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
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.
+23246 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.
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
