Circles
Tangents to Circles
Grade 11
Question:
<p>PA and PB are two tangents drawn from point P to circle of radius 5. A line is drawn from point P which cuts circle at C and D such that PC = 5 and PD = 15 and \(\angle APB = \theta\). Find the area of \(\triangle APB\).</p>
<p>(a) \(\frac{25\sqrt{3}}{2}\)</p>
<p>(b) \(25\sqrt{3}\)</p>
<p>(c) \(\frac{75\sqrt{3}}{2}\)</p>
<p>(d) \(\frac{75\sqrt{3}}{4}\)</p>
Step-by-Step Solution
Key Concept: Use the tangent-radius relationship to find the angle, then calculate the area of the equilateral triangle formed.
<p><strong>Solution:</strong></p><p>Since \(\sin\frac{\theta}{2} = \frac{OA}{OP} = \frac{5}{10} = \frac{1}{2}\)</p><p>Therefore \(\frac{\theta}{2} = 30°\), so \(\theta = 60°\)</p><p>Thus \(\triangle APB\) is an equilateral triangle.</p><p>Area \(= \frac{\sqrt{3}}{4}(5\sqrt{3})^2 = \frac{\sqrt{3}}{4} \cdot 75 = 25\sqrt{3}\)</p>
Correct Answer: B