Areas of Combinations of Plane Figures: Semicircles and Designs for CBSE Class 10

While basic combinations pair squares with simple circular quadrants, advanced board examination questions present intricate, multi-layered geometric artwork: symmetrical floral table covers with hexagonal borders, running tracks flanked by semicircular curves, and crescent-shaped shaded regions trapped between overlapping semicircles and circular quadrants.

In CBSE Class 10 Mathematics, Chapter 11 (Areas Related to Circles), these advanced shaded design problems are standard 4-mark and 5-mark questions in Section D. Solving them requires strategic decomposition of shapes and applying exact algebraic cancellations.

What You Will Learn

• How to decompose complex decorative designs into basic geometrical components
• Problem Type 1: The Round Table Cover with Six Symmetrical Designs (NCERT Classic)
• Problem Type 2: Shaded region trapped between a Circular Quadrant and a Semicircle (The $98\text{ cm}^2$ Classic)
• Problem Type 3: The Athletic Running Track (Straight bounds and semicircular ends)
• Step-by-step proofs and algebraic cancellation shortcuts
• Presentation standards for full marks in 5-mark questions

1. Problem Type 1: The Round Table Cover with Six Symmetrical Designs

Problem Statement (NCERT Question 13 & CBSE Classic):

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    *  <-- Design 1
                               *    / \     / \    *
                              *    F   \ O /   C    *
                               *    \  /   \  /    *
                                *    E-------D    *
                                  *             *
                                      * * * *
                       6 Equal Designs around a Regular Hexagon!

Step-by-Step Solution:

Step 1: Geometric Decomposition

• The table cover is a circle with radius $r = 28\text{ cm}$.
• Connecting center $O$ to all six vertices ($A, B, C, D, E, F$) divides the central circle into 6 equal sectors: $\theta = \frac{360^\circ}{6} = \mathbf{60^\circ}$
• Each of the 6 designs is a minor segment corresponding to a central angle of $60^\circ$!

Step 2: Calculate Area of One Design (One Minor Segment)

$\text{Area of One Design} = \text{Area of Sector } (60^\circ) - \text{Area of Equilateral } \Delta OAB$

1. Area of Sector: $\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 28 \times 28$ $= \frac{1}{6} \times 22 \times 4 \times 28 = \frac{2464}{6} = \mathbf{\frac{1232}{3}\text{ cm}^2 = 410.67\text{ cm}^2}$

2. Area of Equilateral Triangle $\Delta OAB$: $\text{Area}(\Delta OAB) = \frac{\sqrt{3}}{4} r^2 = \frac{1.7}{4} \times 28 \times 28 = 1.7 \times 7 \times 28 = \mathbf{333.2\text{ cm}^2}$

3. Area of One Segment: $\text{Area of 1 Design} = \frac{1232}{3} - 333.2 = \frac{1232 - 999.6}{3} = \frac{232.4}{3}\text{ cm}^2$

Step 3: Calculate Total Area of All 6 Designs

$\text{Total Area} = 6 \times \left(\frac{232.4}{3}\right) = 2 \times 232.4 = \mathbf{464.8\text{ cm}^2}$

Step 4: Calculate Total Cost

$\text{Cost} = \text{Total Area} \times \text{Rate} = 464.8 \times 0.35 = \mathbf{₹\,162.68}$

Therefore, <u>the total cost of making the designs is ₹$162.68$</u>.

2. Problem Type 2: Shaded Region Between a Quadrant and Semicircle (CBSE 4-Mark Classic)

Problem Statement:

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:

Step 1: Analyze the Geometry

1. $ABCP$ is a quadrant of a circle with center $A$ and radius $r = 14\text{ cm}$.
2. In right-angled triangle $\Delta ABC$ ($\angle A = 90^\circ$): $BC = \sqrt{AB^2 + AC^2} = \sqrt{14^2 + 14^2} = \sqrt{2 \times 14^2} = \mathbf{14\sqrt{2}\text{ cm}}$
3. Diameter of semicircle $= BC = 14\sqrt{2}\text{ cm}$. Radius of semicircle $R = \frac{14\sqrt{2}}{2} = \mathbf{7\sqrt{2}\text{ cm}}$.

Step 2: Formulate the Shaded Region Equation

Notice from the diagram: $\mathbf{\text{Shaded Area} = \text{Area of Semicircle on } BC - \text{Area of Segment } BPC}$

Step 3: Calculate Area of Segment $BPC$ of the Quadrant

$\text{Area of Segment } BPC = \text{Area of Quadrant } ABCP - \text{Area of } \Delta ABC$

• $\text{Area of Quadrant} = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 22 \times 28 = \mathbf{154\text{ cm}^2}$.
• $\text{Area of } \Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = \mathbf{98\text{ cm}^2}$. $\text{Area of Segment } BPC = 154 - 98 = \mathbf{56\text{ cm}^2}$

Step 4: Calculate Area of Semicircle on $BC$

$\text{Area of Semicircle} = \frac{1}{2} \pi R^2 = \frac{1}{2} \times \frac{22}{7} \times (7\sqrt{2})^2 = \frac{1}{2} \times \frac{22}{7} \times (49 \times 2) = \frac{1}{2} \times \frac{22}{7} \times 98 = 11 \times 14 = \mathbf{154\text{ cm}^2}$

Step 5: Compute Final Shaded Area

$\text{Shaded Area} = \text{Area of Semicircle} - \text{Area of Segment } BPC$ $\text{Shaded Area} = 154 - 56 = \mathbf{98\text{ cm}^2}$

Important: <u>Notice the astonishing algebraic result! The area of the shaded crescent ($98 ext{ cm}^2$) is EXACTLY EQUAL to the area of the right-angled triangle $\Delta ABC$ (rac{1}{2} imes 14 imes 14 = 98 ext{ cm}^2)! This historical theorem is known as the Lune of Hippocrates.</u>

3. Problem Type 3: The Athletic Running Track (NCERT Classic)

An athletic track has straight parallel lines of length $106\text{ m}$ and semicircular ends with inner diameter $60\text{ m}$. The track is $10\text{ m}$ wide throughout.

       +------------------- 106 m -------------------+
      /                                               \  <-- Semicircular outer curve
     /  +----------------- 106 m -----------------+        |   |                                         |     |
    |   |                                         |     |
     \  +----------------- 106 m -----------------+    /
      \                                               /  <-- Semicircular outer curve
       +------------------- 106 m -------------------+

Key Calculations:

1. Inner Perimeter Distance: $\text{Distance} = 2 \times (\text{Length of Straight}) + 2 \times (\text{Inner Semicircle Arc})$ $\text{Distance} = 2(106) + 2\pi r = 212 + 2 \times \frac{22}{7} \times 30 = 212 + \frac{1320}{7} = \mathbf{\frac{2804}{7}\text{ m} = 400.57\text{ m}}$
2. Total Area of the Track: $\text{Area} = 2 \times (\text{Area of Straight Rectangles}) + \text{Area of Semicircular Ends Ring}$
   • Area of 2 rectangles $= 2 \times (106 \times 10) = \mathbf{2120\text{ m}^2}$.
   • The two semicircular ends combine to form a complete circular ring with inner radius $r = 30\text{ m}$ and outer radius $R = 30 + 10 = 40\text{ m}$: $\text{Ring Area} = \pi(R^2 - r^2) = \frac{22}{7}(40^2 - 30^2) = \frac{22}{7}(1600 - 900) = \frac{22}{7}(700) = \mathbf{2200\text{ m}^2}$
   • Total Track Area: $\text{Total Area} = 2120 + 2200 = \mathbf{4320\text{ m}^2}$

4. Summary and Examination Tips

Exam Tip: In the table cover problem, multiply by $6$ at the very end when using the segment fraction $\frac{232.4}{3}$. Multiplying by $6$ immediately cancels the denominator $3$ ($6/3 = 2$), avoiding recurring decimal errors!

Common Mistake: Forgetting to convert outer radius in track problems. If inner diameter is $60\text{ m}$, inner radius is $30\text{ m}$, and outer radius is $30 + 10 = 40\text{ m}$ (NOT $60 + 10 = 70\text{ m}$!).