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

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

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New post 25 Mar 2018, 09:05
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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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New post 15 Jan 2019, 16:29
AbdurRakib wrote:
Image


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

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New post 14 Aug 2019, 22:37
AbdurRakib wrote:
Image


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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New post 12 Sep 2019, 18:03
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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New post 12 Sep 2019, 18:40
AbdurRakib wrote:
Image


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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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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