Area of Composite Figures

Find area of complex shapes

Area of Composite Figures

A composite figure is a shape made up of two or more basic shapes. To find the area, you break it into simpler shapes, find each area, then add them together!


What Is a Composite Figure?

A composite figure (also called a complex figure) is created by combining basic shapes like:

  • Rectangles
  • Triangles
  • Circles
  • Trapezoids
  • Semicircles

Real-world examples:

  • Floor plans of houses
  • Gardens with multiple sections
  • Swimming pools with different depths
  • Letters and logos

Review: Area Formulas for Basic Shapes

Rectangle: A = length × width = lw

Square: A = side × side = s²

Triangle: A = (1/2) × base × height = (1/2)bh

Parallelogram: A = base × height = bh

Trapezoid: A = (1/2) × (base₁ + base₂) × height = (1/2)(b₁ + b₂)h

Circle: A = πr² (where r = radius)

Semicircle: A = (1/2)πr² (half of a circle)


Strategy for Finding Area of Composite Figures

Step 1: Break the figure into basic shapes

  • Look for rectangles, triangles, circles, etc.
  • Draw lines to separate shapes

Step 2: Find the area of each basic shape

  • Use appropriate formula for each piece
  • Label dimensions carefully

Step 3: Add all the areas together

  • Total Area = Area₁ + Area₂ + Area₃ + ...

Method 1: Addition Method

Break the figure into parts and ADD the areas.

Example 1: L-Shaped Figure

An L-shaped figure can be split into two rectangles:

  • Top rectangle: 8 ft × 3 ft
  • Bottom rectangle: 5 ft × 4 ft

Solution:

Area of top rectangle: A₁ = 8 × 3 = 24 ft²

Area of bottom rectangle: A₂ = 5 × 4 = 20 ft²

Total Area: A = 24 + 20 = 44 ft²

Answer: 44 ft²

Example 2: Rectangle with Triangle on Top

A house-shaped figure: rectangle base with triangular roof

  • Rectangle: 12 m × 8 m
  • Triangle: base = 12 m, height = 5 m

Solution:

Rectangle area: A₁ = 12 × 8 = 96 m²

Triangle area: A₂ = (1/2) × 12 × 5 = 30 m²

Total Area: A = 96 + 30 = 126 m²

Answer: 126 m²

Example 3: Rectangle with Semicircle

A shape with rectangle (10 cm × 6 cm) and semicircle on one end (diameter = 6 cm, radius = 3 cm)

Solution:

Rectangle area: A₁ = 10 × 6 = 60 cm²

Semicircle area: A₂ = (1/2)π(3²) = (1/2)π(9) = 4.5π ≈ 14.14 cm²

Total Area: A = 60 + 14.14 ≈ 74.14 cm²

Answer: About 74.14 cm²


Method 2: Subtraction Method

Sometimes it's easier to find a large shape's area, then SUBTRACT a missing piece!

Example 1: Rectangle with Corner Cut Out

A 10 in × 8 in rectangle with a 3 in × 2 in rectangle cut from one corner.

Solution:

Large rectangle area: A₁ = 10 × 8 = 80 in²

Cut-out rectangle area: A₂ = 3 × 2 = 6 in²

Remaining Area: A = 80 - 6 = 74 in²

Answer: 74 in²

Example 2: Circle with Hole

A circular disc with outer radius 10 cm and a circular hole with radius 4 cm.

Solution:

Outer circle area: A₁ = π(10²) = 100π cm²

Inner circle (hole) area: A₂ = π(4²) = 16π cm²

Remaining Area: A = 100π - 16π = 84π ≈ 263.89 cm²

Answer: About 263.89 cm²

Example 3: Rectangle with Triangle Removed

A 15 ft × 10 ft rectangle with a triangle (base = 6 ft, height = 4 ft) removed from one corner.

Solution:

Rectangle area: A₁ = 15 × 10 = 150 ft²

Triangle area: A₂ = (1/2) × 6 × 4 = 12 ft²

Remaining Area: A = 150 - 12 = 138 ft²

Answer: 138 ft²


Finding Missing Dimensions

Sometimes you need to calculate missing dimensions before finding areas!

Example: T-Shaped Figure

A T-shaped figure:

  • Top horizontal bar: 12 cm × 3 cm
  • Vertical stem: ? cm × 4 cm
  • Total height: 10 cm

Step 1: Find missing dimension Vertical stem height = 10 - 3 = 7 cm

Step 2: Find areas Top bar: 12 × 3 = 36 cm² Vertical stem: 7 × 4 = 28 cm²

Step 3: Add Total = 36 + 28 = 64 cm²

Answer: 64 cm²


Working with Multiple Shapes

Example: Complex Floor Plan

A room has:

  • Main rectangular area: 20 ft × 15 ft
  • Alcove (small rectangle): 8 ft × 5 ft
  • Circular bay window: radius = 4 ft (semicircle)

Solution:

Main room: A₁ = 20 × 15 = 300 ft²

Alcove: A₂ = 8 × 5 = 40 ft²

Bay window (semicircle): A₃ = (1/2)π(4²) = 8π ≈ 25.13 ft²

Total Area: A = 300 + 40 + 25.13 = 365.13 ft²

Answer: About 365 ft²


Real-World Applications

Landscaping

Problem: A garden has a rectangular section (15 m × 10 m) and a triangular section (base = 10 m, height = 6 m). How much area needs to be planted?

Solution: Rectangle: 15 × 10 = 150 m² Triangle: (1/2) × 10 × 6 = 30 m² Total: 150 + 30 = 180 m²

Answer: 180 m²

Painting

Problem: A wall shaped like a house (rectangle with triangle on top) needs paint. Rectangle is 12 ft × 8 ft, triangle has base 12 ft and height 3 ft. One gallon covers 50 ft². How many gallons needed?

Solution: Rectangle: 12 × 8 = 96 ft² Triangle: (1/2) × 12 × 3 = 18 ft² Total: 96 + 18 = 114 ft² Gallons: 114 ÷ 50 = 2.28 → Need 3 gallons

Answer: 3 gallons

Flooring

Problem: An L-shaped room needs tile. One section is 14 ft × 10 ft, the other is 8 ft × 6 ft. Tiles cost $3 per square foot. What's the total cost?

Solution: Section 1: 14 × 10 = 140 ft² Section 2: 8 × 6 = 48 ft² Total area: 188 ft² Cost: 188 × 3=3 = 564

Answer: $564


Using Both Addition and Subtraction

Example: Swimming Pool

A rectangular pool (20 m × 10 m) has a semicircular extension on one end (radius = 5 m), but a rectangular shallow area (4 m × 3 m) is excluded from the total.

Solution:

Main pool: A₁ = 20 × 10 = 200 m²

Semicircular extension (add): A₂ = (1/2)π(5²) = 12.5π ≈ 39.27 m²

Shallow area to exclude (subtract): A₃ = 4 × 3 = 12 m²

Total: A = 200 + 39.27 - 12 = 227.27 m²

Answer: About 227 m²


Tips for Success

Tip 1: Sketch and label

  • Draw the figure neatly
  • Label all dimensions
  • Mark where you're splitting shapes

Tip 2: Look for patterns

  • Can you make rectangles?
  • Do you see triangles?
  • Are there circles or parts of circles?

Tip 3: Choose your method

  • Addition: When adding sections makes sense
  • Subtraction: When removing a piece is easier
  • Sometimes you can use either method!

Tip 4: Track units

  • Keep units consistent (all feet or all meters)
  • Final answer is in square units (ft², m², etc.)

Tip 5: Use formulas correctly

  • Triangle: Don't forget the 1/2!
  • Circle: π ≈ 3.14 or use calculator's π button
  • Check which dimension is base vs. height

Common Mistakes to Avoid

Mistake 1: Counting overlapping areas twice

  • Make sure pieces don't overlap!
  • Each part of the figure should be counted exactly once

Mistake 2: Using wrong dimensions

  • Base and height must be perpendicular
  • For circles, make sure you have radius (not diameter!)

Mistake 3: Forgetting to add/subtract all pieces

  • Count all sections of the composite figure

Mistake 4: Unit errors

  • Mixing feet and inches
  • Forgetting to convert
  • Writing final answer without square units

Mistake 5: Calculation errors with π

  • Semicircle is HALF the circle, don't forget the 1/2
  • Use consistent value for π (3.14 or calculator)

Practice Strategy

Step 1: Identify

  • What basic shapes make up this figure?

Step 2: Decide on method

  • Addition (add pieces) or subtraction (remove pieces)?

Step 3: Find missing dimensions

  • Do you need to calculate any lengths?

Step 4: Calculate each area

  • Use correct formula for each shape
  • Show your work!

Step 5: Combine

  • Add or subtract as needed
  • Include units!

Step 6: Check reasonableness

  • Does your answer make sense?
  • Is it bigger/smaller than you expected?

Different Ways to Break Up the Same Figure

Example: An L-shape can be split multiple ways:

Method 1: Two rectangles (horizontal cut) Method 2: Two rectangles (vertical cut) Method 3: Large rectangle minus missing corner

All methods give the SAME ANSWER! Choose whichever is easiest for you.


Quick Reference

Addition Method: Total Area = A₁ + A₂ + A₃ + ...

Subtraction Method: Total Area = Large Area - Removed Area

Key Formulas:

  • Rectangle: lw
  • Triangle: (1/2)bh
  • Circle: πr²
  • Semicircle: (1/2)πr²

Remember:

  • Break complex into simple
  • Use appropriate formulas
  • Add or subtract as needed
  • Keep track of units!

Summary

Composite figures are made up of multiple basic shapes.

Strategy:

  1. Break into simpler shapes (rectangles, triangles, circles, etc.)
  2. Find area of each piece using appropriate formula
  3. Add areas together (or subtract if removing a piece)

Two Main Methods:

  • Addition: Add all the pieces
  • Subtraction: Large shape minus removed piece

Applications:

  • Floor plans and architecture
  • Landscaping and gardening
  • Painting and flooring
  • Manufacturing and design

Understanding composite figures is essential for real-world problem solving, construction, design, and advanced geometry!

📚 Practice Problems

1Problem 1easy

Question:

Find the area of an L-shaped figure made of two rectangles: one is 5 ft by 3 ft, and the other is 4 ft by 2 ft.

💡 Show Solution

Break into two rectangles and add:

Rectangle 1: A₁ = 5 × 3 = 15 ft² Rectangle 2: A₂ = 4 × 2 = 8 ft²

Total Area = 15 + 8 = 23 ft²

Answer: 23 ft²

2Problem 2easy

Question:

A rectangular garden is 10 m by 8 m with a rectangular pond inside that is 3 m by 2 m. What is the area of the garden NOT covered by the pond?

💡 Show Solution

Use the subtraction method:

Area of garden: A₁ = 10 × 8 = 80 m² Area of pond: A₂ = 3 × 2 = 6 m²

Area NOT covered = 80 - 6 = 74 m²

Answer: 74 m²

3Problem 3medium

Question:

Find the area of a figure that is a rectangle (12 cm by 6 cm) with a semicircle on top (diameter 6 cm). Use π ≈ 3.14.

💡 Show Solution

Break into rectangle + semicircle:

Rectangle: A₁ = 12 × 6 = 72 cm²

Semicircle:

  • Radius = 6/2 = 3 cm
  • Full circle area = πr² = 3.14 × 3² = 3.14 × 9 = 28.26 cm²
  • Semicircle area = 28.26/2 = 14.13 cm²

Total Area = 72 + 14.13 = 86.13 cm²

Answer: 86.13 cm²

4Problem 4medium

Question:

A trapezoid-shaped pool deck has parallel sides of 15 ft and 10 ft, with a height of 8 ft. There is a rectangular hot tub (4 ft by 3 ft) cut out of the deck. What is the remaining deck area?

💡 Show Solution

Step 1: Area of trapezoid deck A = (1/2)(b₁ + b₂)h A = (1/2)(15 + 10)(8) A = (1/2)(25)(8) A = 100 ft²

Step 2: Area of hot tub A = 4 × 3 = 12 ft²

Step 3: Subtract Remaining deck = 100 - 12 = 88 ft²

Answer: 88 ft²

5Problem 5hard

Question:

A figure consists of a square (side 10 in), with a triangle on top (base 10 in, height 6 in) and a semicircle cut out of the bottom (diameter 10 in). Find the total area. Use π ≈ 3.14.

💡 Show Solution

Addition and subtraction method:

Square: A₁ = 10² = 100 in²

Triangle: A₂ = (1/2)(10)(6) = 30 in²

Semicircle (cut out):

  • Radius = 10/2 = 5 in
  • Full circle = π(5²) = 3.14(25) = 78.5 in²
  • Semicircle = 78.5/2 = 39.25 in²

Total = Square + Triangle - Semicircle Total = 100 + 30 - 39.25 = 90.75 in²

Answer: 90.75 in²