Question

The areas of $△ABC$ and $△PQR$ are $144c{m}^2$ and $324c{m}^2$, 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 $△ABC=144c{m}^2$
Area of $△PQR=324c{m}^2$
We know that
$\frac{\text{ Area of }△ABC}{\text{ Area of }△PQR}=\frac{A{D}^2}{P{S}^2}$
$\frac{144}{324}=\frac{A{D}^2}{P{S}^2}\Rightarrow \frac{AD}{PS}=\frac{\sqrt{144}}{\sqrt{324}}\Rightarrow \frac{AD}{PS}=\frac{12}{18}$
$\Rightarrow \frac{AD}{PS}=\frac{2}{3}\therefore \text{ Required ratio }=2:3$