Question

The ratio of the corresponding sides of similar triangles $ABC$ and $A^{\prime }{B}^{\prime }{C}^{\prime }$ is $2:1$. Also, the altitudes $CD$ and $C^{\prime }{D}^{\prime },$ that we have drawn in these triangles are also in the same ratio $2:1$. Then, prove that the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Answer 1

According to the given condition let the length of $CD$ and $C^{\prime }{D}^{\prime }$ be $2x$ and $x$ respectievly.

$\frac{CD}{C^{\prime }{D}^{\prime }}=\frac{2}{1}$
Area of a triangle$=\frac{1}{2}\times base\times altitude$

$\therefore \frac{Areaof\Delta ABC}{Areaof\Delta A^{\prime }{B}^{\prime }{C}^{\prime }}=\frac{\frac{1}{2}\times 42\times 2x}{\frac{1}{2}\times 21\times x}$
$=\frac{42\times 2}{21\times 1}=\frac{84}{21}=\frac{4}{1}={\left(\frac{2}{1}\right)}^2$
$\therefore$ The ratio of the areas of these similar triangles is the square of the ratio of the measures of the corresponding sides.