How to Solve: Area and Perimeter (Composite Shapes, Circles, Sectors, Arcs)

How to Solve: Area and Perimeter (Composite Shapes, Circles, Sectors, Arcs)

Exam Context: GCSE (Edexcel/AQA/OCR) & A-Level Maths Score Impact: 6–10% of your exam paper (typically 2–4 questions, worth 8–12 marks).

? Introduction

"This topic turns a single tricky shape into easy rectangles and circles—saving you 10+ marks on exam day. Miss it, and you’ll lose marks on every question about gardens, pizza slices, or even Olympic running tracks."

? WHAT YOU NEED TO KNOW FIRST

1. Area & perimeter of basic shapes (rectangles, triangles, circles).
2. How to split a shape into simpler parts (composite shapes).
3. Degrees in a circle (360°) and how to find fractions of it.

? KEY VOCABULARY

⚡ FORMULAS TO KNOW

? STEP-BY-STEP METHOD

For Composite Shapes (Area & Perimeter)

1. Split the shape into rectangles, triangles, circles, or sectors.
2. Label all given lengths (even if you have to find them).
3. Find missing lengths using algebra or Pythagoras if needed.
4. Calculate area/perimeter of each part using the correct formula.
5. Add or subtract areas (e.g., subtract a missing semicircle).
6. For perimeter, add all outer edges (including arcs).
7. Check units (cm² for area, cm for perimeter).

For Sectors & Arcs

1. Identify the angle (θ) and radius (r).
2. For area: Use A = (θ/360) × πr².
3. For arc length: Use L = (θ/360) × 2πr.
4. For perimeter of a sector: Add 2r + arc length.
5. Simplify fractions (e.g., 90/360 = ¼).

✏️ WORKED EXAMPLES

Example 1 – Basic Composite Shape

Question: Find the area and perimeter of this shape (all lengths in cm):

   _______________
  |               |
  |               | 4
  |_______        |

          |       |

          | 3     | 5

          |_______|

(A rectangle with a 3×4 rectangle missing from the bottom right.)

Solution: 1. Split into two rectangles:
- Big rectangle: 7 cm × 5 cm (height = 4 + 1 = 5).
- Small rectangle: 3 cm × 4 cm. 2. Area:
- Big rectangle: 7 × 5 = 35 cm².
- Small rectangle: 3 × 4 = 12 cm².
- Total area = 35 – 12 = 23 cm². 3. Perimeter:
- Outer edges: 7 + 5 + 4 + 3 + 1 + 3 = 23 cm.
(Why? Add all sides, including the "step" at the bottom.)

What we did and why: - Split the shape to avoid confusion. - Subtracted the missing part for area. - Added all outer edges for perimeter (no shortcuts!).

Example 2 – Medium (Sector + Rectangle)

Question: A shape is made of a rectangle (8 cm × 6 cm) with a semicircle on top (diameter = 6 cm). Find the area and perimeter.

Solution: 1. Area:
- Rectangle: 8 × 6 = 48 cm².
- Semicircle: ½ × π × (3)² = 4.5π ≈ 14.14 cm².
- Total area = 48 + 14.14 = 62.14 cm². 2. Perimeter:
- Rectangle sides: 8 + 6 + 8 = 22 cm.
- Semicircle arc: ½ × π × 6 = 3π ≈ 9.42 cm.
- Total perimeter = 22 + 9.42 = 31.42 cm.
(Why? Don’t add the diameter—it’s inside the shape!)

What we did and why: - Used diameter = 6 cm → radius = 3 cm. - Added the semicircle’s arc, not its diameter, to the perimeter.

Example 3 – Exam-Style (Sector + Triangle)

Question (A-Level): A sector of angle 40° has radius 10 cm. A triangle is formed by the two radii and the chord. Find: a) The area of the sector. b) The perimeter of the sector. c) The area of the triangle.

Solution: a) Sector area:
A = (40/360) × π × 10² = (1/9) × 100π ≈ 34.91 cm².

b) Perimeter of sector:
- Arc length: (40/360) × 2π × 10 = (20/9)π ≈ 6.98 cm.
- Two radii: 10 + 10 = 20 cm.
- Total perimeter = 20 + 6.98 = 26.98 cm.

c) Triangle area:
- Use ½ab sin C: ½ × 10 × 10 × sin(40°) ≈ 32.14 cm².

What we did and why: - Used sector formulas for (a) and (b). - For (c), remembered the triangle area formula with sine (A-Level only).

❌ Common Mistakes

? EXAM TRAPS