MP Board · Class 10 · Mathematics · Areas Related to CirclesFind the area of the design in the given figure, which is formed by two quadrants of circles of radius 8 cm each, inscribed inside a square of side 8 cm. (Use $\pi = \frac{22}{7}$)
Given Data:
- Side of the square ABCD ($a$) = $8$ cm
- Radius of each quadrant of the circle ($r$) = $8$ cm
- The design is formed by the intersection of two quadrants originating from opposite vertices (say B and D) of the square.
Step 1: Area of the Square\nThe formula for the area of a square is:
$$\text{Area of square} = \text{side}^2$$ $$\text{Area of square} = 8 \times 8 = 64\text{ cm}^2$$
Step 2: Area of One Quadrant\nA quadrant is one-fourth of a full circle. Therefore, its area is given by:
$$\text{Area of one quadrant} = \frac{1}{4} \times \pi r^2$$ $$\text{Area of one quadrant} = \frac{1}{4} \times \frac{22}{7} \times (8)^2$$ $$\text{Area} = \frac{1}{4} \times \frac{22}{7} \times 64 = \frac{22 \times 16}{7} = \frac{352}{7}\text{ cm}^2$$
Step 3: Area of Two Quadrants
$$\text{Area of two quadrants} = 2 \times \frac{352}{7} = \frac{704}{7}\text{ cm}^2$$
Step 4: Understanding the Intersection (The Design Region)\nWhen two quadrants of radius equal to the side of the square are drawn from adjacent vertices, their union covers the square plus an overlapping region. Alternatively, if drawn from opposite corners, the area of the design (shaded region representing the intersection of two quadrants) can be calculated using the principle of inclusion-exclusion:
$$\text{Area of design} = (\text{Area of Sector 1} + \text{Area of Sector 2}) - \text{Area of Square}$$ $$\text{Area of design} = \frac{352}{7} + \frac{352}{7} - 64$$ $$\text{Area of design} = \frac{704}{7} - 64$$ $$\text{Area of design} = \frac{704 - (64 \times 7)}{7}$$ $$\text{Area of design} = \frac{704 - 448}{7} = \frac{256}{7}\text{ cm}^2$$