Find the mean proportion between :

(i) 81 and 121

(ii) 1.8 and 0.2

(iii) $\dfrac{2}{3}$ and $\dfrac{8}{27}$

(iv) 0.32 and 0.08

(v) $\dfrac{1}{25}$ and $\dfrac{1}{16}$

(i) 81 and 121

Mean proportion between 81 and 121

$= \sqrt{81 \times 121} \\[1em] = \sqrt{81} \times \sqrt{121} \\[1em] = 9 \times 11 \\[1em] = 99 \\[1em]$

Hence, the answer is 99

(ii) 1.8 and 0.2

Mean proportion between 1.8 and 0.2

$= \sqrt{1.8 \times 0.2} \\[1em] = \sqrt{0.36} \\[1em] = 0.6$

Hence, the answer is 0.6

(iii) $\dfrac{2}{3}$ and $\dfrac{8}{27}$

Mean proportion between $\dfrac{2}{3}$ and $\dfrac{8}{27}$

$= \sqrt{\dfrac{2}{3} \times \dfrac{8}{27}} \\[1em] = \sqrt{\dfrac{16}{81}} \\[1em] = \dfrac{\sqrt{16}}{\sqrt{81}} \\[1em] = \dfrac{4}{9}$

Hence, the answer is $\dfrac{4}{9}$

(iv) 0.32 and 0.08

Mean proportion between 0.32 and 0.08

$= \sqrt{0.32 \times 0.08} \\[1em] = \sqrt{0.0256} \\[1em] = 0.16$

Hence, the answer is 0.16

(v) $\dfrac{1}{25}$ and $\dfrac{1}{16}$

Mean proportion between $\dfrac{1}{25}$ and $\dfrac{1}{16}$

$= \sqrt{\dfrac{1}{25} \times \dfrac{1}{16}} \\[1em] = \sqrt{\dfrac{1}{400}} \\[1em] = \dfrac{1}{\sqrt{400}} \\[1em] = \dfrac{1}{20}$

Hence, the answer is $\dfrac{1}{20}$