A focused Grade 7 learning session for Area of Composite Shapes using visual models, worked examples, strategy guidance, and a short mastery check.
Student goal: I can solve problems involving area of composite shapes.
Area of Composite Shapes is part of Grade 7 Measurement Geometry. This activity focuses on the central idea students need before they move into independent practice.
Students will build Grade 7 mastery of area of composite shapes using models, equations, and test-style reasoning.
| What to notice | How to use it |
|---|---|
| Numbers, units, labels, diagrams, or graphs | Identify what each part represents before calculating. |
| The relationship in the problem | Connect the model to an equation, comparison, or explanation. |
Use the figures and worked questions in the practice set below as the visual model for this skill.
Read the first practice question slowly. Decide what the question is asking, choose a strategy, solve, and then compare your reasoning with the answer explanation.
For Area of Composite Shapes, a strong worked example should show the setup, the calculation or model, and a final answer that matches the context.
Many Grade 7 test questions use real situations. Look for quantities, units, rates, measures, expressions, or data values that connect the math to the story.
After solving, explain what the answer means in the situation, not just the number you calculated.
I can solve problems involving area of composite shapes.
Use the warmup, guided question, practice set, and exit questions below to check whether this skill is ready or needs more practice.
Use scratch paper. The explanation matters more than simply seeing whether the choice is correct.
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?
Answer C: \(49\) 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\(^2\).
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.
Answer C: \(33\) m\(^2\)
Rectangle: \(6 \times 4 = 24\) m\(^2\); square: \(3 \times 3 = 9\) m\(^2\). Total: \(24 + 9 = 33\) m\(^2\).
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?
Answer D: \(28\) sq units
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.
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?
Answer B: \(96\) 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\(^2\).
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?
Answer D: \(33\) 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\(^2\).
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?
Answer B: \(45\) sq units
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.
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?
Answer A: \(70\) m\(^2\)
Rectangle: \(10 \times 8 = 80\) m\(^2\). Triangle removed: \(\frac{1}{2} \times 10 \times 2 = 10\) m\(^2\). Remaining: \(80 - 10 = 70\) m\(^2\).
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?
Answer C: \(88.26\) 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\(^2\).
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?
Answer C: \(90\) 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\(^2\).
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?
Answer A: \(276\) 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\(^2\).
An L-shaped yard has a vertical rectangle (\(4\) m \(\times\) \(10\) m) and a horizontal rectangle (\(6\) m \(\times\) \(4\) m) that overlap in a \(4\) m \(\times\) \(4\) m square. What is the total area?
Answer A: \(48\) m\(^2\)
Add the two rectangles, but subtract the overlap once (since it was counted twice). Vertical rectangle: \(4 \times 10 = 40\) m\(^2\). Horizontal rectangle: \(6 \times 4 = 24\) m\(^2\). Overlap region: \(4 \times 4 = 16\) m\(^2\). Total: \(40 + 24 - 16 = 48\) m\(^2\).
