Find the square root of:

Solution:

(i) We know:

$\sqrt{\frac{441}{961}}=\frac{\sqrt{441}}{\sqrt{961}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{441}=\sqrt{(3 \times 3) \times(7 \times 7)}=3 \times 7=21$

$\sqrt{961}=\sqrt{31 \times 31}=31$

$\therefore \sqrt{\frac{441}{961}}=\frac{21}{31}$

(ii) We have:

$\sqrt{\frac{324}{841}}=\frac{\sqrt{324}}{\sqrt{841}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{324}=\sqrt{2 \times 2 \times 3 \times 3 \times 3 \times 3}=2 \times 3 \times 3=18$

$\sqrt{841}=\sqrt{29 \times 29}=29$

$\therefore \sqrt{\frac{324}{841}}=\frac{18}{29}$

(iii) By looking at the book's answer key, the fraction should be $\sqrt{4 \frac{29}{49}}$, not $\sqrt{4 \frac{29}{29}}$.

We know:

$\sqrt{4 \frac{29}{49}}=\sqrt{\frac{225}{49}}=\frac{\sqrt{225}}{\sqrt{49}}$

$\sqrt{225}=15$

$\sqrt{49}=7$

$\therefore \sqrt{4 \frac{29}{49}}=\frac{15}{7}$

(iv) We know:

$\sqrt{2 \frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{\sqrt{64}}{\sqrt{25}}=\frac{8}{5}$

(v) We know:

$\sqrt{2 \frac{137}{196}}=\sqrt{\frac{529}{196}}=\frac{\sqrt{529}}{\sqrt{196}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{529}=\sqrt{23 \times 23}=23$

$\sqrt{196}=\sqrt{2 \times 2 \times 7 \times 7}=2 \times 7=14$

$\therefore \sqrt{2 \frac{137}{196}}=\frac{23}{14}$

(vi) We know:

$\sqrt{23 \frac{26}{121}}=\sqrt{\frac{2809}{121}}=\frac{\sqrt{2809}}{\sqrt{121}}$

Now, let us compute the square roots of the numerator and the denominator separately.
'

$\sqrt{121}=11$

$\therefore \sqrt{23 \frac{26}{121}}=\frac{53}{11}$

(vii) We know:

$\sqrt{25 \frac{544}{729}}=\sqrt{\frac{18769}{729}}=\frac{\sqrt{18769}}{\sqrt{729}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{729}=27$

$\therefore \sqrt{25 \frac{544}{729}}=\frac{137}{27}$

(viii) We know:

$\sqrt{75 \frac{46}{49}}=\sqrt{\frac{3721}{49}}=\frac{\sqrt{3721}}{\sqrt{49}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{49}=7$

 

$\therefore \sqrt{75 \frac{46}{49}}=\frac{61}{7}$

(ix) We know:

$\sqrt{3 \frac{942}{2209}}=\sqrt{\frac{7569}{2209}}=\frac{\sqrt{7569}}{\sqrt{2209}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\therefore \sqrt{3 \frac{942}{2209}}=\frac{87}{47}$

(x) We know:

$\sqrt{3 \frac{334}{3025}}=\sqrt{\frac{9409}{3025}}=\frac{\sqrt{9409}}{\sqrt{3025}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\therefore \sqrt{3 \frac{334}{3025}}=\frac{97}{55}$

(xi) We know:

$\sqrt{21 \frac{2797}{3364}}=\sqrt{\frac{73441}{3364}}=\frac{\sqrt{73441}}{\sqrt{3364}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\therefore \sqrt{21 \frac{2797}{3364}}=\frac{271}{58}$

(xii) We know:

$\sqrt{38 \frac{11}{25}}=\sqrt{\frac{961}{25}}=\frac{\sqrt{961}}{\sqrt{25}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{961}=31$

$\sqrt{25}=5$

$\therefore \sqrt{38 \frac{11}{25}}=\frac{31}{5}$

(xiii) We know:

$\sqrt{23 \frac{394}{729}}=\sqrt{\frac{17161}{729}}=\frac{\sqrt{17161}}{\sqrt{729}}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{729}=27$

$\therefore \sqrt{23 \frac{394}{729}}=\frac{131}{27}=4 \frac{23}{27}$

(xiv) We know:

$\sqrt{21 \frac{51}{169}}=\sqrt{\frac{3600}{169}}=\frac{\sqrt{3600}}{169}$

Now, let us compute the square roots of the numerator and the denominator separately.

$\sqrt{3600}=\sqrt{60 \times 60}=60$

$\sqrt{169}=\sqrt{13 \times 13}=13$

$\therefore \sqrt{21 \frac{51}{169}}=\frac{60}{13}=4 \frac{8}{13}$

(xv) We know:

$\sqrt{10 \frac{151}{225}}=\sqrt{\frac{2401}{225}}=\frac{\sqrt{2401}}{\sqrt{225}}$

Now let us compute the square roots of the numerator and the denominator separately.

$\sqrt{2401}=\sqrt{7 \times 7 \times 7 \times 7}=7 \times 7=49$

$\sqrt{225}=\sqrt{3 \times 3 \times 5 \times 5}=3 \times 5=15$

$\therefore \sqrt{10 \frac{151}{225}}=\frac{49}{15}=3 \frac{4}{15}$