RRB Junior Engineer Stage 1 Model Papers 10

Here,
∠ACB = 90°
∠ADC = 90°
∠BDC = 90°
Triangles ACB, ADC and BDC are right angle triangles.
Here, Area of ∆ABC = Area of ∆ADC + Area of ∆BDC
$\Rightarrow \mathit{a}\mathit{b}=\mathit{P}\mathit{c}\Rightarrow \mathit{c}=\frac{\mathit{a}\mathit{b}}{\mathit{p}}$....(1)
Now, In ∆ABC,
${\mathit{C}}^2={\mathit{a}}^2+{\mathit{b}}^2{\left(\frac{\mathit{a}\mathit{b}}{\mathit{p}}\right)}^2={\mathit{a}}^2+{\mathit{b}}^2$
$\Rightarrow \frac{{\mathit{a}}^2{\mathit{b}}^2}{{\mathit{p}}^2}={\mathit{a}}^2+{\mathit{b}}^2$
$\frac{1}{{\mathit{p}}^2}=\frac{1}{{\mathit{a}}^2}+\frac{1}{{\mathit{b}}^2}$