Question

AB and CD are diameters of a circle with centre O and radius 7 cm. If \(\angle BOD = 30^\circ\), then find the area and perimeter of the shaded region.

Answer 1

Given:
- Radius, \(r = 7\) cm.
- \(AB\) and \(CD\) are diameters.
- \(\angle BOD = 30^\circ\).
- \(\angle AOC = \angle BOD = 30^\circ\) (vertically opposite angles).
- Shaded regions: sector \(BOD\) and sector \(AOC\).

Step 1: Area of one sector
Area of one sector: \[\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2\] Substitute: \[\text{Area of sector } BOD = \frac{30^\circ}{360^\circ} \times \pi \times (7)^2 = \frac{1}{12} \times \frac{22}{7} \times 49 = \frac{1}{12} \times 22 \times 7 = \frac{154}{12} = \frac{77}{6} \text{ cm}^2\]

Step 2: Total shaded area
Sector \(AOC\) area: \(\frac{77}{6}\) cm\(^2\).
Total shaded area = Area of sector \(BOD\) + Area of sector \(AOC\): \[2 \times \frac{77}{6} = \frac{77}{3} \text{ cm}^2\] Numerical value: \[\frac{77}{3} \approx 25.67 \text{ cm}^2\]

Step 3: Arc length of one sector
Arc length: \[l = \frac{\theta}{360^\circ} \times 2 \pi r\] Substitute: \[\text{Arc length } BD = \frac{30^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 = \frac{1}{12} \times 2 \times 22 = \frac{44}{12} = \frac{11}{3} \text{ cm}\]

Step 4: Total arc length
Arc length \(AC\): \(\frac{11}{3}\) cm.
Total arc length = \(\frac{11}{3} + \frac{11}{3} = \frac{22}{3}\) cm.

Step 5: Sum of radii
Radii: \(OA = OB = OC = OD = 7\) cm.
Sum of radii = \(7 + 7 + 7 + 7 = 28\) cm.

Step 6: Perimeter of shaded region
\[\text{Perimeter} = \text{Total arc length} + \text{Sum of radii} = \frac{22}{3} + 28 = \frac{22 + 84}{3} = \frac{106}{3} \text{ cm}\] Numerical value: \[\frac{106}{3} \approx 35.33 \text{ cm}\]

Final Answer:
\[\boxed{\text{Area} = \frac{77}{3} \text{ cm}^2, \quad \text{Perimeter} = \frac{106}{3} \text{ cm}}\]