Question

In figure, there are shown sectors of radii 7 cm and 3.5 cm. Find the area and perimeter of the shaded region ABCD. 

Hint:

To find the area of a sector, use the formula \( \frac{\theta}{360^\circ} \times \pi r^2 \) and for the perimeter, sum the arc lengths and straight lines.

Answer 1

Step 1: Understand the given values.}
We are given two sectors with radii \( OA = 7 \, \text{cm} \) and \( OB = 3.5 \, \text{cm} \). The central angle for both sectors is \( 30^\circ \).
Step 2: Find the area of sector \( OAC \).}
The formula for the area of a sector is: \[ \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 \] For sector \( OAC \) with radius \( 7 \, \text{cm} \) and central angle \( 30^\circ \): \[ \text{Area of sector OAC} = \frac{30^\circ}{360^\circ} \times \pi \times 7^2 \] \[ \text{Area of sector OAC} = \frac{1}{12} \times \pi \times 49 = \frac{49\pi}{12} \, \text{cm}^2 \]
Step 3: Find the area of sector \( OBD \).}
For sector \( OBD \) with radius \( 3.5 \, \text{cm} \) and central angle \( 30^\circ \): \[ \text{Area of sector OBD} = \frac{30^\circ}{360^\circ} \times \pi \times 3.5^2 \] \[ \text{Area of sector OBD} = \frac{1}{12} \times \pi \times 12.25 = \frac{12.25\pi}{12} \, \text{cm}^2 \]
Step 4: Find the area of the shaded region ABCD.}
The area of the shaded region \( ABCD \) is the area of sector \( OAC \) minus the area of sector \( OBD \): \[ \text{Area of shaded region} = \frac{49\pi}{12} - \frac{12.25\pi}{12} \] \[ \text{Area of shaded region} = \frac{(49 - 12.25)\pi}{12} = \frac{36.75\pi}{12} \] \[ \text{Area of shaded region} = 3.0625\pi \, \text{cm}^2 \] Using \( \pi \approx 3.14 \): \[ \text{Area of shaded region} \approx 3.0625 \times 3.14 = 9.62 \, \text{cm}^2 \]

Step 5: Find the perimeter of the shaded region ABCD.}
The perimeter of the shaded region is the sum of the lengths of the two arcs \( AB \) and \( BD \), plus the straight line \( AD \). - Length of arc \( AB \) for sector \( OAC \): \[ \text{Length of arc AB} = \frac{30^\circ}{360^\circ} \times 2\pi \times 7 = \frac{1}{12} \times 2\pi \times 7 = \frac{14\pi}{12} = \frac{7\pi}{6} \, \text{cm} \] - Length of arc \( BD \) for sector \( OBD \): \[ \text{Length of arc BD} = \frac{30^\circ}{360^\circ} \times 2\pi \times 3.5 = \frac{1}{12} \times 2\pi \times 3.5 = \frac{7\pi}{12} \, \text{cm} \] - The length of the straight line \( AD \) is the difference between the radii: \[ AD = 7 - 3.5 = 3.5 \, \text{cm} \] Thus, the perimeter is: \[ \text{Perimeter of shaded region} = \frac{7\pi}{6} + \frac{7\pi}{12} + 3.5 \] \[ \text{Perimeter of shaded region} = \frac{14\pi}{12} + \frac{7\pi}{12} + 3.5 = \frac{21\pi}{12} + 3.5 = \frac{7\pi}{4} + 3.5 \] Using \( \pi \approx 3.14 \): \[ \text{Perimeter of shaded region} = \frac{7 \times 3.14}{4} + 3.5 = \frac{21.98}{4} + 3.5 = 5.495 + 3.5 = 8.995 \, \text{cm} \] % Final Answer Final Answer:
The area of the shaded region is approximately \( 9.62 \, \text{cm}^2 \) and the perimeter is approximately \( 9 \, \text{cm} \).