Triangles And Properties Of Triangle Question 25
Question: If, x, y and z are perpendiculars drawn on a, b and c, respectively, then the value of $ \frac{bx}{c}+\frac{cy}{a}+\frac{az}{b} $ will be
Options:
A) $ \frac{a^{2}+b^{2}+c^{2}}{2R} $
B) $ \frac{a^{2}+b^{2}+c^{2}}{R} $
C) $ \frac{a^{2}+b^{2}+c^{2}}{4R} $
D) $ \frac{2(a^{2}+b^{2}+c^{2})}{R} $
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Answer:
Correct Answer: A
Solution:
[a] Let area of triangle be $ \Delta $ , then according to question, $ \Delta =\frac{1}{2}ax=\frac{1}{2}by=\frac{1}{2}cz\therefore \frac{bx}{c}+\frac{cy}{a}+\frac{az}{b} $ $ =\frac{b}{c}( \frac{2\Delta }{a} )+\frac{c}{a}( \frac{2\Delta }{b} )+\frac{a}{b}( \frac{2\Delta }{c} ) $ $ =\frac{2\Delta (b^{2}+c^{2}+a^{2})}{abc} $ $ =\frac{2(a^{2}+b^{2}+c^{2})}{abc}.\frac{abc}{4R}=\frac{a^{2}+b^{2}+c^{2}}{2R} $
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