Question

ABC Is right triangle, right angled at C. If p is the length of the perpendicular from C to AB and a, b, c have the usual meaning, then $\frac{1}{a^2}+\frac{1}{b^2}$ is ____.

• A. $1/p^2$
• B. $2/p^2$
• C. $p^2$
• D. $2p^2$

Answer 1

The correct option is A $1/p^2$

Say $\left|AB\right|=c\left|AC\right|=b\left|BC\right|=a\left|CD\right|=p$

Now, area of $△ABC=\frac{1}{2}\times$base$\times$height

$=\frac{1}{2}\times BC\times AC$

$=\frac{1}{2}ab$........(i)

Area of $△ABC=\frac{1}{2}\times AB\times CD$

$=\frac{1}{2}cp$........(ii)

$\therefore \frac{1}{2}ab=\frac{1}{2}cp=>ab=cp$

Now using Pythagoras theorem,

${AB}^2={BC}^2+{AC}^2=>c^2=a^2+b^2$

Substitute $c=\frac{ab}{p}$

$\therefore \frac{{\left(ab\right)}^2}{p^2}=a^2+b^2=>\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$