Question

In the following figure $△ABC$ is an equilateral triangle and $AC=x$ $cm$. $\overline{AD}$ is median on $\overline{BC},D\in \overline{BC}$. If $AD=y$ $cm$, then $y=..................cm$

• A. $\sqrt{\frac{3}{2}.x}$
• B. $\frac{\sqrt{3}}{2}.x$
• C. $\frac{\sqrt{3x}}{2}$
• D. $\frac{3}{2}.x$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}.x$

$△ABC$ is an equilateral triangle and $AC=x$ cm. Then, $AB=BC=AC=x$
$\overline{AD}$ is median on $\overline{BC},D\in \overline{BC}$
$\therefore D$ is the midpoint of $BC$
$\therefore CD=DB=\frac{x}{2}$

Now, $△ADB$ is a right angled triangle at $D$
By Pythagoras theorem,

$A{C}^2=A{D}^2+C{D}^2$
$⟹y^2+{\left(\frac{x}{2}\right)}^2=x^2$
$⟹y^2=\frac{3}{4}{x}^2$
$⟹y=\frac{\sqrt{3}}{2}x$

Hence, option B is correct.