| ABC, with right angled at C, |
| AB = c, BC =- a, CA = b and |
| CL=p CL is perpendicular to AB. |
| Which of the following is true? |
A) \[\frac{1}{p}=\frac{a}{bc}+\frac{b}{ca}\]
B) \[\frac{1}{p}=\frac{b}{ac}+\frac{c}{ab}\]
C) \[\frac{1}{p}=\frac{c}{ab}+\frac{a}{bc}\]
D) \[\frac{1}{p}=\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\]
Correct Answer: A
Solution :
[a] \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\]. Area of the triangle \[ABC=\frac{1}{2}-BC\text{ }\times \text{ }AC\] = \[\frac{1}{2}\]ab .... (i) Also, area of the triangle \[ABC=\frac{1}{2}\,\,\,CL\text{ }\times \text{ }AB\] = \[\frac{1}{2}\], .pc ....(ii) \[\therefore \] from (i) and (ii) we have, ab =pc or \[\frac{1}{p}=\frac{c}{ab}=\frac{{{c}^{2}}}{c(ab)}\] \[\Rightarrow \frac{1}{p}=\frac{{{a}^{2}}+{{b}^{2}}}{c(ab)}=\frac{1}{c}\left( \frac{a}{b}+\frac{b}{a} \right)\] \[=\frac{a}{bc}+\frac{b}{ac}\]
You need to login to perform this action.
You will be redirected in
3 sec
