Tenth Grade Mathematics: Areas of Combinations of Plane Figures

In architecture, manufacturing, graphic design, and artistic tilework, surfaces are rarely composed of a single, uniform geometric figure. Instead, aesthetic and functional patterns are created by combining, overlapping, and subtracting basic plane figures such as squares, triangles, rectangles, semicircles, sectors, and segments. Solving the areas of these composite shaded regions requires decomposing complex drawings into elementary constituent geometries and applying the fundamental Principle of Superposition.

The Principle of Superposition and Area Decomposition

The area of any bounded composite planar region satisfies the additive property of two-dimensional measure:

Area Addition and Subtraction Axioms:
1. Total Composite Area: If a region R is partitioned into non-overlapping
   sub-regions R1, R2, ..., Rk, then:
                      Area(R) = Area(R1) + Area(R2) + ... + Area(Rk)

2. Shaded Remainder Area: If a smaller geometric figure F2 is removed from an
   enclosing figure F1, the remaining shaded area is:
                      Area(Shaded) = Area(F1) - Area(F2)

The Three-Stage Solving Strategy:
1. Identify the Master Boundary: Determine the outer enclosing geometric polygon
   or circle (e.g., square, circle, or equilateral triangle).
2. Identify the Subtracted Voids: Identify the shapes of the unshaded cavities
   (e.g., quadrants, circular cutouts, or triangular windows).
3. Compute and Subtract: Calculate each individual area using exact standard
   formulas and compute the net shaded difference.

Common Composite Configurations

Pattern 1: Four Quadrants in a Square (The "Square-Circle" Remainder)
A square of side a has four quadrants of radius r = a/2 cut from its corners:
+-------+-------+
| (   ) | (   ) |  Four quadrants of radius a/2 combine to form ONE complete circle
|       |       |  of radius a/2!
+-------+-------+  Shaded Corner Remainder:
| (   ) | (   ) |  Area = Area(Square) - Area(Complete Circle)
+-------+-------+       = a^2 - pi * (a/2)^2 = a^2 * (1 - pi/4)

Pattern 2: Semicircles on the Sides of a Polygon
Semicircles erected outwards on the sides of a right triangle or square.

Pattern 3: Concentric Annular Sectors (Flower Bed and Arch Designs)
Regions bounded between concentric arcs of radii R and r subtending an angle theta:
Area = (theta / 360) * pi * (R^2 - r^2)

Detailed Worked Example: The Four-Leaf Clover Pattern

Find the area of the shaded design enclosed within a square ABCD of side 10 cm, where each side of the square acts as the diameter of a semicircle drawn inside the square. (Use pi = 3.14).

Geometric Insight:
When four semicircles are drawn with the four sides as diameters, they overlap
in the center to form a four-petaled symmetrical cloverleaf:

                 A ----------------- B
                   |   \   |   /   |
                   |     \ | /     |
                   |-------*-------|
                   |     / | \     |
                   |   /   |   \   |
                 D ----------------- C

Let us calculate the area of the unshaded regions first.
Label the four unshaded regions outside the petals as I, II, III, and IV.

Step 1: Notice that Region I + Region III are formed by taking the whole square
        and subtracting the two semicircles drawn on opposite sides AD and BC:
        Area(I + III) = Area(Square) - [Area(Semicircle AD) + Area(Semicircle BC)]
        Since both semicircles have diameter 10 cm (radius r = 5 cm), together
        they form ONE complete circle of radius 5 cm!
        Area(I + III) = 10^2 - pi * 5^2 = 100 - 3.14 * 25 = 100 - 78.5 = 21.5 cm^2.

Step 2: By identical symmetry, Region II + Region IV are formed by subtracting
        the semicircles on AB and CD:
        Area(II + IV) = 100 - 78.5 = 21.5 cm^2.

Step 3: Total unshaded area of the four corner voids:
        Area(Unshaded) = Area(I + III) + Area(II + IV) = 21.5 + 21.5 = 43 cm^2.

Step 4: The shaded area of the four petals is the square minus the unshaded area:
        Area(Shaded Petals) = Area(Square) - Area(Unshaded)
                            = 100 - 43 = 57 cm^2.

The area of the shaded four-leaf design is 57 cm^2.

Chapter Exercises

Exercise 1: Find the area of the shaded region in a square ABCD of side 14 cm, where four identical circles are removed such that each circle touches two sides of the square and two other circles.

Exercise 2: From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut. Find the area of the remaining portion of the square.

Exercise 3: In an equilateral triangle ABC of side 12 cm, an arc of radius 6 cm is drawn with center at each vertex. Find the area of the region enclosed between the three arcs in the center of the triangle. (Use pi = 3.14 and sqrt(3) = 1.732).

Solutions and Explanations

Solution 1: The square ABCD has side length a = 14 cm, so the total area of the square is Area(Square) = 14^2 = 196 cm^2. The four identical circles are packed in a 2x2 grid inside the square. The diameter of each circle is d = 14 / 2 = 7 cm, giving a radius of r = 7/2 = 3.5 cm. The area of one circle is Area = pi * r^2 = (22/7) * (7/2)^2 = (22/7) * (49/4) = 77 / 2 = 38.5 cm^2. The total area of all four circles is: Total Circle Area = 4 * 38.5 = 154 cm^2. The area of the shaded remaining portion is: Area(Shaded) = Area(Square) - Total Circle Area = 196 - 154 = 42 cm^2.

Solution 2: The square has side length 4 cm, so Area(Square) = 4^2 = 16 cm^2. From the four corners, four quadrants of radius r = 1 cm are cut. The combined area of four quadrants is equal to the area of one complete circle of radius 1 cm: Area(4 Quadrants) = pi * (1)^2 = pi cm^2. In the center, a circle of diameter 2 cm (radius r = 1 cm) is cut: Area(Center Circle) = pi * (1)^2 = pi cm^2. Total removed area = pi + pi = 2*pi cm^2. Using pi = 22/7: Removed Area = 2 * (22/7) = 44/7 cm^2. The area of the remaining portion of the square is: Area(Remaining) = 16 - 44/7 = (112 - 44) / 7 = 68 / 7 = 9.71 cm^2.

Solution 3: In equilateral triangle ABC of side a = 12 cm, each interior angle is 60 degrees. The area of the equilateral triangle is: Area(triangle) = (sqrt(3)/4) * a^2 = (sqrt(3)/4) * 12^2 = (sqrt(3)/4) * 144 = 36 * sqrt(3) = 36 * 1.732 = 62.352 cm^2. At each of the three vertices, an arc of radius r = 6 cm subtending 60 degrees is drawn. Each arc forms a sector with central angle 60° and radius 6 cm. Area of one sector = (60 / 360) * pi * r^2 = (1/6) * 3.14 * 6^2 = (1/6) * 3.14 * 36 = 6 * 3.14 = 18.84 cm^2. The total area of all three sectors is: Total Sector Area = 3 * 18.84 = 56.52 cm^2. The shaded central region enclosed between the three arcs is the triangle area minus the three sectors: Area(Enclosed) = 62.352 - 56.52 = 5.832 cm^2.