Question

A chord whose length is equal to the radius of a circle is drawn to divide the circle into two parts. If the radius of the circle is 42 cm, then what is the area of the smaller part (in \( cm^{2} \))?

• A. \( 21^{2}(\frac{\pi}{3}-\frac{\sqrt{3}}{2}) \)
• B. \( 42^{2}(\frac{\pi}{2}-\frac{\sqrt{3}}{4}) \)
• C. \( 21^{2}(\frac{\pi}{6}-\frac{\sqrt{3}}{4}) \)
• D. \( 42^{2}(\frac{\pi}{6}-\frac{\sqrt{3}}{4}) \)

Hint:

For a chord equal to the radius, the segment area is always \( r^2(\frac{\pi}{6} - \frac{\sqrt{3}}{4}) \). Memorizing this form saves derivation time.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks for the area of the minor segment of a circle where the chord length is equal to the radius. The radius is explicitly given as 42 cm. Step 2: Key Formula:
Area of a minor segment = Area of Sector - Area of the corresponding Triangle. \[ \text{Area} = r^2 \left( \frac{\pi \theta}{360^{\circ}} - \frac{\sin \theta}{2} \right) \] Step 3: Detailed Explanation:
1. Find the central angle (\(\theta\)): Since the chord length equals the radius, the triangle formed by the chord and the two radii is an equilateral triangle. Thus, the central angle \(\theta = 60^{\circ} = \frac{\pi}{3}\). 2. Calculate the Area: Substitute \( r = 42 \) and \(\theta = 60^{\circ}\) into the formula. \[ \text{Area} = \text{Area of Sector} - \text{Area of Equilateral Triangle} \] \[ \text{Area} = \left( \frac{60}{360} \pi r^2 \right) - \left( \frac{\sqrt{3}}{4} r^2 \right) \] \[ \text{Area} = \left( \frac{1}{6} \pi r^2 \right) - \left( \frac{\sqrt{3}}{4} r^2 \right) \] \[ \text{Area} = r^2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) \] Substituting \( r = 42 \): \[ \text{Area} = 42^{2}\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \] Step 4: Final Answer:
The correct expression is \( 42^{2}(\frac{\pi}{6}-\frac{\sqrt{3}}{4}) \).