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Can someone solve this?

Can someone solve this?-example-1
User Dan Grahn
by
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1 Answer

7 votes

Answer:

  1. x=7, y=4; arc JKL = 262°
  2. BL = 2
  3. ∠2 = 70°, ∠3 = 80°, ∠4 = 100°
  4. x = 24
  5. 12.22 cm

Explanation:

1A. Opposite angles of an inscribed quadrilateral are supplementary.

∠K = 180° -∠M

7x = 180 -131 = 49

x = 7 . . . . . divide by 7

∠L = 180 -∠J

11x +y = 180 -99 . . . fill in values

11(7) +y = 81 . . . . . . fill in x

y = 4 . . . . . subtract 77

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1B. The arc measure is twice the inscribed angle measure:

arc JKL = 2∠M = 2(131°)

arc JKL = 262°

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2. Tangents from the same point are congruent.

AM = AL = 3

CN = CM = 10 -AM = 7

BL = BN = 9 -CN

BL = 2

__

3. Linear angles are supplementary.

∠2 = 180° -∠1 = 180° -110°

∠2 = 70°

Angles where chords cross are the average of the subtended arcs.

∠3 = (100° +60°)/2

∠3 = 80°

∠4 = 180° -∠3 = 100°

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4. The product of the segments from the crossing point to the circle intersection points is a constant.

12x = 8(x+12)

4x = 96 . . . . subtract 8x

x = 24

__

5. Arc length is given by the formula ...

s = rθ . . . . θ in radians

s = (7 cm)(100°(π/180°)) = (7 cm)(5π/9) ≈ 12.22 cm

User PSGuy
by
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