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# Figure ABCD is a rectangle with sides of length x centimeters and

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Re: Figure ABCD is a rectangle with sides of length x centimeters and [#permalink]
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 centimeters and [#permalink]
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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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Re: Figure ABCD is a rectangle with sides of length x centimeters and [#permalink]
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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 centimeters and [#permalink]
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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 centimeters and [#permalink]
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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 centimeters and [#permalink]
kiranaithal78 wrote:
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

Though this is an old post i am not sure how x^2+y^2=z^2=169 is written as (x+y)(x-y)=169 ; as it is not same as x^2-y^2.
Members plz correct me if i am missing something
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Re: Figure ABCD is a rectangle with sides of length x centimeters and [#permalink]
cruiseav wrote:
kiranaithal78 wrote:
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

Though this is an old post i am not sure how x^2+y^2=z^2=169 is written as (x+y)(x-y)=169 ; as it is not same as x^2-y^2.
Members plz correct me if i am missing something

You are right x^2 + y^2 cannot be written as (x + y)(x - y).

(x + y)(x - y) = x^2 - y^2.
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Re: Figure ABCD is a rectangle with sides of length x centimeters and [#permalink]
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Re: Figure ABCD is a rectangle with sides of length x centimeters and [#permalink]
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