Areas Related to Circles: High-Yield Shaded Region Problems Class 10

In geometry examinations, composite planar area problems are visual puzzles that test your ability to decompose complex artistic figures into standard geometric primitives—sectors, segments, squares, equilateral triangles, and semicircles. In CBSE Class 10 Mathematics, Chapter 11 (Areas Related to Circles) carries $6$ to $8$ marks, featuring heavily in Section C (3 marks) and Section D (4 to 5 marks).

Whether calculating the cost of embroidering the borders of a hexagonal table cover or proving that a shaded crescent between a circular quadrant and a semicircle equals the area of a right triangle, mastering algebraic factorization eliminates tedious multi-digit arithmetic.

In this master guide, we solve the three most famous shaded region problems in the Class 10 syllabus.

What You Will Learn

• Strategic 3-step geometric decomposition framework
• Master Problem 1: The Round Table Cover with Six Symmetrical Designs (NCERT Classic)
• Master Problem 2: The Four Mutually Touching Circles Inscribed in a Square ($42\text{ cm}^2$ problem)
• Master Problem 3: The Lune of Hippocrates (Quadrant and Semicircle Crescent Proof)
• Factorization techniques and avoiding recurring decimal rounding errors

1. The 3-Step Decomposition Framework

    Step 1: Identify Outer Parent Shape ───> Square, Circle, or Sector.
    Step 2: Identify Inner Subtracted Shapes ─> Segments, Quadrants, or Triangles.
    Step 3: Combine Algebraically FIRST ────> Shaded Area = Area(Outer) - Area(Inner).
                                             Factor common π and r before computing numbers!

2. High-Yield Solved Board Examination Problems

Problem 1: The Round Table Cover with Six Symmetrical Designs (5-Mark Heavyweight)

Problem: A round table cover has six equal designs as shown in the figure. If the radius of the cover is $28\text{ cm}$, find the cost of making the designs at the rate of ₹$0.35\text{ per cm}^2$. (Use $\sqrt{3} = 1.7$).

                                      * * * *
                                  *             *
                                *    A-------B    *  <-- Minor Segment Design 1
                               *    / \     / \    *
                              *    F   \ O /   C    *  Radius r = 28 cm
                               *    \  /   \  /    *
                                *    E-------D    *
                                  *             *
                                      * * * *
                         Regular Hexagon inscribed in Circle!
                         Central Angle for each sector θ = 360° / 6 = 60°

Step-by-Step Solution:

1. Analyze the Symmetrical Geometry:
   • The circular cover has radius $r = 28\text{ cm}$.
   • The six designs are identical minor segments formed by a regular hexagon inscribed inside the circle.
   • The central angle for each sector is: $\theta = \frac{360^\circ}{6} = \mathbf{60^\circ}$
2. Area of ONE Design (One Minor Segment): $\text{Area of 1 Design} = \text{Area of Sector } (60^\circ) - \text{Area of Equilateral } \Delta OAB$
   • Since $\theta = 60^\circ$ and $OA = OB = r$, $\Delta OAB$ is an equilateral triangle! $\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 28 \times 28 = \frac{1}{6} \times 22 \times 4 \times 28 = \frac{2464}{6} = \mathbf{\frac{1232}{3}\text{ cm}^2}$ $\text{Area of } \Delta OAB = \frac{\sqrt{3}}{4} r^2 = \frac{1.7}{4} \times 28 \times 28 = 1.7 \times 7 \times 28 = \mathbf{333.2\text{ cm}^2}$
3. Area of ONE Segment: $\text{Area of 1 Design} = \frac{1232}{3} - 333.2 = \frac{1232 - 999.6}{3} = \mathbf{\frac{232.4}{3}\text{ cm}^2}$
4. Total Area of All 6 Designs: $\text{Total Area} = 6 \times \left( \frac{232.4}{3} \right) = 2 \times 232.4 = \mathbf{464.8\text{ cm}^2}$ (Notice how multiplying by 6 cancelled the denominator 3 cleanly!)
5. Calculate Total Cost at ₹$0.35$ per $\text{cm}^2$: $\text{Cost} = 464.8 \times 0.35 = \mathbf{₹\,162.68}$
6. Therefore, <u>the total cost of making the designs is ₹$162.68$</u>.

Problem 2: Four Touching Circles on a Square (The $42 ext{ cm}^2$ Benchmark)

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      /   |  Side a = 14 cm
                         |    +---------------+    |  Radius r = 7 cm
                         |   /     REGION      \   |
                         |  /                   \  |
                         D (Circle) ------------- C (Circle)

Step-by-Step Solution:

1. Analyze Dimensions:
   • Side of square $a = 14\text{ cm}$.
   • Since adjacent circles touch externally: Radius $r = \frac{14}{2} = \mathbf{7\text{ cm}}$.
2. Formulate Strategy: The square contains four corner angles of $90^\circ$ each. $\text{Shaded Area} = \text{Area of Square } ABCD - 4 \times (\text{Area of Quadrant})$
3. Observe the Geometric Combination: Four quadrants of the same radius $r = 7\text{ cm}$ combine to form one complete circle: $4 \times \left( \frac{1}{4}\pi r^2 \right) = \pi r^2$ $\text{Shaded Area} = a^2 - \pi r^2$
4. Substitute Values: $\text{Area of Square} = 14^2 = 196\text{ cm}^2$ $\text{Area of Full Circle} = \frac{22}{7} \times 7 \times 7 = 22 \times 7 = 154\text{ cm}^2$ $\text{Shaded Area} = 196 - 154 = \mathbf{42\text{ cm}^2}$
5. Therefore, <u>the area of the shaded region is $42\text{ cm}^2$</u>.

Problem 3: The Lune of Hippocrates (Quadrant & Semicircle Crescent)

Problem: In the figure, $ABCP$ is a quadrant of a circle of radius $14\text{ cm}$ and a semicircle is drawn with $BC$ as diameter. Find the area of the shaded region.

                           A
                           |                           |                      14 cm |  \  <-- Quadrant Arc BPC
                           |                              +----+
                           B 14 C
                             \  /
                              ~~  <-- Semicircle on BC as diameter (SHADED CRESCENT!)

Step-by-Step Solution:

1. In Right-Angled Triangle $\Delta ABC$ ($\angle A = 90^\circ$): $BC = \sqrt{14^2 + 14^2} = \sqrt{2 \times 14^2} = \mathbf{14\sqrt{2}\text{ cm}}$
   • Diameter of semicircle $= 14\sqrt{2}\text{ cm} \implies$ Radius $R = \mathbf{7\sqrt{2}\text{ cm}}$.
2. Formulate the Shaded Area Equation: $\mathbf{\text{Shaded Area} = \text{Area of Semicircle on } BC - \text{Area of Segment } BPC}$
3. Calculate Component Areas:
   • $\text{Area of Segment } BPC = \text{Area of Quadrant } ABCP - \text{Area of } \Delta ABC$ $\text{Area of Quadrant} = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = \mathbf{154\text{ cm}^2}$ $\text{Area of } \Delta ABC = \frac{1}{2} \times 14 \times 14 = \mathbf{98\text{ cm}^2}$ $\text{Area of Segment } BPC = 154 - 98 = \mathbf{56\text{ cm}^2}$
   • $\text{Area of Semicircle on } BC = \frac{1}{2}\pi R^2 = \frac{1}{2} \times \frac{22}{7} \times (7\sqrt{2})^2 = \frac{1}{2} \times \frac{22}{7} \times 98 = \mathbf{154\text{ cm}^2}$.
4. Calculate Shaded Area: $\text{Shaded Area} = 154 - 56 = \mathbf{98\text{ cm}^2}$

   The Astonishing Result: <u>The area of the shaded crescent ($98 ext{ cm}^2$) is EXACTLY EQUAL to the area of the right-angled triangle $\Delta ABC$ ($98 ext{ cm}^2$)!</u>

3. Summary and Examination Tips

Exam Tip: In the table cover problem, NEVER convert $\frac{1232}{3}$ or $\frac{232.4}{3}$ into rounded decimals! Keep the fraction over $3$; when you multiply by $6$ designs at the end, the fraction cancels completely: $6 \times \frac{232.4}{3} = 2 \times 232.4 = 464.8$!

Common Mistake: In the Lune problem, subtracting the triangle from the semicircle directly. The shaded region is bounded between the semicircle and the circular arc of the quadrant; you must subtract the segment area, not the triangle!