Find the square root of numbers by prime factorization of 81, 625 and 1024

$$81 = 3 \times 27$$ $$27 = 3 \times 9$$ $$9 = 3 \times 3$$

इसलिए, $81 = 3 \times 3 \times 3 \times 3 = 3^4$

$$\sqrt{81} = \sqrt{3 \times 3 \times 3 \times 3} = \sqrt{(3 \times 3) \times (3 \times 3)} = \sqrt{9 \times 9} = \sqrt{9^2} = 9$$ $$\sqrt{81} = \boxed{\text{9}}$$

$$625 = 5 \times 125$$ $$125 = 5 \times 25$$ $$25 = 5 \times 5$$

इसलिए, $625 = 5 \times 5 \times 5 \times 5 = 5^4$

$$\sqrt{625} = \sqrt{5 \times 5 \times 5 \times 5} = \sqrt{(5 \times 5) \times (5 \times 5)} = \sqrt{25 \times 25} = \sqrt{25^2} = 25$$ $$\sqrt{625} = \boxed{\text{25}}$$

$$1024 = 2 \times 512$$ $$512 = 2 \times 256$$ $$256 = 2 \times 128$$ $$128 = 2 \times 64$$ $$64 = 2 \times 32$$ $$32 = 2 \times 16$$ $$16 = 2 \times 8$$ $$8 = 2 \times 4$$ $$4 = 2 \times 2$$

इसलिए, $1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}$

$$\sqrt{1024} = \sqrt{2^{10}} = 2^{\frac{10}{2}} = 2^5$$ $$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$$ $$\sqrt{1024} = \boxed{\text{32}}$$