We are given a circle with center $O$ and radius $7$ cm. $AB$ and $CD$ are diameters, and $\angle BOD = 30^\circ$. We need to find the area and perimeter of the shaded region.
### Step 1: Area of Sector $BOD$
The area of a sector is given by:
$ \displaystyle A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 $
Substituting values:
$ \displaystyle A_{\text{sector}} = \frac{30^\circ}{360^\circ} \times 3.14 \times (7)^2 $
$ \displaystyle = \frac{1}{12} \times 3.14 \times 49 $
$ \displaystyle = \frac{153.86}{12} = 12.82 \text{ cm}^2 $
### Step 2: Area of $\triangle BOD$
Since $\angle BOD = 30^\circ$, $\triangle BOD$ is an isosceles triangle with sides $OB = OD = 7$ cm. The area of a triangle is:
$ \displaystyle A_{\triangle} = \frac{1}{2} \times OB \times OD \times \sin 30^\circ $
Since $\sin 30^\circ = \frac{1}{2}$:
$ \displaystyle A_{\triangle} = \frac{1}{2} \times 7 \times 7 \times \frac{1}{2} $
$ \displaystyle = \frac{49}{4} = 12.25 \text{ cm}^2 $
### Step 3: Shaded Region Area
Shaded area = Area of sector $BOD$ - Area of $\triangle BOD$:
$ \displaystyle A_{\text{shaded}} = 12.82 - 12.25 = 0.57 \text{ cm}^2 $
### Step 4: Perimeter of Shaded Region
Perimeter consists of arc $BD$ and the two line segments $OB$ and $OD$.
Arc length is given by:
$ \displaystyle L_{\text{arc}} = \frac{\theta}{360^\circ} \times 2\pi r $
$ \displaystyle L_{\text{arc}} = \frac{30^\circ}{360^\circ} \times 2 \times 3.14 \times 7 $
$ \displaystyle = \frac{1}{12} \times 43.96 = 3.66 \text{ cm} $
Total perimeter:
$ \displaystyle P_{\text{shaded}} = OB + OD + \text{Arc } BD $
$ \displaystyle = 7 + 7 + 3.66 = 17.66 \text{ cm} $
Area of shaded region = $0.57$ cm$^2$
Perimeter of shaded region = $17.66$ cm.