Question

Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.

Answer 1

Given: Two isosceles triangles have equal vertical angles and their areas are in the ratio of 36:25.

To find: Ratio of their corresponding heights.

Suppose ∆ABC and ∆PQR are two isosceles triangles with $\angle \mathrm{A}=\angle \mathrm{P}$.

Now, AB = AC and PQ = PR

$\therefore \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{PQ}}{\mathrm{PR}}$

In ∆ABC and ∆PQR,

$\angle \mathrm{A}=\angle \mathrm{P}$

$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{PQ}}{\mathrm{PR}}$

∴ ∆ABC $~$∆PQR (SAS Similarity)

Let AD and PS be the altitudes of ∆ABC and ∆PQR, respectively.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

$$\begin{gathered} \therefore \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{PQR}\right)}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^2 \\ \Rightarrow \frac{36}{25}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^2 \\ \Rightarrow \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{6}{5} \end{gathered}$$

Hence, the ratio of their corresponding heights is 6 : 5.