# √(3.5÷2)²+(3.25)² =?.............plz give me an easy way of solving this root

Please find below the solution to the asked query :

$\sqrt{{\left(\frac{3.5}{2}\right)}^{2}+{\left(3.25\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{35}{20}\right)}^{2}+{\left(\frac{325}{100}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{7}{4}\right)}^{2}+{\left(\frac{13}{4}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{49}{16}+\frac{169}{16}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{49+169}{16}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{218}}{4}=\frac{14.76}{4}\phantom{\rule{0ex}{0ex}}=3.69.$

$\sqrt{{\left(\frac{3.5}{2}\right)}^{2}+{\left(3.25\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{35}{20}\right)}^{2}+{\left(\frac{325}{100}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{7}{4}\right)}^{2}+{\left(\frac{13}{4}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{49}{16}+\frac{169}{16}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{49+169}{16}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{218}}{4}=\frac{14.76}{4}\phantom{\rule{0ex}{0ex}}=3.69.$

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