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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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.

Thanks for the help. Did not think about the integer part.

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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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

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)

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.

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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Bunuel


that should be 34 right?
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My solution might be slightly longer but was more intuitive to me:

x-y=7 –> x-7=y

y in x^2+y^2=z^2=169

x^2+(x-7)^2=169
= 2x^2-14x+120=0
=x^2-7x+60=0
=(x-5)(x+12)=0

–> y=12, x=5 –> 2(12+5)=34
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ABCD is a rectangle, so it is made of two right triangle. So, x^2 + y^2 = z^2 . We need to know the perimeter of ABCD, 2(x+y).
1) x - y = 7, but we need the value of x +y. Not suff.
2) z = 13 or, z^2 = 169, x^2 + y^2 =169, (x+y)^2 -2xy =169. We cant find (x+y) from it. Not suff.
Together, 169 = x^2 + (x- 7)^2 or, 2x^2 -14x+49 =169 or, 2x^2 -14x -120 =0, or, x^2 -7x -60 =0, x^2 -12x +5x -60 =0. So x is 12 and y is 5. Sufficient.
C is the answer.

Bunuel if we were told the sides have integral values then B would've been sufficient?
does that mean whenever we are given for eg 5 as a hypotenuse side and told the other sides have integral values, the other sides would be 3 and 4?
or given 25 as a hypotenuse value, then the other sides (if having integral values) would def be 15 and 20?

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