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:

Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

13 Jan 2012, 12:06

Good one:

TRIANGLE RULES: 1. Third side < Sum of other two sides 2. Third side > Difference of other two sides

1. Starting with BC=11, AC=1, we get AB=9. Rule 1 is violated because BC > AB + AC. Keep on going BC = 12, 13... it gets worst. Therefore, the perimeter of the triangle has to be greater than 20. Suff. 2. This one I had to semi-guess. I picked 3-4-5 triangle and found that area < perimeter. So, perimeter has to be > 20. Suff.

D.
_________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

07 May 2012, 21:56

While solving statement-2 in this problem, I realized that – For a right triangle with a given area, perimeter will be minimum when that right triangle is an "isosceles" right triangle.

I feel this property makes sense. Still, can someone confirm it please?

Using this property the problem can be solved as follows: Assume that the given right triangle with area 40, is an isosceles right triangle and then find the perimeter. If the perimeter of isosceles right triangle is greater than 20, that will mean any perimeter for that right triangle is definitely greater than 20.

Lets call base of right triangle = b Lets call height of right triangle = h Area given = 20. ==> (1/2)b.h = 20 ==> b.h = 40

If we assume this right triangle to be isosceles, then b=h= sq.root(40) => hypotenuse = sq.root(80) => perimeter = 2. sq.root(40) + sq.root(80) => This is greater than 20. => So minimum perimeter is more than 20. So any perimeter for this right triangle with the area as 20, will be more than 20. => Statement#2 is sufficient.

Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

09 Jul 2013, 09:55

(1). The Third side is greater than the difference betn the other two sides. Therefore atleast two sides are greater than 10. Therefore perimeter is >20. Sufficient.

(2) 1/2 bh =20 , if the base corresponding to this height is 1 (a smaller value gives an even greater height) , height is 40 , thus other side is greater > 40 (since other side is the hypotenuse in the right triangle formed by this height and base). Thus, perimeter > 20. Sufficient.

Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

10 Dec 2013, 06:26

Is the perimeter of triangle ABC greater than 20?

(1) BC-AC=10. This can be rewritten as BC=10+AC. The sum of any two sides of a triangle must be greater than the third. So, for example, BC + AC must be greater than AB. Because BC is at least 10, AB must be greater than 10 as well. SUFFICIENT.

(2) The area of the triangle is 20. This I solved with a bit of intuition and drawing out but it's best to recognize that for triangles of the same area, perimeter is minimized when the triangle is equilateral. Knowing the area of an equilateral triangle will allow us to figure out a side length and thus, the perimeter: 20 = s^2 (Sqrt3/4) s^2 = 80/(Sqrt 3) s = 6.8 6.8+ 6.8+ 6.8 = 20.4 > 20

So when we solve for a triangle with the minimal possible value for it's given area it's greater than 20. SUFFICIENT.

Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

20 Jun 2014, 03:15

An easier way to understand the calculation of the second statement is Following. Since for a given area an equilateral triangle has the smallest perimeter, Hence for an area of 20 , each side of the equilateral comes out to be 6.8. Hence the minimum perimeter for a triangle with area 20 is 3a=20.4; hence the perimeter has to be greater than 20. Hope it helps

Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

20 Sep 2015, 07:43

sriharimurthy wrote:

Question Stem : Is AB + BC + AC > 20?

St. (1) : BC = AC + 10

Triangle Property : The sum of any two sides of a triangle is always greater than the third.

Since we are given that one of the sides is greater than 10, the sum of the other two sides must also be greater than 10. Hence the perimeter will always be greater than 20. Statement is sufficient.

St. (2) : A = 40

Triangle Property : For triangles with same area, the perimeter is smallest for an equilateral triangle.

Area of equilateral triangle with side x = \(\frac{\sqrt{3}}{4}x^2\)

Therefore, \(\frac{\sqrt{3}}{4}x^2\) = 40

\(x^2 = \frac{160}{\sqrt{3}}\)

Now, in order to speed up calculations, I will assume \(\sqrt{3}\) to be equal to 2.

If the condition is satisfied with \(\sqrt{3}\) equal to 2 then it will definitely be satisfied with the actual value of \(\sqrt{3}\) which is less than 2.

Therefore, \(x^2 = \frac{160}{2}\) = 80

This tells us that x is almost 9. More importantly, it tells us that x is greater than 8. Thus perimeter will be 3*x = 24.

Since this is the minimum perimeter possible (actually it is still less than what the actual minimum would be due to our approximations), we can conclude that the question stem will always be true.

Hence Sufficient.

Answer : D

Another interesting triangle property :For triangles with same perimeter, the area is maximum for an equilateral triangle. (If you think about it, this property goes hand in hand with the one we used in St. 2).

Dear Bunnel, Thanks for your explanation. I think I almost understood the logic but cannot figure out clearly how the triangle property you mentioned "For triangles with same perimeter, the area is maximum for an equilateral triangle" is related to the Statement 2. Can you please explain how I can link the property to the statement 2?

Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

20 Sep 2015, 08:02

andy2whang wrote:

sriharimurthy wrote:

Question Stem : Is AB + BC + AC > 20?

St. (1) : BC = AC + 10

Triangle Property : The sum of any two sides of a triangle is always greater than the third.

Since we are given that one of the sides is greater than 10, the sum of the other two sides must also be greater than 10. Hence the perimeter will always be greater than 20. Statement is sufficient.

St. (2) : A = 40

Triangle Property : For triangles with same area, the perimeter is smallest for an equilateral triangle.

Area of equilateral triangle with side x = \(\frac{\sqrt{3}}{4}x^2\)

Therefore, \(\frac{\sqrt{3}}{4}x^2\) = 40

\(x^2 = \frac{160}{\sqrt{3}}\)

Now, in order to speed up calculations, I will assume \(\sqrt{3}\) to be equal to 2.

If the condition is satisfied with \(\sqrt{3}\) equal to 2 then it will definitely be satisfied with the actual value of \(\sqrt{3}\) which is less than 2.

Therefore, \(x^2 = \frac{160}{2}\) = 80

This tells us that x is almost 9. More importantly, it tells us that x is greater than 8. Thus perimeter will be 3*x = 24.

Since this is the minimum perimeter possible (actually it is still less than what the actual minimum would be due to our approximations), we can conclude that the question stem will always be true.

Hence Sufficient.

Answer : D

Another interesting triangle property :For triangles with same perimeter, the area is maximum for an equilateral triangle. (If you think about it, this property goes hand in hand with the one we used in St. 2).

Dear Bunnel, Thanks for your explanation. I think I almost understood the logic but cannot figure out clearly how the triangle property you mentioned "For triangles with same perimeter, the area is maximum for an equilateral triangle" is related to the Statement 2. Can you please explain how I can link the property to the statement 2?

Thanks in advanace regards Andy

You are not going to get a reply as the post you are quoting is from 2009.

Let me try to answer your question. You are given a particular area in statement 2. Now, based on this value of area is there a property of triangles that you can apply to see whether you do get triangles with perimter >20 and area =20?

As per the property mentioned, of all triangles with EQUAL areas, equilateral triangle will have the smallest perimeter. Thus, side of an equilateral triangle with area of 20

---> \(\frac{\sqrt{3}*a^2}{4} = 20\), where a = side of the equilateral triangle.

----> a = (approx.) 6.79 ---> Perimeter (minimum of the triangles possible with area = 20 ) = 3*a=20.3 > 20.

Thus, when the minimum perimeter is 20.3, then all the other possible traingles will have the perimeter > 20. Thus this statement is sufficient.

Thus if the perimeter of the equilateral triangle is 20, then each side of the triangle = 20/3.

Thus, the area of such an equilateral triangle = \(\frac{\sqrt{3}*a^2}{4} = 20\) = \(\frac{\sqrt{3}*(20/3)^2}{4} = =173/9 < 180/9 =20\). Thus we see that with perimeter 20 , the smallest area is <20.

Thus, if we are given an area of 20 , then the perimeter of the smallest triangle (=an equilateral triangle) MUST be >20.
_________________

Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

27 Sep 2015, 04:33

For triangles with same area, the perimeter is smallest for an equilateral triangle. please Bunuel clarify this property. I do not grasp this concept so far. I memorize it but I can not apply it my self in similar question and when I should. thank you in advance for your great help.

Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

27 Sep 2015, 04:44

hatemnag wrote:

For triangles with same area, the perimeter is smallest for an equilateral triangle. please Bunuel clarify this property. I do not grasp this concept so far. I memorize it but I can not apply it my self in similar question and when I should. thank you in advance for your great help.

The concept IS what you mentioned.

For all the triangles given to you with a fixed area, the equilateral triangle will have the smallest perimeter.

This is a very unique property of equilateral triangles.

Let's say in a PS question, you are given that area of a triangle is 20 square units, what is the smallest perimeter of this triangle ?

You will be able to use the property above to see that for a triangle with a given area to have the smallest perimeter, it must be an equilateral triangle. Based on this you can use the formula

\(\frac{\sqrt{3}*a^2}{4} = 20\) ----> calculate 'a' and then for perimeter= 3a

This is how you will be able to apply this property.

Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink]

Show Tags

30 Sep 2015, 08:52

i got answer using another approach which may not be correct always. statement 2 :

1/2 * b * h = 20 b*h =40 possible combinations 10*4,20*2, 8*5 etc. in these combinations let's consider base as 10,20,8. and we know the angle property that sum of 2 sides is greater than the 3rd side, so in the above combinations sum of other 2 sides must be greater than 10 and 20. in 3rd case since height is 5, using common sense we can conclude that sum of other two sides must be greater than 12.

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...