Find the mean proportion between :

(i) 28 and 63

(ii) 2.5 and 0.9

(iii) 6.25 and 1.6

(iv) $\Big(\sqrt{26} - \sqrt{17}\Big)$ and $\Big(\sqrt{26} + \sqrt{17}\Big)$

(v) 6 + 3 $\sqrt{3}$ and 8 − 4 $\sqrt{3}$

(i) Given,

28 and 63

Let mean proportional between 28 and 63 be x.

28 : x :: x : 63

$\Rightarrow \dfrac{28}{x} = \dfrac{x}{63} \\[1em] \Rightarrow x^2 = 28 \times 63 \\[1em] \Rightarrow x^2 = 1764 \\[1em] \Rightarrow x = \sqrt{1764} = 42$

Hence, the mean proportional is 42.

(ii) Given,

2.5 and 0.9

Let mean proportional between 2.5 and 0.9 be x.

2.5 : x :: x : 0.9

$\Rightarrow \dfrac{2.5}{x} = \dfrac{x}{0.9} \\[1em] \Rightarrow x^2 = 2.5 \times 0.9 \\[1em] \Rightarrow x^2 = 2.25 \\[1em] \Rightarrow x = \sqrt{2.25} = 1.5$

Hence, the mean proportional is 1.5.

(iii) Given,

6.25 and 1.6

Let mean proportional between 6.25 and 1.6 be x.

6.25 : x :: x : 1.6

$\Rightarrow \dfrac{6.25}{x} = \dfrac{x}{1.6} \\[1em] \Rightarrow x^2 = 6.25 \times 1.6 \\[1em] \Rightarrow x^2 = 10 \\[1em] \Rightarrow x = \sqrt{10}$

Hence, the mean proportional is $\sqrt{10}$.

(iv) Given,

$\Big(\sqrt{26} - \sqrt{17}\Big)$ and $\Big(\sqrt{26} + \sqrt{17}\Big)$

Let mean proportional between $\Big(\sqrt{26} - \sqrt{17}\Big)$ and $\Big(\sqrt{26} + \sqrt{17}\Big)$ be x

$\Big(\sqrt{26} - \sqrt{17}\Big):x::x:\Big(\sqrt{26} + \sqrt{17}\Big)$

$\Rightarrow \dfrac{\sqrt{26} - \sqrt{17}}{x} = \dfrac{x}{\sqrt{26} + \sqrt{17}} \\[1em] \Rightarrow x^2 = (\sqrt{26} - \sqrt{17}) \times (\sqrt{26} + \sqrt{17}) \\[1em]$

Hence, the mean proportional is 3.

(v) Given,

6 + $3\sqrt{3}$ and 8 − $4\sqrt{3}$.

Let mean proportion between 6 + $3\sqrt{3}$ and 8 − $4\sqrt{3}$ be x.

6 + 3 $\sqrt{3} : x :: x : 8 − 4 \sqrt{3}$

$\Rightarrow \dfrac{6 + 3\sqrt{3}}{x} = \dfrac{x}{8 − 4\sqrt{3}} \\[1em] \Rightarrow x^2 = (6 + 3\sqrt{3}) \times (8 − 4\sqrt{3}) \\[1em] \Rightarrow x^2 = 48 - 24\sqrt{3} + 24\sqrt{3} - 36 \\[1em] \Rightarrow x^2 = 12 \\[1em] \Rightarrow x = 2\sqrt3$

Hence, the mean proportional is $2\sqrt3$.