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# In the figure shown,PQRSTU is a regular polygon with sides of lenght x

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In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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23 Jun 2016, 11:47
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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)
[Reveal] Spoiler: OA

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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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23 Jun 2016, 13:49

The equilateral triangle inside a regular hexagon of side x will have it's side as $$x*\sqrt{3}$$. Therefore perimeter is 3x*$$\sqrt{3}$$
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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23 Jun 2016, 19:19
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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)

Hi,

two ways this Q can be done..

(I) Logic....
It can be clearly seen the Triangle is an equilateral triangle ........... Each side joins edges of two adjacent sides of a regular hexagon...
so our ANS should be a multiple of 3........ ONLY D and C are left...
Now D gives side as $$x\sqrt{3}$$ and C as $$x\sqrt{3}/2 \approx{x*1.7/2}$$.....
Now the side of equilateral triangle cannot be less than the side of hexagon.... so C is also out
ans D...

(II) Method...

hexagon has each angle as 120.....
Total sum of angles = (n-2)*180 = 4*180...
each angle = $$\frac{4*180}{6} = 120$$..

take any triangle formed by two sides x and third side as the side of internal triangle..
so it becomes ISOSCELES triangle with equal sides as x and central angle = 120.... so other two angles are (180-120)/2 = 30..
Draw a perpendicular on third side from central angle, it will meet the THIRD side in center and will form a 30-60-90 triangle....
here opposite to $$90^{\circ}$$= x, so opposite $$60^{\circ}$$ = $$x*\sqrt{3}/2$$, which is nothing but HALF of third side..
so third side = $$2*x*\sqrt{3}/2 = x\sqrt{3}$$...
therefore PERIMETER = $$3*x\sqrt{3}$$
D
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In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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21 Aug 2016, 04:00
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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.
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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24 Aug 2016, 04:06
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.
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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24 Aug 2016, 04:15
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?
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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24 Aug 2016, 04:27
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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.
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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24 Aug 2016, 05:23
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.
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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25 Dec 2016, 06:23
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
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In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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11 Jan 2017, 10:19
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!
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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05 Apr 2017, 23:46

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
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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09 Oct 2017, 05:15
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...
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In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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09 Oct 2017, 13:09
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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 [ 10.32 KiB | Viewed 4664 times ]

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

$$3x\sqrt{3}$$

Hope it helps.
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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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12 Oct 2017, 16:58
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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.

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Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x [#permalink]

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18 Oct 2017, 20:18
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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

$$3x\sqrt{3}$$

Hope it helps.
Re: In the figure shown,PQRSTU is a regular polygon with sides of lenght x   [#permalink] 18 Oct 2017, 20:18
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