Question:
In a triangle \(ABC\), the internal bisector of angle \(A\) intersects \(BC\) at \(D\). If \(AB = 15\) cm, \(AC = 20\) cm, and \(BC = 14\) cm, find the length of \(BD\).
एक त्रिभुज \(ABC\) में कोण \(A\) का आंतरिक समद्विभाजक \(BC\) को \(D\) पर काटता है। यदि \(AB = 15\) सेमी, \(AC = 20\) सेमी और \(BC = 14\) सेमी है, तो \(BD\) की लंबाई ज्ञात कीजिए।
Options:
- 6 cm
- 8 cm
- 7 cm
- 5.5 cm
- 6 सेमी
- 8 सेमी
- 7 सेमी
- 5.5 सेमी
Correct Answer: a)
Explanation:
By Angle Bisector Theorem:
\(\frac{BD}{DC} = \frac{AB}{AC} = \frac{15}{20} = \frac{3}{4}\).
Since \(BC = BD + DC = 14\) cm:
\(BD = \frac{3}{3 + 4} \times 14 = \frac{3}{7} \times 14 = 6\) cm.
\(\frac{BD}{DC} = \frac{AB}{AC} = \frac{15}{20} = \frac{3}{4}\).
Since \(BC = BD + DC = 14\) cm:
\(BD = \frac{3}{3 + 4} \times 14 = \frac{3}{7} \times 14 = 6\) cm.
कोण समद्विभाजक प्रमेय के अनुसार:
\(\frac{BD}{DC} = \frac{AB}{AC} = \frac{15}{20} = \frac{3}{4}\)
\(BD = \frac{3}{7} \times 14 = 6\) सेमी।
\(\frac{BD}{DC} = \frac{AB}{AC} = \frac{15}{20} = \frac{3}{4}\)
\(BD = \frac{3}{7} \times 14 = 6\) सेमी।