Question

ABC is a right triangle, right angled at C. If p is the length of perpendicular from C to AB & a, b, c have usual meaning, then prove that
(i) pc = ab
(ii) $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

(i) Area of $\Delta ABC=\frac{1}{2}BC\times AC$

$=\frac{1}{2}\times a\times b=\frac{ab}{2}$3

$area=\frac{1}{2}\times CD\times AB$

$\frac{1}{2}\times p\times c=\frac{pc}{2}$

$\frac{ab}{2}=\frac{pc}{2}$

$ab=pc......\left(1\right)$

(ii) In $\Delta ABC$

$A{B}^2=A{C}^2+B{C}^2$

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

$\frac{1}{p^2}=\frac{b^2+a^2}{a^2+b^2}$

$\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$