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

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dave13 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)

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 ? dave13 :
niks18 , apparently able
to get a word in edgewise addressed what to do
with a "2" that is now "1."

I'm hoping a third ratio is also clear, see below.
That ratio also must be divided by 2.

sadikabid27 , maybe this discussion or this post will help.
See below for a link to 30-60-90 triangle overview.

dave13 , yes, we create two identical 30-60-90 triangles at vertex U.

So let's take niks18 's answer and line up the ratios.

Normal = $$30: 60 : 90$$
Normal = $$1 : \sqrt{3} : 2$$
Normal = $$x : x\sqrt{3} : 2x$$

But we can't call long sides PU and UT "2x" or "2."
Both are hexagon sides.
Hexagon side length is not $$2x$$ or $$2$$
It is $$x$$. Or, if you're thinking in numeric multiples, $$1$$

2x or 2 got divided by 2.
What you do to one part of a ratio,
you must do to all parts.

Divide all ratio parts by 2
$$30 : 60 : 90$$
Adjusted = $$\frac{1}{2} : \frac{\sqrt{3}}{2} : 1$$
Adjusted = $$\frac{x}{2} : \frac{x\sqrt{3}}{2} : x$$

Length of ONE side of the big triangle =
$$\frac{x\sqrt{3}}{2} + \frac{x\sqrt{3}}{2} = x\sqrt{3}$$

Hope that helps. P.S. For a quick review of 30-60-90 triangles as well as an interesting proof see HERE.
dave13 - last thing . . . one part of the answer you quoted may have confused you.
I wrote Regular hexagon, 120 degrees at each vertex, equilateral triangle
Highlight refers to the big triangle in the center, not the smaller right triangles we create.

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

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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)

Attachment:
The attachment 1494909854_591a839e50d63.png is no longer available

A hexagon is made up of 6 equilateral triangles (see drawing attached).
Each side of triangle PRT corresponds to 2* (The height of each equilateral triangle), so the perimeter will be 6*(height of each equilateral triangle) = (6$$x\sqrt{3}$$)/2 or D
Attachments hex.png [ 29.91 KiB | Viewed 453 times ]

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

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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)

Attachment:
1494909854_591a839e50d63.png

An Engineer's method~

sinA/A=sinB/B=sinC/C

let the side of the isosc. triangle PQR be "y",
So, sin30/x = sin120/y
or, y = (3)^0.5 * x [since, sin30=1/2 & sin120=(3)^0.5/2]

Hence, perimeter = 3*y = 3*x*(3)^0.5 [since, triangle PRT is equilat.]
OA=D
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Re: In the figure shown,PQRSTU is a regular polygon with sides of length x  [#permalink]

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The 30-60-90 triangle can be determined by 180(n-2)/n = 120, then splitting this isoceles triangle in 1/2 to find the base.
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Re: In the figure shown,PQRSTU is a regular polygon with sides of length x  [#permalink]

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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)

Attachment:
1494909854_591a839e50d63.png

Side of the triangle = $$x\sqrt{3}$$
Perimeter of the triangle =$$3 x \sqrt{3}$$

IMO D

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E-mail : kinshook.chaturvedi@gmail.com Re: In the figure shown,PQRSTU is a regular polygon with sides of length x   [#permalink] 12 Sep 2019, 18:40

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