we are dealing with a 5,12,13 triangle? Hence B would be sufficient to answer the question.

The point is that a right triangle with hypotenuse 13, doesn't mean that we have (5, 12, 13) right triangle. If we are told that the lengths of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 13 would be (5, 12, 13). Or in other words: $$x^2+y^2=13^2$$ DOES NOT mean that $$x=5$$ and $$y=12$$. Certainly this is one of the possibilities but definitely not the only one. In fact $$x^2+y^2=13^2$$ has infinitely many solutions for $$x$$ and $$y$$ and only one of them is $$x=5$$ and $$y=12$$.

For example: $$x=1$$ and $$y=\sqrt{168}$$ or $$x=2$$ and $$y=\sqrt{165}$$...

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

Hope it's clear.
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Re: Figure ABCD is a rectangle with sides of length x centimete [#permalink]

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03 Feb 2014, 05:52
Thanks for the help. Did not think about the integer part.
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Re: Figure ABCD is a rectangle with sides of length x centimete [#permalink]

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15 Jul 2014, 02:13
How can we use the both statement to find the answer?
I marked B and Bunuel explained how it is wrong. My second answer would be E
I mean I know 12-5 = 7 and that with z =13 so we can say that three sides "can" be 5,12,13 but can we say that's the only possible value?
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Figure ABCD is a rectangle with sides of length x centimete [#permalink]

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15 Jul 2014, 02:35
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b2bt wrote:
How can we use the both statement to find the answer?
I marked B and Bunuel explained how it is wrong. My second answer would be E
I mean I know 12-5 = 7 and that with z =13 so we can say that three sides "can" be 5,12,13 but can we say that's the only possible value?

Given that $$x - y = 7$$ and $$z = 13$$. We also know that $$x^2 + y^2 = z^2 = 169$$. We want to find the perimeter which is $$2x+2y = 2(x+y)$$.

Square $$x - y = 7$$ --> $$x^2 - 2xy + y^2 = 49$$ --> $$169 - 2xy = 49$$ --> $$2xy = 120$$.

Add $$2xy = 120$$ to$$x^2 + y^2 = 169$$: $$x^2 + 2xy + y^2 = 169 + 120$$ --> $$(x+y)^2 = 289$$ --> $$x+y = 17$$ --> $$2(x+y) = 24$$.

Hope it's clear.
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Re: Figure ABCD is a rectangle with sides of length x centimete [#permalink]

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22 Jul 2014, 09:31
I looked at z=13 as
x^2+y^2=z^2=169
(x+y)(x-y)=169
(x+y)=169/7..... as from 1. we know that x-y=7

hence perimeter is 2(x+y)= 2(169/7)....

Hence C
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Re: Figure ABCD is a rectangle with sides of length x centimete [#permalink]

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Re: Figure ABCD is a rectangle with sides of length x centimete [#permalink]

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24 Dec 2015, 15:00
I knew there's a trick with B. since we are not told that x and y are integers, B alone is not sufficient.
we have statement 1:
x-y=7
thus x^2+y^2 -2xy=49. so not sufficient.

statement 2 alone:
z=13, well, z^2 = x^2 + y^2.
we have 169=x^2 + y^2.
not sufficient.

we need to find the perimeter, or 2(x+y).

we are told that x^2+y^2-2xy = 49
now we have
169=x^2 + y^2.
combine both:
169-2xy=49
120=2xy
60=xy.

now:
xy=60
x^2+y^2=169.

combine both:
x^2+y^2 + 2xy = 169+120
(x+y)^2 = 289.
now, this is sufficient, as we can find x+y, and thus find 2(x+y)
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Re: Figure ABCD is a rectangle with sides of length x centimete [#permalink]

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20 Sep 2016, 08:11
C is correct. Here's why:

(1) x-y =7 --> NOT SUFFICIENT (we don't have any values to arrive at the area)

(2) z = 13 --> NOT SUFFICIENT (this is a good thing to know, but there are many combinations of x and y that could get us to 13^2 if we use the pythagorean theorem)

Together (1) + (2) - We can rewrite (1) to be x = 7+y for one side and the other will be y; square both, add together, and set equal to 13^2 and solve for y.
Re: Figure ABCD is a rectangle with sides of length x centimete   [#permalink] 20 Sep 2016, 08:11
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