Question

The areas of $\triangle ABC$ and $\triangle PQR$ are $144 cm ^2$ and $324 cm ^2$, respectively. If $AD$ and $PS$ are the corresponding altitudes of the triangles $ABC$ and $PQR$, respectively, then the ratio of the lengths of $AD$ and PS is

• A. $2: 3$
• B. $4: 9$
• C. $3: 2$
• D. $9: 4$

Answer 1

Answer: (A)

Given, area of $\triangle ABC =144 cm ^2$
Area of $\triangle PQR =324 cm ^2$
We know that
$\frac{\text { Area of } \triangle ABC }{\text { Area of } \triangle PQR }=\frac{ AD ^2}{ PS ^2} $
$\frac{144}{324}=\frac{ AD ^2}{ PS ^2} \Rightarrow \frac{ AD }{ PS }=\frac{\sqrt{144}}{\sqrt{324}} \Rightarrow \frac{ AD }{ PS }=\frac{12}{18} $
$\Rightarrow \frac{ AD }{ PS }=\frac{2}{3} \therefore \text { Required ratio }=2: 3$