Triangles ABC and PQR are similar. AB = 3 cm and the matching side PQ = 5 cm…

The areas of similar triangles are in the ratio of the squares of their corresponding sides: $\dfrac{\text{ar}(ABC)}{\text{ar}(PQR)} = \left(\dfrac{AB}{PQ}\right)^2$. The same holds for corresponding altitudes. When $DE \parallel BC$, $\triangle ADE \sim \triangle ABC$, and $\dfrac{\text{ar}(ADE)}{\text{ar}(ABC)} = \left(\dfrac{AD}{AB}\right)^2$.