# In this q what have we dine to PD in the last step. Is it rationalisation ?

Yes, in the last step rationalization is done,

$PD=\frac{8}{\sqrt{3}-1}\phantom{\rule{0ex}{0ex}}Rationalizing,\phantom{\rule{0ex}{0ex}}PD=\frac{8}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}PD=\frac{8\times \left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\phantom{\rule{0ex}{0ex}}\left[\mathbf{U}\mathbf{sin}\mathbf{g}\mathbf{}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\left(\mathbf{a}\mathbf{-}\mathbf{b}\right)\left(\mathbf{a}\mathbf{+}\mathbf{b}\right)\right]\phantom{\rule{0ex}{0ex}}PD=\frac{8\left(\sqrt{3}+1\right)}{3-1}\phantom{\rule{0ex}{0ex}}PD=\frac{8\left(\sqrt{3}+1\right)}{2}=4\left(\sqrt{3}+1\right)m.$

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