Practice problems on sectors and circles with detailed mathematical solutions.
In the figure below, arc \( \overset{\frown}{AB} \) has length equal to twice the radius \( r \) of the circle. Find the area of sector \( OAB \) in terms of \( r \).
A sector has a central angle of \( 40^\circ \) and an area of \( 20 \text{ cm}^2 \). Calculate the arc length of the sector.
In the figure below, \( \overset{\frown}{AB} \) and \( \overset{\frown}{DC} \) are arcs of concentric circles with center \( O \). The perimeter of figure \( ABCD \) is 22 cm. Calculate:
a) Angle \( \angle AOB \) (in radians)
b) The area of figure \( ABCD \)
The arc length formula is:
\[ \text{arc length} = r \cdot \theta \]
Given \( \text{arc length} = 2r \):
\[ 2r = r \cdot \theta \implies \theta = 2 \text{ radians} \]
The sector area formula is:
\[ \text{Area} = \frac{1}{2} r^2 \theta \]
Substituting \( \theta = 2 \):
\[ \text{Area} = \frac{1}{2} r^2 \cdot 2 = r^2 \]
Thus, the area of sector \( OAB \) is \( r^2 \).
The sector area formula is:
\[ \text{Area} = \frac{1}{2} r^2 \theta \]
Convert \( 40^\circ \) to radians: \( 40^\circ \times \frac{\pi}{180} = \frac{2\pi}{9} \).
Given \( \text{Area} = 20 \):
\[ 20 = \frac{1}{2} r^2 \cdot \frac{2\pi}{9} \]
\[ 20 = \frac{\pi r^2}{9} \implies r^2 = \frac{180}{\pi} \implies r = \sqrt{\frac{180}{\pi}} \]
The arc length formula is:
\[ s = r \theta = \sqrt{\frac{180}{\pi}} \cdot \frac{2\pi}{9} \]
\[ s = \frac{2\pi}{9} \cdot \sqrt{\frac{180}{\pi}} = \frac{2}{9} \sqrt{180\pi} \]
\[ s = \frac{2}{9} \sqrt{36 \cdot 5\pi} = \frac{2}{9} \cdot 6 \sqrt{5\pi} = \frac{4\sqrt{5\pi}}{3} \]
Numerically: \( s \approx 5.3 \text{ cm} \) (to 1 decimal place).
Let \( \theta \) (in radians) be \( \angle AOB \). The perimeter is:
\[ \overset{\frown}{AB} + 6 + \overset{\frown}{DC} + 6 = 22 \]
Using arc length formulas:
\[ \overset{\frown}{AB} = 2\theta, \quad \overset{\frown}{DC} = 8\theta \]
Thus:
\[ 2\theta + 6 + 8\theta + 6 = 22 \]
\[ 10\theta + 12 = 22 \implies 10\theta = 10 \implies \theta = 1 \text{ radian} \]
Area of figure \( ABCD \) = Area of large sector − Area of small sector:
\[ \text{Area} = \frac{1}{2} \cdot 8^2 \cdot 1 - \frac{1}{2} \cdot 2^2 \cdot 1 \]
\[ \text{Area} = 32 - 2 = 30 \text{ cm}^2 \]