Question

∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio $\frac{\mathrm{A}\left(∆\mathrm{MNT}\right)}{\mathrm{A}\left(∆\mathrm{QRS}\right)}$.

Answer 1

The areas of two similar triangles are proportional to the squares of their corresponding altitudes.
$$\begin{gathered} \therefore \frac{\mathrm{A}\left(∆\mathrm{MNT}\right)}{\mathrm{A}\left(∆\mathrm{QRS}\right)}={\left(\frac{5}{9}\right)}^2 \\ =\frac{25}{81} \end{gathered}$$