Question

Two isosceles triangles have equal angles and their areas are in the ratio 16 : 25. The ratio of their corresponding heights is

(a) 4 : 5
(b) 5 : 4
(c) 3 : 2
(d) 5 : 7

Answer 1

Given: Two isosceles triangles have equal vertical angles and their areas are in the ratio of 16:25.

To find: Ratio of their corresponding heights.

Let ∆ABC and ∆PQR be two isosceles triangles such that $\angle \mathrm{A}=\angle \mathrm{P}$. Suppose AD ⊥ BC and PS ⊥ QR .

In ∆ABC and ∆PQR,

$$\begin{gathered} \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}} \\ \angle \mathrm{A}=\angle \mathrm{P} \\ \therefore ∆\mathrm{ABC}~∆\mathrm{PQR}\left(\mathrm{SAS}\mathrm{similarity}\right) \end{gathered}$$

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

Hence,

$$\begin{gathered} \frac{\mathrm{Ar}\left(∆\mathrm{ABC}\right)}{\mathrm{Ar}\left(∆\mathrm{PQR}\right)}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^2 \\ \Rightarrow \frac{16}{25}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^2 \\ \Rightarrow \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{4}{5} \end{gathered}$$

Hence we got the result as