Question 26

If $$a \colon b \colon c = \frac{1}{3} \colon \frac{1}{6} \colon \frac{1}{9}$$, then the value of $$\frac{a^2 + b^2}{b^2 - c^2}$$ is

Multiplying the ratio by 18 gives $$a \colon b \colon c = 6 \colon 3 \colon 2$$. Then $$\frac{a^2 + b^2}{b^2 - c^2} = \frac{36 + 9}{9 - 4} = \frac{45}{5} = 9$$.

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