Areas of Combinations of Plane Figures: Squares and Circles for CBSE Class 10

In pure mathematics, we calculate the area of standard geometric shapes—circles, sectors, squares, and triangles—in isolation. However, in art, textile embroidery, tile paving, and landscape architecture, shapes rarely exist alone. They are combined into intricate composite designs where circles are inscribed in squares, quadrants are carved from corners, or overlapping circular arcs create shaded motifs.

In CBSE Class 10 Mathematics, Chapter 11 (Areas Related to Circles), combinations of plane figures represent the most visually engaging and common 3-mark and 4-mark board examination problems. Solving them requires analyzing which standard geometric shapes have been added or subtracted.

What You Will Learn

• The universal strategic principle for composite shaded region problems
• Problem Type 1: The Grazing Horse in a Square Grass Field (NCERT Classic)
• Problem Type 2: Four corner quadrants and a central circle cut from a square
• Problem Type 3: Four mutually touching circles circumscribed by a square
• Step-by-step algebraic methods to eliminate cumbersome multi-step multiplications
• Presentation standards to ensure full marks

1. The Universal Strategy for Shaded Regions

Every shaded area problem is a puzzle of geometric addition or subtraction:

    [ Area of Shaded Region ] = [ Area of Enclosing Outer Figure ] - [ Area of Unshaded Inner Figures ]
                                              OR
    [ Area of Shaded Region ] = [ Sum of Individual Overlapping Components ] - [ Double-Counted Overlaps ]

3-Step Problem Solving Rule:

1. Identify the Parent Geometric Shapes: Look past the complex shaded pattern to identify the basic shapes (squares, rectangles, circles, quadrants, triangles).
2. Determine the Operation (Add or Subtract): Write out a single plain-English equation (e.g., $\text{Shaded Area} = \text{Area of Square} - 4 \times \text{Area of Quadrant}$).
3. Factor Algebraically Before Calculating: Group common factors like $\pi$ or $r^2$ before performing arithmetic. This prevents compounding rounding errors!

2. High-Yield Solved Board Examination Problems

Solved Example 1: The Grazing Horse Problem (NCERT Classic)

Problem: A horse is tied to a peg at one corner of a square-shaped grass field of side $15\text{ m}$ by means of a $5\text{ m}$ long rope. Find: (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were $10\text{ m}$ long instead of $5\text{ m}$. (Use $\pi = 3.14$).

             Corner Peg P -------------------- 15 m --------------------+
                         | \                                            |
                     5 m |   \  <-- Grazing Area (Quadrant of radius 5m)|
                         |     \                                        |
                         +-------+                                      |
                         |                                              |
                         |                 Square Grass Field           |
                         +----------------------------------------------+

Solution:

1. Analyze the Geometry:

   • The field is a square, so the corner angle where the horse is tethered is strictly a right angle: $\theta = 90^\circ$.
   • The grazing boundary forms a quadrant of a circle (a sector of central angle $90^\circ$) with radius equal to the length of the rope.

2. (i) Grazing Area with $5\text{ m}$ Rope ($r_1 = 5\text{ m}$): $\text{Area}_1 = \frac{90^\circ}{360^\circ} \times \pi r_1^2 = \frac{1}{4} \times 3.14 \times 5^2$ $\text{Area}_1 = \frac{1}{4} \times 3.14 \times 25 = \frac{78.5}{4} = \mathbf{19.625\text{ m}^2}$

3. (ii) Grazing Area with $10\text{ m}$ Rope ($r_2 = 10\text{ m}$): $\text{Area}_2 = \frac{1}{4} \times \pi r_2^2 = \frac{1}{4} \times 3.14 \times 10^2 = \frac{314}{4} = \mathbf{78.5\text{ m}^2}$

4. Calculate the Increase in Grazing Area: $\text{Increase in Area} = \text{Area}_2 - \text{Area}_1 = 78.5 - 19.625 = \mathbf{58.875\text{ m}^2}$ (Alternatively: $\frac{1}{4}\pi(r_2^2 - r_1^2) = \frac{1}{4} \times 3.14 \times (100 - 25) = \frac{1}{4} \times 3.14 \times 75 = 58.875\text{ m}^2$).

5. Therefore:

   • <u>(i) The initial grazing area is $19.625\text{ m}^2$</u>.
   • <u>(ii) The increase in grazing area is $58.875\text{ m}^2$</u>.

Solved Example 2: Corner Quadrants Cut from a Square (NCERT Classic)

Problem: From each corner of a square of side $4\text{ cm}$, a quadrant of a circle of radius $1\text{ cm}$ is cut and also a circle of diameter $2\text{ cm}$ is cut as shown in the figure. Find the area of the remaining portion of the square. (Use $\pi = 22/7$).

                         +-----+-------------------+-----+
                         | Q1  |                   | Q2  |  <-- Quadrants (r = 1 cm)
                         +-----+                   +-----+
                         |               O               |
                         |         (Circle d = 2)        |  <-- Circle (r = 1 cm)
                         +-----+                   +-----+
                         | Q4  |                   | Q3  |
                         +-----+-------------------+-----+

Solution:

1. Dimensions Given:

   • Side of square $a = 4\text{ cm}$.
   • Radius of each of the 4 corner quadrants: $r = 1\text{ cm}$.
   • Diameter of central circle $d = 2\text{ cm} \implies$ Radius $r = 1\text{ cm}$.

2. Formulate the Strategy: $\text{Remaining Shaded Area} = \text{Area of Square} - [4 \times (\text{Area of Quadrant}) + \text{Area of Central Circle}]$

3. Calculate Component Areas:

   • $\text{Area of Square} = a^2 = 4 \times 4 = \mathbf{16\text{ cm}^2}$.
   • Notice that four quadrants of the same radius combine to form one complete circle: $4 \times \left(\frac{1}{4} \pi r^2\right) = \pi r^2$
   • The central circle also has radius $r = 1\text{ cm}$, so its area is $\pi r^2$.
   • Total removed area: $\text{Removed Area} = \pi r^2 + \pi r^2 = 2\pi r^2 = 2 \times \frac{22}{7} \times 1^2 = \mathbf{\frac{44}{7}\text{ cm}^2}$

4. Calculate Remaining Area: $\text{Remaining Area} = 16 - \frac{44}{7} = \frac{16 \times 7 - 44}{7} = \frac{112 - 44}{7} = \mathbf{\frac{68}{7}\text{ cm}^2 = 9.71\text{ cm}^2}$

5. Therefore, <u>the area of the remaining portion of the square is $\frac{68}{7}\text{ cm}^2$ (or $9.71\text{ cm}^2$)</u>.

Solved Example 3: Four Touching Circles on a Square

Problem: $ABCD$ is a square of side $14\text{ cm}$. With centers $A, B, C,$ and $D$, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region enclosed between the four circles.

                         A (Circle) ------------- B (Circle)
                         |  \                   /  |
                         |   \     SHADED      /   |
                         |    +---------------+    |
                         |   /     REGION      \   |
                         |  /                   \  |
                         D (Circle) ------------- C (Circle)

Solution:

1. Analyze the Geometry:

   • Side of square $AB = 14\text{ cm}$.
   • Since adjacent circles touch each other externally, the radius of each circle is half the side of the square: $r = \frac{14}{2} = \mathbf{7\text{ cm}}$
   • The square contains four corners of $90^\circ$ each.
   • Therefore, the region inside the square that is covered by the circles consists of four quadrants of radius $7\text{ cm}$.

2. Formulate Strategy: $\text{Shaded Area} = \text{Area of Square } ABCD - [4 \times \text{Area of a Quadrant}]$ $\text{Shaded Area} = \text{Area of Square } ABCD - \text{Area of One Full Circle}$

3. Compute Numerical Values:

   • $\text{Area of Square} = 14^2 = \mathbf{196\text{ cm}^2}$.
   • $\text{Area of Full Circle} = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 22 \times 7 = \mathbf{154\text{ cm}^2}$.

4. Calculate Shaded Area: $\text{Shaded Area} = 196 - 154 = \mathbf{42\text{ cm}^2}$

5. Therefore, <u>the area of the shaded region is $42\text{ cm}^2$</u>.

3. Summary and Examination Tips

Exam Tip: Whenever four identical quadrants are removed, always write: "Four quadrants of radius $r$ combine to form one complete circle of area $\pi r^2$". This simplifies calculations immediately!

Common Mistake: Calculating the horse's grazing area using the entire square ($15^2 = 225$). The length of the side of the field ($15 ext{ m}$) is extra information provided to show that the $5 ext{ m}$ or $10 ext{ m}$ rope fits inside the field; the horse cannot graze beyond the length of its rope!