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 × 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:
- Break into simpler shapes (rectangles, triangles, circles, etc.)
- Find area of each piece using appropriate formula
- 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
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