Question

Find the value to three places of decimals of each of the following. It is given that $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236$ and $\sqrt{10}=3.162$.

(i) $\frac{2}{\sqrt{3}}$

(ii) $\frac{3}{\sqrt{10}}$

(iii) $\frac{\sqrt{5}+1}{\sqrt{2}}$

(iv) $\frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$

(v) $\frac{2+\sqrt{3}}{3}$

(vi) $\frac{\sqrt{2}-1}{\sqrt{5}}$

Answer 1

$\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236and\sqrt{10}=3.162$

(i) $\frac{2}{\sqrt{3}}=\frac{2\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\frac{2\sqrt{3}}{3}=\frac{2\times 1.732}{3}=\frac{3.464}{3}=1.154$

(ii) $\frac{3}{\sqrt{10}}=\frac{3\times \sqrt{10}}{\sqrt{10}\times \sqrt{10}}=\frac{3\sqrt{10}}{10}$$=\frac{3\left(3.162\right)}{10}=\frac{9.486}{10}=0.9486$

(iii) $\frac{\sqrt{5}+1}{\sqrt{2}}=\frac{(\sqrt{5}+1)\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}$$=\frac{\sqrt{10}+\sqrt{2}}{2}=\frac{3.162+1.414}{2}=\frac{4.576}{2}=2.288$

(iv) $\frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\frac{(\sqrt{10}+\sqrt{15})\sqrt{2}}{\sqrt{2}\times \sqrt{2}}$$=\frac{\sqrt{20}+\sqrt{30}}{2}=\frac{\sqrt{4\times 5}+\sqrt{10}\times \sqrt{3}}{2}=\frac{2\sqrt{5}+\sqrt{10}\times \sqrt{3}}{2}=\frac{2\left(2.236\right)+3.162\times 1.732}{2}=\frac{4.472+5.476}{2}=\frac{9.948}{2}=4.974$

(v) $\frac{2+\sqrt{3}}{3}=\frac{2+1.732}{3}=\frac{3.732}{3}=1.244$

(vi) $\frac{\sqrt{2}-1}{\sqrt{5}}=\frac{(\sqrt{2}-1)\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$$=\frac{\sqrt{10}-\sqrt{5}}{5}=\frac{3.162-2.236}{5}=\frac{0.926}{5}=0.185$