Question 78

If $$3 \sqrt[]{7}+\sqrt{343}=19.21$$, then find the value of $$\sqrt{252}+20\sqrt{7}$$

Solution

Given, $$3\sqrt{7}+\sqrt{343}=19.21$$

$$=$$> $$3\sqrt{7}+\sqrt{7^3}=19.21$$

$$=$$> $$3\sqrt{7}+7\sqrt{7}=19.21$$

$$=$$> $$10\sqrt{7}=19.21$$

$$=$$> $$\sqrt{7}=\frac{\ 19.21}{10}$$

$$\therefore\ \sqrt{252}+20\ \sqrt{7}=\sqrt{16\times7\ }+20\sqrt{7}=4\sqrt{7}+20\sqrt{7}=24\sqrt{7}=24\times\ \frac{\ 19.21}{10}=49.9$$

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