Question 59

It is given that $$\triangle ABC \sim \triangle PRQ$$ and that Area ABC : Area PRQ = 16: 169. If AB=x, AC=y, BC =z (all in cm), then PQ is equal to:

Solution

Ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides.

so,

$$16:169=y^{2}:(PQ)^{2}$$

$$\frac{4^{2}}{13^{2}}=\frac{y^{2}}{(PQ)^{2}}$$

$$\frac{4}{13}=\frac{y}{PQ}$$

$$PQ=\frac{13y}{4}$$

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