Question:
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.
Solution:
Let the two triangles be ABC and DEF with altitudes AP and DQ, respectively.
It is given that $\triangle A B C \sim \triangle D E F$.
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
$\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D E F)}=\frac{(A P)^{2}}{(D Q)^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D E F)}=\frac{6^{2}}{9^{2}}$
$=\frac{36}{81}$
$=\frac{4}{9}$
Hence, the ratio of their areas is 4 : 9