Area of Composite and Shaded Figures - Sixth Grade Mathematics

A composite figure is a complex geometric shape composed of two or more fundamental polygons and curves, whose total area is determined by additive decomposition or subtractive removal.

The Strategy of Decomposition

Real-world objects rarely consist of solitary standard rectangles or triangles. Architectural floor plans, machine parts, and land parcels are composite figures.

The Two Master Strategies:
1. Additive Method (Decomposition):
   Divide the irregular figure into simpler, non-overlapping geometric shapes
   (rectangles, triangles, trapezoids, semicircles), calculate the area of each part,
   and SUM them together:
   Total Area = Area_1 + Area_2 + Area_3 + ...

2. Subtractive Method (Bounding Box / Negative Space):
   Enclose the irregular shape inside a larger standard outer boundary,
   and SUBTRACT the empty or unshaded areas:
   Shaded Area = Total Outer Area - Unshaded Inner Area

Additive Decomposition Walkthrough

Consider an L-shaped living room floor plan:

L-Shaped Room Floor Plan:
                 10 m
        +-------------------+
        |                   |
        |      Region A     | 6 m
        |                   |
        +---------+---------+
        |         |
        | Region B| 4 m
        |         |
        +---------+
            4 m

Split horizontally into two rectangles:
Region A (Top): Length = 10 m, Width = 6 m.
                Area A = 10 * 6 = 60 m^2.

Region B (Bottom): Length = 4 m, Width = 4 m.
                   Area B = 4 * 4 = 16 m^2.

Total Area = Area A + Area B = 60 + 16 = 76 m^2.

Notice that splitting it vertically would yield: Left rectangle: 4 m * 10 m = 40 m^2. Right rectangle: (10 - 4) m * 6 m = 6 m * 6 m = 36 m^2. Total Area = 40 + 36 = 76 m^2. Both decomposition methods arrive at the identical answer.

Composite Shapes with Triangles and Semicircles

Many composite figures combine polygons with curves or triangles, such as a basketball court key or a house profile.

House Facade Profile:
                     /\
                    /  \  Roof (Triangle)
                   /    \  Base = 12 m, Height = 5 m
                  +------+
                  |      |
                  |      |  Walls (Rectangle)
                  |      |  Width = 12 m, Height = 8 m
                  +------+

Area of Triangle (Roof) = (1 / 2) * b * h = (1 / 2) * 12 * 5 = 30 m^2.
Area of Rectangle (Walls) = 12 * 8 = 96 m^2.
Total Area = 30 + 96 = 126 m^2.

Subtractive Method for Shaded Regions

When calculating shaded borders, photo frames, or circular holes drilled into metal plates, subtract the inner hole from the outer area.

Square Plate with Circular Hole:
         +-----------------------+
         |       14 cm           |
         |         .-''''-.      |
         |       .'        '.    |
         |      /     r=3    \   | 14 cm
         |     |       *      |  |
         |      \            /   |
         |       '.        .'    |
         |         '-....-'      |
         +-----------------------+

Step 1: Calculate Outer Area (Square):
        Area_outer = s^2 = 14^2 = 196 cm^2.

Step 2: Calculate Inner Cutout Area (Circle with r = 3 cm, pi approx. 3.14):
        Area_inner = pi * r^2 = 3.14 * 3^2 = 3.14 * 9 = 28.26 cm^2.

Step 3: Subtract inner from outer:
        Shaded Area = 196 - 28.26 = 167.74 cm^2.

Exercises

Exercise 1. An L-shaped garden is composed of a 12-meter by 5-meter rectangle joined to an 8-meter by 4-meter rectangle. What is the total area of the garden?

Exercise 2. A playground has the shape of a rectangle 20 meters long and 10 meters wide, with a semicircle attached to one of the 10-meter ends. Using pi approx. 3.14, find the total area of the playground.

Exercise 3. A rectangular metal sheet measures 15 inches by 20 inches. A technician cuts out a rectangular hole measuring 6 inches by 8 inches from the center. What is the area of the remaining metal sheet?

Exercise 4. A square picture frame has an outer side length of 16 inches. The opening for the photograph is a square with side length 10 inches. What is the area of the frame border surrounding the photo?

Exercise 5. A triangular gable on a barn has a base of 16 feet and a height of 9 feet. In the center of the gable is a rectangular ventilation window measuring 3 feet by 4 feet. Find the area of wood siding required to cover the gable excluding the window.

Solutions

Solution to Exercise 1. Calculate the area of each rectangular section: Area 1 = 12 * 5 = 60 m^2; Area 2 = 8 * 4 = 32 m^2. Total Area = 60 + 32 = 92 square meters.

Solution to Exercise 2. The playground consists of a rectangle and a semicircle: Area of rectangle = 20 * 10 = 200 m^2. The attached semicircle has a diameter of 10 meters, so its radius is r = 10 / 2 = 5 meters. Area of semicircle = (1/2) * pi * r^2 = (1/2) * 3.14 * 5^2 = (1/2) * 3.14 * 25 = (1/2) * 78.5 = 39.25 m^2. Total Area = 200 + 39.25 = 239.25 square meters.

Solution to Exercise 3. Outer area of metal sheet = 15 * 20 = 300 in.^2. Area of cutout hole = 6 * 8 = 48 in.^2. Remaining metal area = 300 - 48 = 252 square inches.

Solution to Exercise 4. Outer square area = 16^2 = 256 in.^2. Inner photo opening area = 10^2 = 100 in.^2. Area of the frame border = 256 - 100 = 156 square inches.

Solution to Exercise 5. Total area of the triangular gable = (1/2) * b * h = (1/2) * 16 * 9 = 8 * 9 = 72 ft^2. Area of rectangular window = 3 * 4 = 12 ft^2. Wood siding area needed = 72 - 12 = 60 square feet.