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

Question

https://gmatclub.com/forum/download/file.php?id=22245 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 Untitled.png

Bunuel · Accepted Answer

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.

Mattd · Answer

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.

Mattd · Answer

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

b2bt · Answer

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?

Bunuel · Answer

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.

kiranaithal78 · Answer

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

mvictor · Answer

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)

cruiseav · Answer

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

Bunuel · Answer

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

(x + y)(x - y) = x^2 - y^2.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards