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Let $$\triangle$$ ABC ~ $$\triangle$$ DEF and their areas be, respectively, 64 $$cm^2$$and 121 $$cm^2$$. If EF = 15.4 cm, find BC.

Given, $$\triangle$$ ABC ~ $$\triangle$$ DEF,
Area of $$\triangle$$ ABC = 64 $$cm^2$$
Area of $$\triangle$$ DEF = 121 $$cm^2$$
EF = 15.4 cm

$${{Area of \triangle ABC} \over {Area of \triangle DEF}} = {{AB^2} \over {DE^2}}$$

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
$$\Rightarrow$$ $$\frac{AC^2}{DF^2} = \frac{BC^2}{EF^2}$$
$$\Rightarrow$$ $$\frac{64}{121} = \frac{BC^2}{EF^2}$$
$$\Rightarrow$$ $$(\frac{8}{11})^2 = (\frac{BC}{15.4})^2$$
$$\Rightarrow$$ $$\frac{8}{11} = \frac{BC}{15.4}$$
$$\Rightarrow$$ BC = $$8×\frac{15.4}{11}$$
$$\Rightarrow$$ BC = 8 × 1.4
$$\Rightarrow$$ BC = 11.2 cm