ABC is a right angled triangled, right angled at C and P is the length of the perpendicular from C on AB. If a, b and c are the length of the sides BC, CA and AB respectively, then

Solution(By Examveda Team)

According to question,
ACB is a right angle triangle
∴ area of ΔACB

$$\frac{1}{2}$$ × AC × BC = $$\frac{1}{2}$$ × AB × PC
$$\frac{1}{2}$$ × b × a = $$\frac{1}{2}$$ × c × p
c = $$\frac{{ab}}{p}$$ . . . . . . . (i)
By using pythagoras theorem
AB² = AC² + BC²
c² = b² + a² . . . . . . . . . . (ii)
Put the value of C in equation (ii)
$$\eqalign{ & {\left( {\frac{{ab}}{p}} \right)^2} = {a^2} + {b^2} \cr & \Rightarrow \frac{{{a^2}{b^2}}}{{{p^2}}} = {a^2} + {b^2} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{{{a^2}}}{{{a^2}{b^2}}} + \frac{{{b^2}}}{{{a^2}{b^2}}} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr} $$

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Alternate :
From figure,
$$\eqalign{ & P = \frac{{ab}}{c} \cr & P = \frac{{ab}}{{\sqrt {{a^2} + {b^2}} }}\,\left( {\because {a^2} + {b^2} = {c^2}} \right) \cr & {P^2} = \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}} \cr & \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr} $$