In Q12, two similar triangles have $BC = 3$ cm, $EF = 8$ cm, and area $\triangle ABC = 3$ sq cm. Then area $\triangle DEF = ?$

For similar triangles the area ratio is the square of the side ratio: $\frac{\triangle DEF}{\triangle ABC} = \frac{EF^{2}}{BC^{2}} = \frac{8^{2}}{3^{2}} = \frac{64}{9}$. So area $\triangle DEF = 3 \times \frac{64}{9} = \frac{64}{3}$ sq cm.