Trigonometric Equations Question 356
Question: If $ p_1,p_2,p_3 $ are altitudes of a triangle $ ABC $ from the vertices $ A,B,C $ and $ \Delta $ the area of the triangle, then $ p_1^{-2}+p_2^{-2}+p_3^{-2} $ is equal to
Options:
A) $ \frac{a+b+c}{\Delta } $
B) $ \frac{a^{2}+b^{2}+c^{2}}{4{{\Delta }^{2}}} $
C) $ \frac{a^{2}+b^{2}+c^{2}}{{{\Delta }^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- We have $ \frac{1}{2}ap_1=\Delta ,\frac{1}{2}bp_2=\Delta ,\frac{1}{2}cp_3=\Delta $
Þ $ p_1=\frac{2\Delta }{a},p_2=\frac{2\Delta }{b},p_3=\frac{2\Delta }{c} $ \ $ \frac{1}{p_1^{2}}+\frac{1}{p_2^{2}}+\frac{1}{p_3^{2}}=\frac{a^{2}+b^{2}+c^{2}}{4{{\Delta }^{2}}} $ .
BETA
