Using our knowledge of sectors, we will find the perimeters and areas of composite figures (figures combining circles, semicircles, sectors, and straight-line figures). Let us learn common patterns.
Semicircles and Quadrants
A semicircle with radius r (central angle 180°):
・Arc length = πr
・Area = πr²/2
A quadrant with radius r (central angle 90°):
・Arc length = πr/2
・Area = πr²/4
Area of a Cut-Out Figure
For problems in which a smaller figure is cut from a larger figure, use:
Area ＝ area of the whole − area of the cut-out part
This is a pattern solved by addition and subtraction.
Example ①
A figure is made by cutting four quadrants of radius 2 cm from the four corners of a square with side length 8 cm. Find the area of the remaining part. Leave π as it is.
Solution:
Area of square = 8×8 = 64 cm².
Total area of the four quadrants = π×2²×4×(90/360) = 4π × 1 = 4π cm².
Therefore, area = 64 − 4π (cm²)
(The four quadrants together are equivalent to one circle with radius 2 cm → π×4 = 4π.)
A figure is made by cutting four quadrants of radius 2 cm from the four corners of a square with side length 8 cm. Find the area of the remaining part. Leave π as it is.
Solution:
Area of square = 8×8 = 64 cm².
Total area of the four quadrants = π×2²×4×(90/360) = 4π × 1 = 4π cm².
Therefore, area = 64 − 4π (cm²)
(The four quadrants together are equivalent to one circle with radius 2 cm → π×4 = 4π.)
Do Not Forget the Perimeter
Even for a cut-out figure, the perimeter is “the total length of the outer lines.” Divide it into straight sections and curved sections (arcs), then add them together.
Example ②
Find the perimeter of a sector with radius 6 cm and central angle 90°. Leave π as it is.
Solution:
Perimeter ＝ two radii + arc.
Two radii = 6×2 = 12 cm.
Arc = 2π×6×(90/360) = 12π × 1/4 = 3π cm.
Therefore, 12 + 3π (cm)
Find the perimeter of a sector with radius 6 cm and central angle 90°. Leave π as it is.
Solution:
Perimeter ＝ two radii + arc.
Two radii = 6×2 = 12 cm.
Arc = 2π×6×(90/360) = 12π × 1/4 = 3π cm.
Therefore, 12 + 3π (cm)
Example ③
Find the perimeter of a semicircle with diameter 10 cm (leave π as it is).
Solution:
Perimeter of a semicircle ＝ diameter + arc (half the circumference).
Diameter = 10 cm, arc = π×10/2 = 5π cm.
Therefore, 10 + 5π (cm)
Find the perimeter of a semicircle with diameter 10 cm (leave π as it is).
Solution:
Perimeter of a semicircle ＝ diameter + arc (half the circumference).
Diameter = 10 cm, arc = π×10/2 = 5π cm.
Therefore, 10 + 5π (cm)
Key Points
- Perimeter of a sector = two radii + arc (do not forget the radii).
- Perimeter of a semicircle = diameter + arc
- For composite figures, use “whole − cut-out” or “sum of divided parts.”
- Four quadrants together make one circle (the total central angle is 360°).
Practice Problems
- Find the perimeter of a sector with radius 4 cm and central angle 60° (leave π as it is).
- Find the area of a semicircle with radius 5 cm (leave π as it is).
- From a square with side length 6 cm, two quadrants of radius 6 cm centered at opposite vertices are cut away, leaving a leaf-shaped figure. Find the area of the remaining part.
※ Strictly speaking, the two quadrants overlap, so it may become “square − two quadrants + overlapping part (leaf shape),” but here find the remainder obtained by subtracting the sum of the two quadrants from the square.
Answers and Explanations
- Answer: 8 + 4π/3 cm
Explanation: Two radii = 4×2 = 8. Arc = 2π×4×60/360 = 4π/3. Total 8 + 4π/3. - Answer: 25π/2 cm²
Explanation: Area of semicircle = πr²/2 = π×25/2 = 25π/2. - Answer: 36 − 18π cm² (remaining area; however, this becomes negative in the calculation, so another interpretation is required depending on the conditions)
Explanation: Square area 36 cm², total area of two quadrants = π×36/4×2 = 18π cm². Since the result is negative, note that the two quadrants overlap; in reality, this develops into a leaf-shaped figure problem. At the Grade 7 level, remain with the basic pattern.