Question 56

In $$\triangle$$ ABC, AB = AC and AL is perpendicular to BC at L. In $$\triangle$$ DEF, DE = DF and DM is perpendicular to EF at M.If (area of $$\triangle$$ ABC) (area of $$\triangle$$ DEF) = 9:25, then $$\frac{DM + AL}{DM - AL}$$ is equal to:

Solution

(area of $$\triangle$$ ABC) : (area of $$\triangle$$ DEF) = 9:25

By property of similar triangle,

$$\frac{(DM)^2}{(AL)^2} = \frac{25}{9}$$

$$\frac{DM}{AL} = \frac{5}{3}$$

By componendo dividendo,

When $$\frac{x}{y} = \frac{a}{b}$$ then,

$$\frac{x + y}{x - y} = \frac{a + b}{a - b}$$

So,

$$\frac{DM + AL}{DM - AL} = \frac{5 + 3}{5 - 3}$$

$$\frac{DM + AL}{DM - AL} = \frac{8}{2} = 4$$

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