What is the ratio of the mean proportional between 14.4 and 3.6 and the third proportional of 5 and 4?
Let the mean proportional between 14.4 and 3.6 = $$x$$
$$=$$> Â $$x^2=14.4\times3.6$$
$$=$$> Â $$x^2=\frac{144}{10}\times\frac{36}{10}$$
$$=$$> Â $$x=\frac{72}{10}$$
Let the third proportional of 5 and 4 = $$y$$
$$=$$> Â $$4^2=5\times y$$
$$=$$> Â $$y=\frac{16}{5}$$
$$\therefore\ $$Required ratio$$=\frac{72}{10}:\frac{16}{5}=\frac{72}{10}:\frac{32}{10}=9:4$$
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