Last visit was: 27 Apr 2024, 07:54 It is currently 27 Apr 2024, 07:54

Close
GMAT Club Daily Prep
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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Difficulty: 555-605 Levelx   Geometryx                        
Show Tags
Hide Tags
Senior Manager
Senior Manager
Joined: 11 May 2014
Status:I don't stop when I'm Tired,I stop when I'm done
Posts: 474
Own Kudos [?]: 38845 [293]
Given Kudos: 220
Location: Bangladesh
Concentration: Finance, Leadership
GPA: 2.81
WE:Business Development (Real Estate)
Send PM
Most Helpful Reply
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11182
Own Kudos [?]: 31984 [93]
Given Kudos: 291
Send PM
Senior SC Moderator
Joined: 22 May 2016
Posts: 5330
Own Kudos [?]: 35499 [47]
Given Kudos: 9464
Send PM
General Discussion
User avatar
Manager
Manager
Joined: 23 Apr 2015
Posts: 234
Own Kudos [?]: 510 [13]
Given Kudos: 36
Location: United States
Concentration: General Management, International Business
WE:Engineering (Consulting)
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
6
Kudos
7
Bookmarks
Answer is D:

The equilateral triangle inside a regular hexagon of side x will have it's side as \(x*\sqrt{3}\). Therefore perimeter is 3x*\(\sqrt{3}\)
Intern
Intern
Joined: 20 Jul 2014
Posts: 17
Own Kudos [?]: 153 [4]
Given Kudos: 2
Location: India
GMAT 1: 710 Q46 V41
GPA: 3.16
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
4
Bookmarks
Hi, can someone please shed some light on why the triangle inside is equilateral? I understood why the 2 sides are equal since all angles and sides are equal, but what is the concept relating to the third side being equal as well?

Additionally, if it is an equilateral triangle, why do we assume the side to be \(\sqrt{3}x\)?

Thank you.
Math Expert
Joined: 02 Sep 2009
Posts: 92952
Own Kudos [?]: 619397 [2]
Given Kudos: 81611
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
2
Kudos
Expert Reply
Neeraj91 wrote:
Hi, can someone please shed some light on why the triangle inside is equilateral? I understood why the 2 sides are equal since all angles and sides are equal, but what is the concept relating to the third side being equal as well?

Additionally, if it is an equilateral triangle, why do we assume the side to be \(\sqrt{3}x\)?

Thank you.


The triangle is equilateral because the distances between any two alternate vertices of PQRSTU are the same - PR = RT = TP.
Intern
Intern
Joined: 20 Jul 2014
Posts: 17
Own Kudos [?]: 153 [0]
Given Kudos: 2
Location: India
GMAT 1: 710 Q46 V41
GPA: 3.16
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
Bunuel wrote:
Neeraj91 wrote:
Hi, can someone please shed some light on why the triangle inside is equilateral? I understood why the 2 sides are equal since all angles and sides are equal, but what is the concept relating to the third side being equal as well?

Additionally, if it is an equilateral triangle, why do we assume the side to be \(\sqrt{3}x\)?

Thank you.


The triangle is equilateral because the distances between any two alternate vertices of PQRSTU are the same - PR = RT = TP.


Hi Bunuel,

Thanks for the reply. So just to make sure I understood the concept, in any regular polygon of n sides, not only are all the sides equal, the distances between ANY two vertices are equal. Would I be correct in framing it so?
Math Expert
Joined: 02 Sep 2009
Posts: 92952
Own Kudos [?]: 619397 [4]
Given Kudos: 81611
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
1
Kudos
2
Bookmarks
Expert Reply
Neeraj91 wrote:
Bunuel wrote:
Neeraj91 wrote:
Hi, can someone please shed some light on why the triangle inside is equilateral? I understood why the 2 sides are equal since all angles and sides are equal, but what is the concept relating to the third side being equal as well?

Additionally, if it is an equilateral triangle, why do we assume the side to be \(\sqrt{3}x\)?

Thank you.


The triangle is equilateral because the distances between any two alternate vertices of PQRSTU are the same - PR = RT = TP.


Hi Bunuel,

Thanks for the reply. So just to make sure I understood the concept, in any regular polygon of n sides, not only are all the sides equal, the distances between ANY two vertices are equal. Would I be correct in framing it so?


No. I'm saying that in this particular case the distances between any two alternate vertices of PQRSTU are the same.

It should be very easy to see: triangles PQR, RST and TUP are congruent (their two sides and the angle between them are equal), thus their third sides also must be equal.
Intern
Intern
Joined: 20 Jul 2014
Posts: 17
Own Kudos [?]: 153 [0]
Given Kudos: 2
Location: India
GMAT 1: 710 Q46 V41
GPA: 3.16
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
Thank you. By congruency test, I understood why they have to be equal theoretically. Visually and logically, it seems obvious they are equal, but I was not sure how to make that conclusion without an underlying theory.
avatar
Intern
Intern
Joined: 06 Dec 2016
Posts: 1
Own Kudos [?]: 2 [0]
Given Kudos: 0
Location: India
Concentration: General Management, Technology
GMAT 1: 590 Q45 V26
GPA: 3.7
WE:Engineering (Military & Defense)
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
Consider Triangle PQR
(a) It is an isosceles triangle hence two smaller angles are equal and = (180-120)/2=30 degree.
(b) Third side of the triangle PQR = cosine projections of 'x' = 2*x*cos 30 = (2)X(x)*(Sqrt (3)/2)=(Sqrt 3)x
(c) Same logic holds for other 2 sides, and hence perimeter = 3*(Sqrt 3)x
Intern
Intern
Joined: 22 Sep 2016
Posts: 13
Own Kudos [?]: 12 [0]
Given Kudos: 12
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
But if we see PQR as an isosceles triangle than the sides are x:x:x(sqrt 2). So thats why I came to the solution 3 times x(sqrt 2) = 3x(sqrt2)
Could somebody please explain where I did a mistake?

EDIT: Sorry, I got it now.
The isosceles triangles needs a right triangle!
Manager
Manager
Joined: 31 Jan 2017
Posts: 55
Own Kudos [?]: 93 [3]
Given Kudos: 25
Location: India
Concentration: Strategy, Leadership
GMAT 1: 680 Q49 V34
GPA: 4
WE:Project Management (Energy and Utilities)
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
1
Kudos
2
Bookmarks
Answer : [D]

PQRSTU is a regular polygon with sides of length x. Its area = 6* (√3/4) *x^2 [since a regular polygon is made of 6 equilateral triangles of side x]

If you can visualize, triangle PRT has half the area of the polygon, [joining vertices P,R,T to the centre]

Area of triangle PRT with side "a" = (√3/4)*a ^2 = (1/2) * area of PQRSTU = (1/2) * 6* (√3/4) *x^2 = 3 * x^2

therefore a= √3 * x

perimeter of triangle PRT = 3 *a = 3 *√3 * x

Alternative Note: The equilateral triangle inside a regular hexagon of side x will have it's side as x ∗ √3
avatar
Intern
Intern
Joined: 30 Oct 2014
Posts: 14
Own Kudos [?]: 2 [0]
Given Kudos: 20
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
I am completely missing something on this one:

Once you split the Equilateral triangle into two 30-60-90 triangles, in terms of x the sides should be x-x(sqrt 3)-2x.

The hypotenuse would be: 2x which is one side of the equilateral. So perimeter for the equilateral would 2x+2x+2x.

Obviously I am wrong, but something just isn't sticking with this one for some reason...
Target Test Prep Representative
Joined: 04 Mar 2011
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Posts: 3043
Own Kudos [?]: 6281 [16]
Given Kudos: 1646
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
11
Kudos
5
Bookmarks
Expert Reply
AbdurRakib wrote:


In the figure shown,PQRSTU is a regular polygon with sides of lenght x.What is the perimeter of triangle PRT in terms of x?

A. (\(x\sqrt{3}\))/2
B. \(x\sqrt{3}\)
C. (3\(x\sqrt{3}\))/2
D. 3\(x\sqrt{3}\)
E. 4\(x\sqrt{3}\)


Since we have a regular polygon with 6 sides, each vertex has an angle of 180(6 - 2)/6 = 360/6 = 120 degrees.

If we drop a height from S to RT, which we can call SV, we see that we have created two 30-60-90 right triangles. In particular, SRV is one of them, with side SR as the hypotenuse, side SV opposite the 30-degree angle, and side RV opposite the 60-degree angle. Since SR = x, SV = x/2 (the shortest side is ½ of the hypotenuse in a 30-60-90 right triangle) and RV = x√3/2 (the side opposite the 60-degree angle is √3 times the shortest side).

We can also see that RT is twice RV; thus, RT = 2(x√3/2) = x√3. Since triangle PRT is an equilateral triangle and RT = x√3, the perimeter of triangle PRT is 3(x√3) = 3x√3.

Answer: D
avatar
Intern
Intern
Joined: 30 Oct 2014
Posts: 14
Own Kudos [?]: 2 [1]
Given Kudos: 20
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
1
Kudos
Boom. Got it. Thanks for that write up!

genxer123 wrote:
AbdurRakib wrote:

In the figure shown,PQRSTU is a regular polygon with sides of lenght x.What is the perimeter of triangle PRT in terms of x?

A. (\(x\sqrt{3}\))/2
B. \(x\sqrt{3}\)
C. (3\(x\sqrt{3}\))/2
D. 3\(x\sqrt{3}\)
E. 4\(x\sqrt{3}\)

OG Q 2017(Book Question: 145)

Quote:
Roosterbooster wrote: I am completely missing something on this one

Attachment:
hhhhhhh.png

Roosterbooster , I think it is easier to draw the right 30-60-90 triangles from the angle of the hexagon. See diagram.

If you divide the triangle into six (not two) congruent 30-60-90 triangles after reading what is below, it will work (but I think drawing from vertex U is easier)

You cannot solve for a side of the triangle very easily by dividing the triangle with one median. Try all three medians. Then your scale is easier to correlate with the side of the hexagon.

I think you also just forgot to divide the ratio by 2. Easy mistake. :-)

Regular hexagon, 120 degrees at each vertex, equilateral triangle:

Because this shape is a regular hexagon, each vertex is 120 degrees. (4 * 180 = 720, and 720 degrees/6 angles = 120 degrees per angle)

Also because the shape is a regular hexagon, the distance between two alternating vertices is equal. PR = RT = TP. The triangle is equilateral.

If you draw a perpendicular bisector from one vertex to the side of the triangle, as I have at vertex U, you have two 30-60-90 triangles, and you can find the length of the triangle's side.

Bisect a vertex, there are two 30-60-90 right triangles, but sides have been scaled down

The 30 - 60 - 90 triangle side ratio IS \(x: x\sqrt{3}: 2x\)

Each part of the ratio has been scaled down by half; the ratio between and among sides remains the same.

The triangle's sides are defined by the hexagon's side length of \(x\).

The side of the hexagon is the side opposite the triangle's right angle. Work backwards. Label that side \(x\).

\(x\) is still opposite the right angle, but it has been scaled down

If \(2x\) has been scaled down to \(x\), you must divide all else by 2.

You can divide the equilateral triangle into six congruent right 30-60-90 triangles, and
The side labeled \(\frac{x\sqrt{3}}{2}\) will still be opposite a 60-degree angle --

but you cannot label the side length opposite the 90 degree angle "\(2x\)" no matter where you draw the right triangles

It is not \(2x\). Defined by the hexagon, it is \(x\)

I compensated.

\(\frac{2x}{2} = x\) - that is the side opposite the right angle

If I divide one part of a ratio by 2, I must divide all terms by 2, so

\(\frac{x}{2}\) = the side opposite the 30 degree angle

Finally, \(\frac{x\sqrt{3}}{2}\) is the side opposite the 60 degree angle (if that side is based on \(x\), \(x\) is now \(\frac{x}{2}\), so \(\frac{x}{2}\) gets multiplied by \(\sqrt{3}\))

Find the side length of the equilateral triangle, then its perimeter

From the diagram: each side of the equilateral triangle has length

\(\frac{x\sqrt{3}}{2}\) + \(\frac{x\sqrt{3}}{2}\) = \(x\sqrt{3}\)

The perimeter? \(x\sqrt{3}\) + \(x\sqrt{3}\) + \(x\sqrt{3}\), OR

\(3x\sqrt{3}\)

Answer D

Hope it helps.
Manager
Manager
Joined: 10 Sep 2014
Posts: 61
Own Kudos [?]: 32 [2]
Given Kudos: 417
Location: Bangladesh
GPA: 3.5
WE:Project Management (Manufacturing)
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
2
Kudos
Bunuel could you please simplify the application of 30-60-90 triangle here? Really confused. :(
VP
VP
Joined: 09 Mar 2016
Posts: 1160
Own Kudos [?]: 1017 [0]
Given Kudos: 3851
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
generis wrote:
AbdurRakib wrote:

In the figure shown,PQRSTU is a regular polygon with sides of lenght x.What is the perimeter of triangle PRT in terms of x?

A. (\(x\sqrt{3}\))/2
B. \(x\sqrt{3}\)
C. (3\(x\sqrt{3}\))/2
D. 3\(x\sqrt{3}\)
E. 4\(x\sqrt{3}\)

OG Q 2017(Book Question: 145)

Quote:
Roosterbooster wrote: I am completely missing something on this one

Attachment:
hhhhhhh.png

Roosterbooster , I think it is easier to draw the right 30-60-90 triangles from the angle of the hexagon. See diagram.

If you divide the triangle into six (not two) congruent 30-60-90 triangles after reading what is below, it will work (but I think drawing from vertex U is easier)

You cannot solve for a side of the triangle very easily by dividing the triangle with one median. Try all three medians. Then your scale is easier to correlate with the side of the hexagon.

I think you also just forgot to divide the ratio by 2. Easy mistake. :-)

Regular hexagon, 120 degrees at each vertex, equilateral triangle:

Because this shape is a regular hexagon, each vertex is 120 degrees. (4 * 180 = 720, and 720 degrees/6 angles = 120 degrees per angle)

Also because the shape is a regular hexagon, the distance between two alternating vertices is equal. PR = RT = TP. The triangle is equilateral.

If you draw a perpendicular bisector from one vertex to the side of the triangle, as I have at vertex U, you have two 30-60-90 triangles, and you can find the length of the triangle's side.

Bisect a vertex, there are two 30-60-90 right triangles, but sides have been scaled down

The 30 - 60 - 90 triangle side ratio IS \(x: x\sqrt{3}: 2x\)

Each part of the ratio has been scaled down by half; the ratio between and among sides remains the same.

The triangle's sides are defined by the hexagon's side length of \(x\).

The side of the hexagon is the side opposite the triangle's right angle. Work backwards. Label that side \(x\).

\(x\) is still opposite the right angle, but it has been scaled down

If \(2x\) has been scaled down to \(x\), you must divide all else by 2.

You can divide the equilateral triangle into six congruent right 30-60-90 triangles, and
The side labeled \(\frac{x\sqrt{3}}{2}\) will still be opposite a 60-degree angle --

but you cannot label the side length opposite the 90 degree angle "\(2x\)" no matter where you draw the right triangles

It is not \(2x\). Defined by the hexagon, it is \(x\)

I compensated.

\(\frac{2x}{2} = x\) - that is the side opposite the right angle

If I divide one part of a ratio by 2, I must divide all terms by 2, so

\(\frac{x}{2}\) = the side opposite the 30 degree angle

Finally, \(\frac{x\sqrt{3}}{2}\) is the side opposite the 60 degree angle (if that side is based on \(x\), \(x\) is now \(\frac{x}{2}\), so \(\frac{x}{2}\) gets multiplied by \(\sqrt{3}\))

Find the side length of the equilateral triangle, then its perimeter

From the diagram: each side of the equilateral triangle has length

\(\frac{x\sqrt{3}}{2}\) + \(\frac{x\sqrt{3}}{2}\) = \(x\sqrt{3}\)

The perimeter? \(x\sqrt{3}\) + \(x\sqrt{3}\) + \(x\sqrt{3}\), OR

\(3x\sqrt{3}\)

Answer D

Hope it helps.


generis hello :) i have one awesome question :-) one thing i dont get why are you dividing by two ? by drawing a perpendiculat line we create a right triangle 30 60 90, we are not dividing right triangle, we create it :? :) no ? :)

Any idea niks18 ? :)
Retired Moderator
Joined: 25 Feb 2013
Posts: 895
Own Kudos [?]: 1528 [0]
Given Kudos: 54
Location: India
GPA: 3.82
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
dave13 wrote:

generis hello :) i have one awesome question :-) one thing i dont get why are you dividing by two ? by drawing a perpendiculat line we create a right triangle 30 60 90, we are not dividing right triangle, we create it :? :) no ? :)

Any idea niks18 ? :)


Hi dave13

I did not get what you mean by dividing by 2? What is being divided by 2 and in which context you are referring?

Secondly the explanations by generis are so detailed that there is hardly anything left for anybody to add to it.
VP
VP
Joined: 09 Mar 2016
Posts: 1160
Own Kudos [?]: 1017 [0]
Given Kudos: 3851
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
niks18 wrote:
dave13 wrote:

generis hello :) i have one awesome question :-) one thing i dont get why are you dividing by two ? by drawing a perpendiculat line we create a right triangle 30 60 90, we are not dividing right triangle, we create it :? :) no ? :)

Any idea niks18 ? :)


Hi dave13

I did not get what you mean by dividing by 2? What is being divided by 2 and in which context you are referring?

Secondly the explanations by generis are so detailed that there is hardly anything left for anybody to add to it.



Hi niks18, I agree Generis explanation is great the only thing i didnt get is why all three \(x: x\sqrt{3}: 2x\)
sides we divide by 2 ... if we divide an isolessces triangle into two right triangles than we need need to divide only one part x by 2 as for axample it is explained in this video ...https://www.khanacademy.org/math/geomet ... t-triangle

so yeah thats my confusion, anyways thanks for trying to understand me :) your patience is appreciated :)
Retired Moderator
Joined: 25 Feb 2013
Posts: 895
Own Kudos [?]: 1528 [1]
Given Kudos: 54
Location: India
GPA: 3.82
Send PM
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
1
Kudos
dave13 wrote:
niks18 wrote:
dave13 wrote:

generis hello :) i have one awesome question :-) one thing i dont get why are you dividing by two ? by drawing a perpendiculat line we create a right triangle 30 60 90, we are not dividing right triangle, we create it :? :) no ? :)

Any idea niks18 ? :)


Hi dave13

I did not get what you mean by dividing by 2? What is being divided by 2 and in which context you are referring?

Secondly the explanations by generis are so detailed that there is hardly anything left for anybody to add to it.



Hi niks18, I agree Generis explanation is great the only thing i didnt get is why all three \(x: x\sqrt{3}: 2x\)
sides we divide by 2 ... if we divide an isolessces triangle into two right triangles than we need need to divide only one part x by 2 as for axample it is explained in this video ...https://www.khanacademy.org/math/geomet ... t-triangle

so yeah thats my confusion, anyways thanks for trying to understand me :) your patience is appreciated :)


Hi dave13

I got your query now. it's a simple ratio operation. As you mentioned that the triangle is a 30-60-90 triangle, hence its sides will be in ratio \(1:\sqrt{3}:2\)

Now take a look at generis solution. Side opposite to 90 angel is UT and UT=x

so we have \(2=x\), hence \(1=\frac{x}{2}\) & \(\sqrt{3}=\frac{x}{2}*\sqrt{3}\). so you got the lengths of your sides now.
GMAT Club Bot
Re: In the figure shown,PQRSTU is a regular polygon with sides of length x [#permalink]
 1   2   
Moderators:
Math Expert
92952 posts
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne