Question

Two sides of a triangle are to have lengths '$a$' cm & '$b$' cm. If the triangle is to have the maximum area, then the length of the median from the vertex containing the sides '$a$' and '$b$' is

• A. $\frac{1}{2}\sqrt{a^2+b^2}$
• B. $\frac{2a+b}{3}$
• C. $\sqrt{\frac{a^2+b^2}{2}}$
• D. $\frac{a+2b}{3}$

Answer 1

The correct option is A $\frac{1}{2}\sqrt{a^2+b^2}$

from the fig:
$CD$ is a median of $△ABC$
Area $\Delta =\frac{ab\sin C}{2}$
$\Delta$ is maximum when $C=\frac{\pi }{2}$
Therefore, $△ACB$ is a right angled triangle.
${AB}^2=a^2+b^2$
By Apollonius's thm.: $a^2+b^2=2({CD}^2+{\left(\frac{AB}{2}\right)}^2)$
$\Rightarrow a^2+b^2=2({CD}^2+\frac{a^2+b^2}{4})$
$\Rightarrow CD=\frac{\sqrt{a^2+b^2}}{2}$
Ans: A