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AbdurRakib
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Updated on: 22 Sep 2019, 15:46
Originally posted by AbdurRakib on 23 Jun 2016, 12:47.
Last edited by MikeScarn on 22 Sep 2019, 15:46, edited 1 time in total.
Last edited by MikeScarn on 22 Sep 2019, 15:46, edited 1 time in total.
Fixed Typo
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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In the figure shown,PQRSTU is a regular polygon with sides of length 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 [ 5.75 KiB | Viewed 80146 times ]
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23 Jun 2016, 20:19
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AbdurRakib
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
09 Oct 2017, 14:09
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AbdurRakib
Quote:
Attachment:
hhhhhhh.png [ 10.32 KiB | Viewed 78210 times ]
• Dividing the equilateral triangle in the center? Doable. Not easy.
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.
Find one side length of the hexagon to find perimeter
• Start by dropping an altitude from one vertex to the side of the internal equilateral triangle
What we know:
-- it is a regular hexagon
-- 120° at each vertex
Because this shape is a regular hexagon, each vertex is 120 degrees.
Total interior degrees of a polygon = 180(n-2) where n = number of sides, here, 6
Total interior degrees of this hexagon: (4 * 180) = 720°
Degree measure per vertex angle: 720 degrees/6 angles = 120 degrees per angle
-- The triangle PQR is equilateral
Because the shape is a regular hexagon, the distance between two alternating vertices is equal.
Thus PR = RT = TP. The triangle is equilateral.
• Next step: Notice 120° at each vertex. We are always looking for 30-60-90 or 45-45-90 triangles
∆ PTU is isosceles: PU = UT
Drop an altitude from vertex U.
That altitude is a perpendicular bisector
-- Now there are two 60° angles at vertex U.
-- The altitude is perpendicular. It creates two 90° angles.
-- ∆ PTU 's other two angles each must = 30°
120° + small ∠ + small ∠ = 180°
Two small angles = (180° - 120° = 60° total
Each small angle = 30°
• From the two 30-60-90 triangles we can find the length of the triangle's side.
There are two 30-60-90 right triangles, but their sides have been scaled down
-- a 30-60-90 triangle has sides in ratio \(x: x\sqrt{3}: 2x\)
-- the ratio of sides holds, but each part of the ratio must be divided by 2
Why divide the ratio by 2?
Because the sides of ∆ PTU are defined by the hexagon's side length,
and the hexagon's side length, we are told, \(x\)
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.
The side opposite the 90° is usually \(2x\)
If \(2x\) has been scaled down to \(x\), we must divide all else by 2.
[Your way: 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 we cannot label the side length opposite the 90 degree angle "\(2x\)"
no matter where we draw the right triangles
It is not \(2x\). Defined by the hexagon, it is \(x\)
• Divide all three parts of the ratio
-- \(\frac{2x}{2} = x\) - 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
(normally the side opposite the 30° angle is \(x\))
Finally, \(\frac{x\sqrt{3}}{2}\) is the side opposite the 60 degree angle
(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.
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