Arc Length and Sector Area Worksheet Answer Key – Geometry

Problem 1 Arc Length in Degrees

s = 40360 · 2π(9) = 19 · 18π = 2π.

2π ≈ 6.28.

2π cm ≈ 6.28 cm

Problem 2 Sector Area in Degrees

Area = 75360 · π(12)² = 524 · 144π = 30π.

30π ≈ 94.25.

30π in² ≈ 94.25 in²

Problem 3 Arc Length of a Wide Angle

s = 132360 · 2π(15) = 1130 · 30π = 11π ≈ 34.56.

≈ 34.6 ft

Problem 4 Central Angle from an Arc

Circumference = 48π. Fraction = 10π48π = 524.

Angle = 524 · 360° = 75°.

75°

Problem 5 Degrees to Radians

Multiply by π180.

(a) 135 · π180 = 3π4. (b) 210 · π180 = 7π6. (c) 80 · π180 = 4π9.

(a) 3π/4 (b) 7π/6 (c) 4π/9

Problem 6 Radians to Degrees

Multiply by 180π.

(a) 5π6 · 180π = 150°. (b) 3π10 · 180π = 54°. (c) 2 · 180π ≈ 114.59°.

(a) 150° (b) 54° (c) ≈ 114.6°

Problem 7 Radian Measure from Arc and Radius

θ = sr = 148 = 1.75 radians.

In degrees: 1.75 · 180π ≈ 100.27°.

1.75 rad ≈ 100.3°

Problem 8 Arc Length with Radians

s = rθ = 6.5 · 2.4 = 15.6.

15.6 m

Problem 9 Sector Area with Radians

Area = ½ r²θ = ½ · 100 · 1.2 = 60.

60 ft²

Problem 10 Radius from a Sector Area

120360 · πr² = 54π, so 13r² = 54 and r² = 162.

r = √162 = 9√2 ≈ 12.73.

r = 9√2 ≈ 12.7 in

Problem 11 Radius from an Arc Length

84360 = 730. So 730 · 2πr = 7π, giving 14πr30 = 7π.

r = 7 · 30 ÷ 14 = 15.

r = 15 cm

Problem 12 Perimeter of a Sector

Arc = 45360 · 2π(8) = 18 · 16π = 2π ≈ 6.28.

Perimeter = 8 + 8 + 2π = 16 + 2π ≈ 22.28.

≈ 22.28 m

Problem 13 Similar Sectors

Arc lengths: 16 · 2π(5) = 5π3 and 16 · 2π(15) = 5π. Ratio 5π ÷ (5π/3) = 3.

Areas: 16 · 25π = 25π6 and 16 · 225π = 75π2. Ratio = 9.

The radius ratio is 3; arcs scale by 3 and areas by 3² = 9.

Arcs 3 : 1; areas 9 : 1

Problem 14 Area of a Circular Segment

Sector area = 60360 · 100π ≈ 52.36.

Triangle area = ½ · 10 · 10 · sin 60° ≈ 43.3.

Segment = 52.36 − 43.3 = 9.06.

≈ 9.06 cm²

Problem 15 Pizza Slice

Radius = 9 in. Each slice has central angle 360° ÷ 8 = 45°.

Area = 18 · 81π ≈ 31.81. Crust = 18 · 18π = 2.25π ≈ 7.07.

Area ≈ 31.81 in²; crust ≈ 7.07 in

Problem 16 Windshield Wiper

(a) Tip path = 110360 · 2π(24) = 1136 · 48π = 44π3 ≈ 46.08 in.

(b) The blade clears the sector of radius 24 minus the sector of radius 6: 1136 · π(24² − 6²) = 1136 · 540π = 165π ≈ 518.36 in².

(a) ≈ 46.1 in (b) ≈ 518.4 in²

Problem 17 Lawn Sprinkler

Area = 220360 · π(14²) = 1118 · 196π = 1,078π9 ≈ 376.29.

Curved edge = 1118 · 28π = 154π9 ≈ 53.76.

Area ≈ 376.3 ft²; edge ≈ 53.8 ft

Problem 18 Bicycle Wheel

(a) s = rθ = 13 · 2.5 = 32.5 in.

(b) θ = s ÷ r = 10013 ≈ 7.69 radians, about 1.22 turns.

(a) 32.5 in (b) ≈ 7.69 rad

Problem 19 Minute Hand

35 minutes is 3560 of a turn: 210°.

Tip distance = 210360 · 2π(9) = 712 · 18π = 10.5π ≈ 32.99.

Area = 712 · 81π = 47.25π ≈ 148.44.

210°; ≈ 33 in; ≈ 148.4 in²

Problem 20 Pie Chart Slice

Angle = 0.35 · 360° = 126°.

Area = 0.35 · π(25) = 8.75π ≈ 27.49. Curved edge = 0.35 · 10π = 3.5π ≈ 11.00.

126°; ≈ 27.49 cm²; ≈ 11.00 cm