Junior high entrance exam / Plane Figures (Area and Angles) 3 / 4

Circles, Sectors, and Composite Figures

Learn the formulas for the area and arc length of circles and sectors, as well as how to solve problems involving composite figures.

Formulas for Sectors

Arc length = 2π r × (central angle/360)
Area = π r² × (central angle/360)

Example 1
Find the area and arc length of a sector with a radius of 6 cm and a central angle of 90°. (π=3.14)

Solution:
Area: 3.14×6×6×(90÷360)=3.14×36×0.25=28.26cm²
Arc length: 2×3.14×6×(90÷360)=2×3.14×6×0.25=9.42cm

Composite Figures

For complex figures, think in terms of “adding” or “subtracting.”

Example 2 (Semicircle and Triangle)
A circle (with a diameter of 10 cm) is inscribed in a square with sides of 10 cm. Find the area of the part left after subtracting the circle from the square. (π=3.14)

Solution:
Area of the square: 10×10=100cm²
Area of the circle: 3.14×5×5=78.5cm²
Remaining area: 100-78.5=21.5cm²
Key Points
  • Sector: multiply the area or circumference of the whole circle by “central angle/360”
  • Composite figures have three patterns: “subtract,” “add,” and “rearrange”
  • Area-preserving transformations: use transformations that change the shape without changing the area

Practice Problems

  1. Find the area of a sector with a radius of 4 cm and a central angle of 120°. (π=3.14)
  2. Find the area of a figure formed by cutting out quarter-circles with a radius of 1 cm from the four corners of a rectangle measuring 6 cm vertically and 8 cm horizontally. (π=3.14)

Answers and Explanations

  1. Answer: Approximately 16.75cm²
    Explanation: Area of a sector=πr²×(central angle÷360)=3.14×4×4×(120÷360)=3.14×16×(1/3)=50.24÷3≈16.75cm²(Exactly, 3.14×16/3≒16.75cm²).
  2. Answer: 44.86cm²
    Explanation: Area of the rectangle: 6×8=48cm². The total area of the four quarter-circles=one circle (radius 1 cm): 3.14×1×1=3.14cm². Remaining area: 48-3.14=44.86cm².

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