Find the third proportional to :

(i) 9 and 6

(ii) $2\dfrac{2}{3}$ and 4

(iii) 1.6 and 2.4

(iv) (2 + $\sqrt{3}$) and (5 + 4 $\sqrt{3}$)

(v) $\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big)$ and $\sqrt{a^{2} + b^{2}}$

(i) Given,

9 and 6

Let third proportional to 9 and 6 be x.

⇒ 9 : 6 = 6 : x

⇒ $\dfrac{9}{6} = \dfrac{6}{x}$

⇒ x = $\dfrac{6^2}{9} = \dfrac{36}{4}$

⇒ x = 4.

Hence, the third proportional is 4.

(ii) Given,

$2\dfrac{2}{3}$ and 4

Let third proportional to $\dfrac{8}{3}$ and 4 be x

$\dfrac{8}{3}$ : 4 = 4 : x

$\Rightarrow \dfrac{4}{x} = \dfrac{\dfrac{8}{3}}{4} \\[1em] \Rightarrow \dfrac{4}{x} = \dfrac{2}{3} \\[1em] \Rightarrow x = \dfrac{3}{2} \times 4 \\[1em] \Rightarrow x = 6.$

Hence, the third proportional is 6.

(iii) Given,

1.6 and 2.4

Let third proportional to 1.6 and 2.4 be x.

1.6 : 2.4 = 2.4 : x

⇒ $\dfrac{1.6}{2.4} = \dfrac{2.4}{x}$

⇒ x = $\dfrac{(2.4)^2}{1.6} = \dfrac{5.76}{1.6}$

⇒ x = 3.6

Hence, the third proportional is 3.6.

(iv) Given,

(2 + $\sqrt{3}$) and (5 + 4 $\sqrt{3}$)

Let third proportional to (2 + $\sqrt{3}$) and (5 + 4 $\sqrt{3}$) be x.

$(2 + \sqrt{3}) : (5 + 4 \sqrt{3}) = (5 + 4\sqrt{3}) : x$

Thus,

$\Rightarrow \dfrac{(2 + \sqrt{3})}{(5 + 4 \sqrt{3})} = \dfrac{(5 + 4 \sqrt{3})}{x} \\[1em] \Rightarrow x = \dfrac{(5 + 4 \sqrt{3})^2}{(2 + \sqrt{3})} \\[1em] = \dfrac{5^2 + 2(5)(4\sqrt3) + (4\sqrt3)^2}{(2 + \sqrt{3})} \\[1em] = \dfrac{25 + 40\sqrt3 + 16 \times 3}{(2 + \sqrt{3})} \\[1em] = \dfrac{25 + 40\sqrt3 + 48}{(2 + \sqrt{3})} \\[1em] = \dfrac{73 + 40\sqrt3}{(2 + \sqrt{3})}$

Multiplying numerator and denominator by $(2 - \sqrt{3})$, we get :

$= \dfrac{(73 + 40\sqrt3)(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \\[1em] = \dfrac{146 - 73\sqrt3 + 80\sqrt{3} - 40(\sqrt3)^2}{2^2 - (\sqrt{3})^2} \\[1em] = \dfrac{146 + 7\sqrt3 - 120}{4 - 3} \\[1em] = 26 + 7\sqrt3.$

Hence, the third proportional is $26 + 7\sqrt3$.

(v) Given,

$\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big)$ and $\sqrt{a^{2} + b^{2}}$

Let third proportional to $\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big) \text{ and } \sqrt{a^{2} + b^{2}}$ be x.

$\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big): \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} + b^{2}}:x$

$\dfrac{\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big)}{\sqrt{a^{2} + b^{2}}} = \dfrac{\sqrt{a^{2} + b^{2}}}{x} \\[1em] x = \dfrac{(\sqrt{a^2 + b^2})^2}{\Big(\dfrac{a}{b} + \dfrac{b}{a}\Big)} \\[1em] = \dfrac{a^2 + b^2}{\dfrac{a^2 + b^2}{ab}} \\[1em] = (a^2 + b^2) \times \dfrac{ab}{a^2 + b^2} \\[1em] = ab.$

Hence, the third proportional is ab.