Area of composite figures (see also 5.4) Knowledge-Based Study

Math · Geometry · Math · Perimeter, Area & Surface Area

Math topic explanations on Area of composite figures (see also 5.4).

Study notes

  1. What is a Composite Figure?

    A composite figure is a geometric shape made up of two or more basic two-dimensional shapes, such as rectangles, squares, triangles, parallelograms, trapezoids, or semicircles.

    To calculate the total area of a composite figure, we break the complex shape down into simpler components whose areas we can calculate directly using standard geometric formulas.

    Key Formulas for Component Shapes

    • Rectangle: Area=length×width=l×w\text{Area} = \text{length} \times \text{width} = l \times w
    • Square: Area=side2=s2\text{Area} = \text{side}^2 = s^2
    • Triangle: Area=12×base×height=12bh\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} b h
    • Trapezoid: Area=12(b1+b2)h\text{Area} = \frac{1}{2} (b_1 + b_2) h
    • Parallelogram: Area=b×h\text{Area} = b \times h
    • Semicircle: Area=12πr2\text{Area} = \frac{1}{2} \pi r^2

    Method 1: Decomposing into Simpler Parts

    The most common way to determine the area of a composite figure is by splitting it into non-overlapping regions and adding their areas together.

    Step-by-Step Strategy:

    1. Divide: Draw auxiliary lines to separate the figure into familiar, basic shapes.
    2. Find Missing Dimensions: Use the given side lengths to determine any unknown base, height, or radius values.
    3. Calculate Individual Areas: Apply the appropriate formula to each individual part.
    4. Sum the Areas: Add the calculated areas together to find the total area.

    Worked Example: L-Shaped Polygon

    Consider an L-shaped region with a vertical height of 10 cm, an overall base width of 8 cm, an upper horizontal edge of 3 cm, and a right vertical edge of 4 cm.

    • We can split this figure vertically into two rectangles:
      • Rectangle 1 (left): width = 3 cm, height = 10 cm
        Area1=3×10=30 cm2\text{Area}_1 = 3 \times 10 = 30\text{ cm}^2
      • Rectangle 2 (right): width = 8−3=5 cm8 - 3 = 5\text{ cm}, height = 4 cm
        Area2=5×4=20 cm2\text{Area}_2 = 5 \times 4 = 20\text{ cm}^2
    • Total Area: Area=Area1+Area2=30+20=50 cm2\text{Area} = \text{Area}_1 + \text{Area}_2 = 30 + 20 = 50\text{ cm}^2.

    Method 2: Subtracting Unwanted Regions

    Sometimes, it is faster to place the composite figure inside a larger, regular bounding shape and subtract the areas of the regions that are not part of the figure.

    Step-by-Step Strategy:

    1. Enclose: Construct a larger bounding rectangle or square that completely contains the figure.
    2. Calculate Bounding Area: Find the area of the entire outer shape.
    3. Identify the Cutouts: Determine the dimensions and areas of the empty corners or holes.
    4. Subtract: Subtract the empty areas from the total bounding area.

    Worked Example: Re-evaluating the L-Shape

    Using the same L-shaped figure from the previous lesson:

    • Bounding Rectangle: base = 8 cm, height = 10 cm
      Total Area=8×10=80 cm2\text{Total Area} = 8 \times 10 = 80\text{ cm}^2
    • Cutout Corner: base = 8−3=5 cm8 - 3 = 5\text{ cm}, height = 10−4=6 cm10 - 4 = 6\text{ cm}
      Cutout Area=5×6=30 cm2\text{Cutout Area} = 5 \times 6 = 30\text{ cm}^2
    • Actual Area: Area=80−30=50 cm2\text{Area} = 80 - 30 = 50\text{ cm}^2.

    Both methods yield the exact same result. You can select whichever method requires fewer calculations.

    Figures with Semicircles and Triangles

    Composite figures frequently combine straight-edged shapes with curved components or inclined roofs.

    Worked Example: A Rectangular Wall with a Triangular Roof

    A figure is composed of a rectangle with a width of 12 m and a height of 7 m, topped by an attached triangle whose base is the top edge of the rectangle (12 m) and whose height is 5 m.

    1. Area of the rectangular base:
      Arearect=12×7=84 m2\text{Area}_{\text{rect}} = 12 \times 7 = 84\text{ m}^2
    2. Area of the triangular top:
      Areatri=12×12×5=30 m2\text{Area}_{\text{tri}} = \frac{1}{2} \times 12 \times 5 = 30\text{ m}^2
    3. Total composite area:
      Areatotal=84+30=114 m2\text{Area}_{\text{total}} = 84 + 30 = 114\text{ m}^2

    Common Pitfalls to Avoid

    • Double-counting shared boundaries: Remember that interior dividing lines are not boundaries of the total perimeter and must not introduce extra areas.
    • Diameter vs. Radius: When a semicircle is attached to an edge, that edge represents the diameter. Always divide by 2 to get the radius before using 12πr2\frac{1}{2}\pi r^2.
    • Consistent Units: Ensure all linear dimensions are in the same unit before performing multiplications.