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