Practice problems on sectors and circles with detailed mathematical solutions.
Problems & Solutions
Problem 1: Area of Sector with Arc Length in Terms of Radius
Problem Statement: 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 \(r\).
Solution:
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 \(r^2\).
Problem 2: Arc Length from Central Angle and Area
Problem Statement: A sector has a central angle of \(40^\circ\) and an area of \(20 \, \text{cm}^2\). Calculate the arc length of the sector.
Solution:
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).
Problem 3: Concentric Circles Perimeter and Area
Problem Statement: 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 \(ABCD\)
Solution:
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:
\[ \text{Area} = \frac{1}{2} \cdot 8^2 \cdot 1 - \frac{1}{2} \cdot 2^2 \cdot 1 \] \[ \text{Area} = 32 - 2 = 30 \, \text{cm}^2 \]