Answer:
- x=7, y=4; arc JKL = 262°
- BL = 2
- ∠2 = 70°, ∠3 = 80°, ∠4 = 100°
- x = 24
- 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
__
1B. The arc measure is twice the inscribed angle measure:
arc JKL = 2∠M = 2(131°)
arc JKL = 262°
__
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°
__
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