Question

ABC is a right triangle, right-angled at C. If BC = a,CA = b, AB = c and p be the length of perpendicular from C on AB, then
(i) $cp=ab$
(ii) $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

• A. True
• B. False

Answer 1

The correct option is A True

Let $CD\perp AB$. Then, $CD=p$
$\therefore$ Area of $\Delta ABC=\frac{1}{2}(Base\times Height)$
$=\frac{1}{2}(AB\times CD)=\frac{1}{2}cp$
Also,
Area of $\Delta ABC=\frac{1}{2}(BC\times AC)=\frac{1}{2}ab$
$\therefore \frac{1}{2}cp=\frac{1}{2}ab$
$\Rightarrow cp=ab$
(ii) Since $\Delta ABC$ is a right triangle, right angled at C.
$\therefore {AB}^2={BC}^2+{AC}^2$
$\Rightarrow c^2=a^2+b^2$
$\Rightarrow {\left(\frac{ab}{p}\right)}^2=a^2+b^2[\because cp=ab\Rightarrow c=\frac{ab}{p}]$
$\Rightarrow \frac{a^2{b}^2}{p^2}=a^2+b^2$
$\Rightarrow \frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2}$
$\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$