Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

This implies that \(l*b = 60\) where l and b are length and breadth respectively. This could mean a million possibilities. So insufficient and we are down to BCE.

Statement 2: Length of diagonal = 13.

This implies that \(\sqrt{l^2 + b^2} = 13\)

So at this point, I am a little confused about one tiny detail.

Scenario 1: We can assume that all sides of the rectangle are integers, in which case Statement alone is sufficient since the only two integers whose sum of squares is 13 are 12 and 5. So we are done.

Scenario 2: We cannot assume that all sides are integers, in which case statement two is insufficient and we use statement 1 with statement 2 to figure out the answer.

b = \(\frac{60}{l}\)

Substituting that into statement 2, we get:

\(\sqrt{l^2 + (60/l)^2} = 13\)

This can be solved for a value of l and hence we get C as our answer. What's the OA?

Last edited by whiplash2411 on 26 Jun 2010, 20:54, edited 1 time in total.

(2) The length of a diagonal of rectangle R is 13.

I wanna discuss this one in detail.

Let the sides be \(x\) and \(y\). Question: \(P=2(x+y)=?\)

(1) \(area=xy=60\). Clearly insufficient. (2) \(x^2+y^2=13^2=169\). Also insufficient.

(1)+(2) Square P --> \(P^2=4(x+y)^2=4(x^2+2xy+y^2)\) as from (1) \(xy=60\) and from (2) \(x^2+y^2=169\), then \(P^2=4(x^2+2xy+y^2)=4(169+120)=4*289\) --> \(P=\sqrt{4*289}=2*17=34\). Sufficient.

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

What about \(13-\frac{13}{\sqrt{2}}-\frac{13}{\sqrt{2}}\) ?
_________________

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

Posted from my mobile device

5-12-13 is a Pythagorean triple. What is Pythagorean triple?

Triangles that have their sides in the ratio of whole numbers (integers) are called Pythagorean Triples. The smallest one is 3:4:5. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.

A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). Such a triple is commonly written \((a, b, c)\), and a well-known example is \((3, 4, 5)\). If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\). There are 16 primitive Pythagorean triples with c ≤ 100: (3, 4, 5)(5, 12, 13)(7, 24, 25)(8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97).

BUT:

Knowing that hypotenuse equals to 13 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 5:12:13. Or in other words: if \(a^2+b^2=13^2\) DOES NOT mean that \(a=5\) and \(b=12\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=13^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=5\) and \(b=12\).

For example: \(a=1\) and \(b=\sqrt{168}\) or \(a=2\) and \(b=\sqrt{165}\) or \(a=4\) and \(b=\sqrt{153}\) or \(a=2.5\) and \(b=\sqrt{162.75}\)...

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.

No this is wrong. a^2+b^2 = 13^2 holds true for many values of (a,b)

You cannot deduce a and b to be 12,5

what if a^2 = 6 and b^=7, it was not even given that they have integral value. Even if it was given a and b are integers you still have to think whether any integral pair a,b exists which is unique.

no doubt 2+2 = 4, but you can not deduce 4 = 2+2 only...it can be 3+1....This reasoning will help you in CR - cause and effect.
_________________

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

Posted from my mobile device

ans c I am no Math Genius ..but if you deduce this a^2 +b^2=C^2..then it doesn't mean that triplet a,b, c are pythgorean triplet.

But if a right angle triangle is given ..then yes a,b,c satisfy pythagoras theorem and if c is given then you can find a and b.

OA is "C", but why??? It should be "B". We know the 13-12-5 principle. If hypotenuse is 13, then the other two sides can be determined. Why not "B"?????

Posted from my mobile device

Dude,

Not Every Red car is Ferrari ...

Just kidding...

So every triplet is not 13-12-5
_________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

Re: What is the perimeter of rectangle R? [#permalink]

Show Tags

27 Sep 2015, 08:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...