ABC is a triangle right angled at B. D is a point on AC such that BD is perpendicular to AC. If AB = p and BC = \sqrt{3}p, then what is BD equal to?
- A. p\sqrt{3}
- B. p\sqrt{2}
- C. \frac{\sqrt{3}p}{2} ✓
- D. \frac{\sqrt{3}p}{4}
Correct Answer: C. \frac{\sqrt{3}p}{2}
Explanation
In right \Delta ABC, AC = \sqrt{p^2 + (\sqrt{3}p)^2} = 2p. The area of the triangle is \frac{1}{2} \times p \times \sqrt{3}p = \frac{1}{2} \times 2p \times BD. Solving this gives BD = \frac{\sqrt{3}p}{2}.
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