Arranging $3\sqrt{9}, 4\sqrt{20}, 6\sqrt{25}$ in ascending order

Question

How to arrange $3\sqrt{9}, 4\sqrt{20}, 6\sqrt{25}$ in ascending order?

Answer 1

HINT: use that $$3\sqrt{9}=9,4\sqrt{20}=8\sqrt{5},6\sqrt{25}=30$$

Answer 2

$3\sqrt{9}=\sqrt{9}\sqrt{9}=\sqrt{81}$

$4\sqrt{20}=\sqrt{16}\sqrt{20}=\sqrt{320}$

$6\sqrt{25}=\sqrt{36}\sqrt{25}=\sqrt{900}$

Now, $81<320<900$, Then $\sqrt{81}<\sqrt{320}<\sqrt{900}$

Hence, $3\sqrt{9}<4\sqrt{20}<6\sqrt{25}\space\space\space\blacksquare$