Question

$$\begin{gathered} △\mathrm{ABC}\mathrm{is}\text{right-angled at C}.\mathrm{Let}\mathrm{BC}=a,\mathrm{CA}=b\text{and}\mathrm{AB}=c\mathrm{and}\mathrm{let}\text{'}p\text{'}\mathrm{be}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{perpendicular}\mathrm{from}\mathrm{C}\mathrm{on}\mathrm{AB}. \\ \mathrm{Prove}\mathrm{tha}\text{t}\left(\mathrm{i}\right)cp=ab \\ \left(\mathrm{ii}\right)\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} \end{gathered}$$

Answer 1

First, we draw a figure of ΔABC in which $\angle C$ = 90°, BC = a, CA = b, AB = c and p is the length of the perpendicular from C on AB.

(i)
We know that the area of a triangle = $\frac{1}{2}$ × Base × Height
Thus,
A (ΔABC) = $\frac{1}{2}$ × AB × CD = $\frac{1}{2}$ × c × p = $\frac{1}{2}$cp …(1)
Also,
A (ΔABC) = $\frac{1}{2}$ × AC × BC = $\frac{1}{2}$× b × a = $\frac{1}{2}$ ab …(2)
From equations (1) and (2), we get:
$\frac{1}{2}$cp = $\frac{1}{2}$ab $\Rightarrow$ cp = ab

Hence proved.

(ii)
On using Pythagoras' Theorem in Δ ABC, we get:
AB² = AC² + BC²
$\Rightarrow$ c² = b² + a²
$\Rightarrow$ c² = a² + b²
Since cp = ab or c =$\frac{ab}{p}$ ,
$$\begin{gathered} {\left(\frac{ab}{p}\right)}^2=a^2+b^2 \\ \Rightarrow \frac{a^2{b}^2}{p^2}=a^2+b^2 \\ \Rightarrow \frac{1}{p^2}=\frac{a^2+b^2}{a^2{b}^2} \\ \Rightarrow \frac{1}{p^2}=\frac{a^2}{a^2{b}^2}+\frac{b^2}{a^2{b}^2} \\ \Rightarrow \frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2} \\ \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} \end{gathered}$$

Hence proved.