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

Guest Jun 14, 2020

#1**0 **

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.

Guest Jun 14, 2020

#3**0 **

1) EF || GH ---> arc(EH) = arc(FG) ---> x^{2} - 2x = 56 - 3x

x^{2} + 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 P_{1}P_{3}P_{5}P_{7}P_{9}P_{11} (with the center of the circle = O)

Since it is a regular hexagon, angle(P_{1}OP_{3}) = 60^{o}

OP_{1} = OP_{3} = P_{1}P_{3} = 1

-- Draw the diameter of the circle that goes through P_{2}.

This bisects angle(P_{1}OP_{3}) and also bisects P_{1}P_{3} -- call this point X.

-- P_{1}X · XP_{3} = P_{2}X · XP_{8} (P_{8} is the other end of the diameter from P_{2})

P_{1}X = XP_{3} = 1/2

Let P_{2}X = x ---> XP_{8} = 2 - x

P_{1}X · XP_{3} = P_{2}X · XP_{8} ---> (1/2)(1/2) = (x)(2 - x)

1/4 = 2x - x^{2}

x^{2} - 2x - 1/4 = 0

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

Now, look at triangle P_{1}XP_{2} -- it is a right triangle with P_{1}P_{2} the hypotenuse

Using the Pythagorean Theorem: (P_{1}P_{2})^{2} = (1/2)^{2} + ( 2 - sqrt(3) )^{2}

---> P_{1}P_{2} = 2 - sqrt(3)

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

geno3141 Jun 16, 2020

#4**0 **

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