Introduction
Area of Composite Shapes is an important Grade 7 math skill because students are moving from simple answers toward explaining how the math works.
In this lesson, students use models, real questions, worked examples, practice problems, and two online quizzes to build confidence with area of composite shapes.
What Is Area of Composite Shapes?
Area of Composite Shapes means measuring how much flat space a figure covers by using equal-sized square units.
The goal is not only to get the answer. Students should be able to show the idea, explain the strategy, and check whether the answer makes sense.
Understanding Area of Composite Shapes
Before solving, students should slow down and decide what each number, shape, unit, or label represents.
- Use square units that cover the figure without gaps or overlaps.
- Count rows and columns when the unit squares are arranged in an array.
- Connect repeated addition to multiplication when finding area.
- Break complex figures into smaller rectangles when that makes the work clearer.
Visual Models
Visual Model 1
Question: A composite figure made of two rectangles stacked vertically (both \(7\) units wide and \(2\) units tall) is shown above. What is the total area?
- A. \(14\) sq units
- B. \(18\) sq units
- C. \(24\) sq units
- D. \(28\) sq units
Why it works: Decompose into two rectangles: First rectangle \(7 \times 2 = 14\) sq units. Second rectangle \(7 \times 2 = 14\) sq units. Add them: \(14 + 14 = 28\) sq units.
Answer: \(28\) sq units
Visual Model 2
Question: A composite figure consists of a rectangle (\(8\) units \(\times\) \(5\) units) joined with a triangle (base \(2\) units, height \(5\) units). What is the total area?
- A. \(40\) sq units
- B. \(45\) sq units
- C. \(50\) sq units
- D. \(55\) sq units
Why it works: Decompose into a rectangle and a triangle. Rectangle: \(8 \times 5 = 40\) sq units. Triangle: \(\frac{1}{2} \times 2 \times 5 = 5\) sq units. Add: \(40 + 5 = 45\) sq units.
Answer: \(45\) sq units
Worked Examples
Example 1
Question: A pentagon-shaped field can be decomposed into a rectangle (base \(8\) units, height \(4\) units) and a triangle (base \(8\) units, height \(2\) units). What is the total area?
- A. \(24\) sq units
- B. \(32\) sq units
- C. \(40\) sq units
- D. \(48\) sq units
- Rectangle: \(8 \times 4 = 32\) sq units.
- Triangle: \(\frac{1}{2} \times 8 \times 2 = 8\) sq units.
- Total: \(32 + 8 = 40\) sq units.
Answer: \(40\) sq units
Example 2
Question: A composite figure shows a house shape: a rectangle (\(10\) units \(\times\) \(5\) units) with a triangular roof (base \(6\) units, height \(3\) units). What is the total area?
- A. \(50\) sq units
- B. \(59\) sq units
- C. \(68\) sq units
- D. \(80\) sq units
- Decompose the house shape into a rectangle and a triangular roof.
- Rectangle: \(10 \times 5 = 50\) sq units.
- Triangle: \(\frac{1}{2} \times 6 \times 3 = 9\) sq units.
- Add: \(50 + 9 = 59\) sq units.
Answer: \(59\) sq units
Example 3
Question: A pentagon-shaped region can be decomposed into a trapezoid (parallel sides \(12\) units and \(12\) units, height \(6\) units) and a triangle (base \(6\) units, height \(3\) units). What is the total area?
- A. \(50\) sq units
- B. \(63\) sq units
- C. \(72\) sq units
- D. \(81\) sq units
- The pentagon is already decomposed.
- Trapezoid: \(\frac{1}{2}(12+12) \times 6 = 72\) sq units.
- Triangle (top): \(\frac{1}{2} \times 6 \times 3 = 9\) sq units.
- Add them: \(72 + 9 = 81\) sq units.
Answer: \(81\) sq units
Real-World Word Problems
Problem 1
Question: A garden is shaped like an L. It consists of a \(6\) m by \(4\) m rectangle and a \(3\) m by \(3\) m square connected at a corner. Find the total area of the garden.
- A. \(12\) m\(^2\)
- B. \(21\) m\(^2\)
- C. \(33\) m\(^2\)
- D. \(45\) m\(^2\)
Why it works: Rectangle: \(6 \times 4 = 24\) m\(^2\); square: \(3 \times 3 = 9\) m\(^2\). Total: \(24 + 9 = 33\) m\(^2\).
Answer: \(33\) m\(^2\)
Problem 2
Question: A rectangular garden is \(10\) m wide and \(8\) m tall. A triangular section is removed from the bottom, with the triangle's base being \(10\) m and height \(2\) m. What is the remaining area?
- A. \(70\) m\(^2\)
- B. \(80\) m\(^2\)
- C. \(90\) m\(^2\)
- D. \(100\) m\(^2\)
Why it works: Rectangle: \(10 \times 8 = 80\) m\(^2\). Triangle removed: \(\frac{1}{2} \times 10 \times 2 = 10\) m\(^2\). Remaining: \(80 - 10 = 70\) m\(^2\).
Answer: \(70\) m\(^2\)
Common Mistakes
- Counting only the outside squares instead of all squares inside the figure.
- Leaving gaps or overlaps when using unit squares.
- Multiplying side lengths before checking whether the figure is a rectangle.
- Forgetting to write square units with an area answer.
Strategy Tips
- Trace the rectangle or figure before counting.
- Use rows and columns to organize unit squares.
- Write an equation after the model makes sense.
- Check whether the answer needs square units.
Practice Questions
Question 1
A composite figure is made of a rectangle (\(8\) ft by \(5\) ft) with a square (\(3\) ft by \(3\) ft) attached to one side. What is the total area?
- A. \(40\) ft\(^2\)
- B. \(45\) ft\(^2\)
- C. \(49\) ft\(^2\)
- D. \(56\) ft\(^2\)
Question 2
A floor plan shows a large rectangle \((12 \text{ ft} \times 9 \text{ ft})\) with a rectangular cutout in one corner \((4 \text{ ft} \times 3 \text{ ft})\). What is the area of the remaining floor?
- A. \(83\) ft\(^2\)
- B. \(96\) ft\(^2\)
- C. \(108\) ft\(^2\)
- D. \(120\) ft\(^2\)
Question 3
A trapezoid-like composite shape is formed by a \(6\) m by \(4\) m rectangle with a triangle (\(6\) m base, \(3\) m height) on top. What is the combined area?
- A. \(18\) m\(^2\)
- B. \(27\) m\(^2\)
- C. \(30\) m\(^2\)
- D. \(33\) m\(^2\)
Question 4
A track is shaped like a rectangle with two semicircles on the shorter ends. The rectangle is \(10\) m long and \(6\) m wide. The semicircles have a diameter of \(6\) m. (Use \(\pi \approx 3.14\).) What is the total area?
- A. \(60\) m\(^2\)
- B. \(74.13\) m\(^2\)
- C. \(88.26\) m\(^2\)
- D. \(94.26\) m\(^2\)
Question 5
A composite shape is formed by placing a right triangle (legs \(5\) cm and \(12\) cm) next to a \(12\) cm \(\times\) \(5\) cm rectangle. What is the total area?
- A. \(30\) cm\(^2\)
- B. \(60\) cm\(^2\)
- C. \(90\) cm\(^2\)
- D. \(150\) cm\(^2\)
Question 6
A parking lot is shaped like a large rectangle (\(20\) yd \(\times\) \(15\) yd) with two rectangular sections removed from opposite corners (each \(4\) yd \(\times\) \(3\) yd). What is the remaining area?
- A. \(276\) yd\(^2\)
- B. \(288\) yd\(^2\)
- C. \(300\) yd\(^2\)
- D. \(312\) yd\(^2\)
Full Answer Explanations Click to show all answers and explanations
Question 1
Answer: \(49\) ft\(^2\)
Decompose into two shapes: Rectangle area \(=8\times5=40\) ft\(^2\); square area \(=3\times3=9\) ft\(^2\). Add the areas: \(40+9=49\) ft\(^2\).
Question 2
Answer: \(96\) ft\(^2\)
Start with the large rectangle: \(12 \times 9 = 108\) ft\(^2\). Subtract the cutout rectangle: \(4 \times 3 = 12\) ft\(^2\). Remaining area: \(108 - 12 = 96\) ft\(^2\).
Question 3
Answer: \(33\) m\(^2\)
Decompose into a rectangle and a triangle. Rectangle: \(6 \times 4 = 24\) m\(^2\). Triangle: \(\frac{1}{2} \times 6 \times 3 = 9\) m\(^2\). Add: \(24 + 9 = 33\) m\(^2\).
Question 4
Answer: \(88.26\) m\(^2\)
Rectangle: \(10 \times 6 = 60\) m\(^2\). Two semicircles = 1 full circle with radius \(3\) m: \(\pi r^2 = 3.14 \times 9 = 28.26\) m\(^2\). Total: \(60 + 28.26 = 88.26\) m\(^2\).
Question 5
Answer: \(90\) cm\(^2\)
Decompose into a rectangle and a right triangle. Rectangle: \(12 \times 5 = 60\) cm\(^2\). Right triangle: \(\frac{1}{2} \times 5 \times 12 = 30\) cm\(^2\). Add: \(60 + 30 = 90\) cm\(^2\).
Question 6
Answer: \(276\) yd\(^2\)
Start with the large rectangle: \(20 \times 15 = 300\) yd\(^2\). Subtract the two removed corner sections: \(2 \times (4 \times 3) = 24\) yd\(^2\). Remaining area: \(300 - 24 = 276\) yd\(^2\).
Connection to Standards
This lesson supports Grade 7 math expectations for reasoning, modeling, problem solving, and explaining answers clearly. It connects classroom skills to the kind of questions students see on state math assessments.
Summary
Area of Composite Shapes becomes easier when students connect the question to a model, use clear steps, and explain why the answer fits.
GOLDEN RULE
Area means every square unit inside the figure.
