Q10:

From a circular card sheet of radius $14$ cm, two circles of radius $3.5$ cm and a rectangle of length $3$ cm and breadth $1$cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take $\pi = \frac{22}{7}$)

Step 1: Given Variables & Initial Setup

To determine the area of the remaining card sheet, we must first establish the dimensions of the original geometric figure and the components that are removed. The problem provides the following parameters:

• Radius of the original large circular sheet, $R = 14 \text{ cm}$
• Radius of the two smaller removed circles, $r = 3.5 \text{ cm}$
• Length of the removed rectangle, $l = 3 \text{ cm}$
• Breadth of the removed rectangle, $b = 1 \text{ cm}$
• Constant for Pi, $\pi = \frac{22}{7}$

Step 2: Visual Representation of the Geometric Configuration

The following scale-accurate diagram illustrates the original circular sheet with the two smaller circles and the rectangle removed. [The diagram uses a scale factor of $10\times$ for visual clarity, where $14 \text{ cm}$ is represented by $140 \text{ units}$].

Step 3: Calculating the Area of the Original Circular Sheet

The area of a circle is given by the formula $A = \pi r^2$. Applying this to the large circular sheet:

$A_{\text{total}} = \pi R^2$
$A_{\text{total}} = \frac{22}{7} \times (14)^2$
$A_{\text{total}} = \frac{22}{7} \times 196$
$A_{\text{total}} = 22 \times 28$
$A_{\text{total}} = 616 \text{ cm}^2$

Step 4: Calculating the Area of the Removed Components

[Per the geometric principle of area addition, we must calculate the individual areas of all removed disjoint regions and sum them].

1. Area of the Two Small Circles:
First, we find the area of one small circle. To simplify the calculation, we can express the radius $3.5 \text{ cm}$ as the fraction $\frac{7}{2} \text{ cm}$.

$A_{\text{small\_circle}} = \pi r^2$
$A_{\text{small\_circle}} = \frac{22}{7} \times \left(\frac{7}{2}\right)^2$
$A_{\text{small\_circle}} = \frac{22}{7} \times \frac{49}{4}$
$A_{\text{small\_circle}} = \frac{22 \times 7}{4} = \frac{154}{4} = 38.5 \text{ cm}^2$

Since two identical circles are removed, their combined area is:

$A_{\text{both\_circles}} = 2 \times 38.5 \text{ cm}^2 = 77 \text{ cm}^2$

2. Area of the Rectangle:
The area of a rectangle is the product of its length and breadth ($A = l \times b$).

$A_{\text{rectangle}} = 3 \text{ cm} \times 1 \text{ cm} = 3 \text{ cm}^2$

3. Total Removed Area:
Summing the areas of the removed components yields:

$A_{\text{removed}} = A_{\text{both\_circles}} + A_{\text{rectangle}}$
$A_{\text{removed}} = 77 \text{ cm}^2 + 3 \text{ cm}^2 = 80 \text{ cm}^2$

Step 5: Calculating the Area of the Remaining Sheet

[By the axiom of area subtraction, the area of a region remaining after removing sub-regions is equal to the area of the original region minus the total area of the removed sub-regions].

$A_{\text{remaining}} = A_{\text{total}} - A_{\text{removed}}$
$A_{\text{remaining}} = 616 \text{ cm}^2 - 80 \text{ cm}^2$
$A_{\text{remaining}} = 536 \text{ cm}^2$

Final Solution: The area of the remaining sheet is $536 \text{ cm}^2$.