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# two geometry questions help please

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So I looked for help with these questions before and found answers but some with an explantion that confused me or just wrong answers. I want to know how to do the problem so I can do them by meself later... here are my two questions.

1. Trapezoid HGFE is inscribed in a circle, with EF $$\parallel$$ GH. If arc GH is 70 degrees, arc EH is $$x^2 - 2x$$ degrees, and arc FG is 56 - 3x degrees, where x > 0, find arc EPF, in degrees.

I already found the answer 205 from a previous answer but that was inncorrect with no explantion.

2. A regular dodecagon $$P_1 P_2 P_3 \dotsb P_{12}$$ is inscribed in a circle with radius 1. Compute
$$(P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2.$$

(The sum includes all terms of the form $$(P_i P_j)^2,$$ where $$1 \le i < j \le 12.$$)

I have found the answer on this website to be 42 but there answer is wrong and I also do not yet know what $$sin$$ is in math.

If someone could help me that would be greatly appreciated.

~ Wolf

Jun 14, 2020

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1. Arc EPF works out to 210 degrees.

2. There are 12 digaonals that have a length of P_1 P_2, which from the Sine Law, is sin (15 degrees).  There are 12 diagonals that have a legnth of P_1 P_3, which from the Sine Law, is sin (30 degrees).  We can appy the same reasoning to the other diagonals, which gives us a total sum of

(12 sin 15)^2 + (12 sin 30)^2 + (12 sin 45)^2 + (12 sin 60)^2 + (12 sin 75)^2 + (12 sin 90)^2 + (12 sin 105)^2 + (12 sin 120)^2 + (12 sin 135)^2 + (12 sin 150)^2 + (12 sin 175)^2 = 864.

Since we have double-counted, the answer is 864/2 = 432.

Jun 14, 2020
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hi! so umm I had already found those answers before and those were incorrect... also how did you get 210 and is there a way to solve problem 2 without sin? sorry for not mentioning that sooner..

~ Wolf

Guest Jun 14, 2020
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1)  EF || GH   --->   arc(EH) = arc(FG)   --->   x2 - 2x  =  56 - 3x

x2 + x - 56  =  0

(x + 8)(x - 7)  =  0

x  =  7                      (can't be negative)

I don't know how to use this value for x to find arc(EPF) because I don't know where P is.

2)  Because this is a regular dodecagon, each side is congruent to every other side.

Since you haven't had any trig, try this:

Within the regular dodecagon,

--  Draw the regular hexagon  P1P3P5P7P9P11  (with the center of the circle = O)

Since it is a regular hexagon, angle(P1OP3)  =  60o

OP1  =  OP3 =  P1P3  =  1

--  Draw the diameter of the circle that goes through P2.

This bisects angle(P1OP3) and also bisects P1P3  --  call this point X.

--  P1X · XP3  =  P2X · XP8        (P8 is the other end of the diameter from P2)

P1X  =  XP3  =  1/2

Let P2X  =  x   --->   XP8  =  2 - x

P1X · XP3  =  P2X · XP8   --->   (1/2)(1/2)  =  (x)(2 - x)

1/4  =  2x - x2

x2 - 2x - 1/4  =  0

By the quadratic formula:  x  =  ( 2 - sqrt(3) ) / 2

Now, look at triangle P1XP2  --  it is a right triangle with P1P2 the hypotenuse

Using the Pythagorean Theorem:  (P1P2)2  =  (1/2)2 + ( 2 - sqrt(3) )2

--->   P1P2  =  2 - sqrt(3)

To find your answer, you'll need to square this and multiply by 12.

Jun 16, 2020
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Ok! so as I was following along with your second answer I squared and multiplied I got the answer $$84-43\sqrt{3}$$. Is this correct or did I do the math wrong? For question one I did not relise until later that I forgot to include an image... here is the image

Thank you so much for helping me!!!

~Wolf

Guest Jun 16, 2020