4. quadrilateral ropq is inscribed in a circle. show all work on no credit. find x and y. what is the measure of each angle of quadrilateral efgh? what is the measure of arc hf?

Step-by-step explanation:

a) The sum of opposite angles of an inscribed quadrilateral is 180°. This lets us use angles E and G to solve for x:

  (x+15) + (2x) = 180

  3x + 15 = 180 . . .simplify

  x +5 = 60 . . . . . divide by 3

  x = 55 . . . . . . . . subtract 5

Similarly, we can use angles F and H to solve for y:

  (3y -60) + (y) = 180

  4y -60 = 180 . . . . simplify

  y -15 = 45 . . . . . . divide by 4

  y = 60 . . . . . . . . . add 15

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b) Then the measures of the angles are ...

  G = 2x = 2·55 = 110

  E = 180 -G = 70

  H = y = 60

  F = 180 -H = 120

The angle measures are ...

  m∠E = 70°, m∠F = 120°, m∠G = 110°, m∠H = 60°

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c) short arc HF is intercepted by inscribed angle E, so the arc will have twice the measure of the angle.

  arc HF = 2·m∠E = 140°

Comment on the problem

Throughout, the only relation being used is that the measure of an arc is twice the measure of the inscribed angle intercepting it. For opposite angles of the quadrilateral, the sum of the two intercepted arcs is 360° (the whole circle), so the sum of the two angles is 180°.

d) 48

step-by-step explanation:

16,32,48,64,80,96,112,128,144,160,176,192,208,224,240,256,272,288,304,320,336