Area and Perimeter of Composite Two-Dimensional Figures - Seventh Grade Mathematics

Composite two-dimensional figures combine multiple polygons and circular curves, analyzed by decomposing regions into additive components or subtracting interior voids.

Decomposing Complex Geometric Shapes

A composite figure is a geometric shape composed of two or more fundamental polygons (such as rectangles, triangles, parallelograms, and trapezoids) and circular regions (circles, semicircles, and quadrants).

               TWO ANALYTICAL STRATEGIES
   Strategy 1: Additive Method (Summing Sub-Regions)
               Total Area = Area_1 + Area_2 + ... + Area_n

   Strategy 2: Subtractive Method (Carving Voids from a Bounding Box)
               Total Area = Bounding Box Area - Void Area(s)

Selecting the most efficient strategy depends on whether the figure is built from joining parts together or by cutting shapes out of a larger solid form.

The Additive Method: Decomposition and Summation

Consider a basketball court key formed by a rectangle joined to a semicircular top.

               ADDITIVE EXAMPLE: BASKETBALL KEY
                     ...---...
                  .-'         '-.   <-- Semicircle (radius r = 6 ft)
                .'               '.
               +-------------------+
               |                   |
               |     Rectangle     | Height = 19 ft
               |                   | Width  = 12 ft
               +-------------------+

Step 1: Calculate the area of the rectangular region: Area of rectangle = length * width = 19 * 12 = 228 square feet.

Step 2: Calculate the area of the semicircular region: Diameter is twelve feet, so radius r is six feet. Area of semicircle = (1/2) * pi * r^2 = 0.5 * 3.14 * 6^2 = 0.5 * 3.14 * 36 = 56.52 square feet.

Step 3: Sum the individual areas: Total Area = 228 + 56.52 = 284.52 square feet.

The Subtractive Method: Frames and Hollow Regions

When an irregular shape consists of a solid outer boundary with an interior cutout or window, subtract the area of the inner shape from the outer shape.

               SUBTRACTIVE EXAMPLE: CIRCULAR RING (WASHER)
                           ...---...
                        .-'   ...   '-.
                      .'    .'   '.    '.   Outer Radius R = 10 cm
                     /     /   *   \     \  Inner Radius r = 6 cm
                      '.    '.   .'    .'
                        '-.   '''   .-'
                           '''---'''

Step 1: Calculate the area of the entire outer circle: Outer Area = pi * R^2 = 3.14 * 10^2 = 3.14 * 100 = 314 square centimeters.

Step 2: Calculate the area of the inner circular hole: Inner Area = pi * r^2 = 3.14 * 6^2 = 3.14 * 36 = 113.04 square centimeters.

Step 3: Subtract the void from the outer region: Shaded Ring Area = 314 - 113.04 = 200.96 square centimeters.

Determining Perimeter of Composite Figures

A frequent error in geometry is adding the perimeters of the individual component shapes to find the composite perimeter. Interior partition lines that separate internal sections are NOT part of the outer boundary.

               PERIMETER OF COMPOSITE SHAPE
   Rule: ONLY sum the lengths of segments forming the EXTERNAL boundary.
         Internal partition lines must be strictly excluded.

In the basketball key example above: The perimeter consists of: Left vertical side: 19 ft Bottom horizontal side: 12 ft Right vertical side: 19 ft Curved semicircular arc: (1/2) * pi * d = 0.5 * 3.14 * 12 = 18.84 ft The internal twelve-foot boundary separating the rectangle from the semicircle is completely ignored. Total Perimeter = 19 + 12 + 19 + 18.84 = 68.84 feet.

Exercises

Work through each problem carefully in your notebook before reading the provided solutions.

Exercise 1. An L-shaped room can be divided into two non-overlapping rectangles: Rectangle A measures twelve feet by eight feet, and Rectangle B measures six feet by four feet. Calculate the total floor area of the room.

Exercise 2. A wooden picture frame has outer dimensions of twenty inches by twenty-four inches. A rectangular opening for the photograph measures sixteen inches by twenty inches. Find the surface area of the wooden frame.

Exercise 3. A playground field is shaped like a rectangle fifty meters long and thirty meters wide, with semicircles attached to both of its shorter ends. Calculate the total area of the field using 3.14 for pi.

Exercise 4. A metal plate is fabricated in the shape of a square with side lengths of ten centimeters. A circular hole of diameter four centimeters is drilled through the center. Calculate the area of the remaining metal plate using 3.14 for pi.

Solutions

Solution to Exercise 1. Apply the additive method by calculating the area of each sub-rectangle. Area of Rectangle A equals twelve times eight, which is ninety-six square feet. Area of Rectangle B equals six times four, which is twenty-four square feet. Add the two areas: ninety-six plus twenty-four equals one hundred twenty square feet. The total floor area of the room is one hundred twenty square feet.

Solution to Exercise 2. Apply the subtractive method. Calculate the total area of the outer bounding rectangle: twenty times twenty-four equals four hundred eighty square inches. Calculate the area of the inner photograph cutout: sixteen times twenty equals three hundred twenty square inches. Subtract the cutout from the total area: four hundred eighty minus three hundred twenty equals one hundred sixty square inches. The surface area of the wooden frame is one hundred sixty square inches.

Solution to Exercise 3. The field consists of one central rectangle and two identical semicircles at the ends. Two semicircles of identical diameter combine to form one complete circle. The diameter of the semicircles matches the thirty-meter width of the rectangle, so radius r is fifteen meters. Calculate the area of the rectangle: fifty meters times thirty meters equals one thousand five hundred square meters. Calculate the area of the complete circle: pi times fifteen squared equals 3.14 times two hundred twenty-five, which equals seven hundred six point five square meters. Sum the components: one thousand five hundred plus seven hundred six point five equals two thousand two hundred six point five square meters.

Solution to Exercise 4. Apply the subtractive method. The outer shape is a square of side ten centimeters: area of square equals ten times ten, which equals one hundred square centimeters. The circular cutout has a diameter of four centimeters, so its radius r is two centimeters. Area of circle equals pi times two squared: 3.14 times four equals twelve point fifty-six square centimeters. Subtract the circle from the square: one hundred minus twelve point fifty-six equals eighty-seven point forty-four square centimeters. The remaining area of the metal plate is eighty-seven point forty-four square centimeters.