Question

The base of an isosceles triangle is $\sqrt{2}$. The medians to the legs intersect each other at right angles. The area of the triangle is

• A. $1.5$
• B. $2$
• C. $2.5$
• D. $3.5$
• E. $4$

Answer 1

The correct option is A $1.5$

In right $△OBC$, using Pythagoras theorem,

${OB}^2+{OC}^2={\left(\sqrt{2}\right)}^2\Rightarrow 2{OB}^2=2[\because OB=OC]$

$\Rightarrow {OB}^2=1\Rightarrow OB=\sqrt{1}=1$

Now, $\frac{OB}{ON}=\frac{2}{1}$ [Medians divide itself in $2:1$]

$ON=\frac{OB}{2}=\frac{1}{2}$, again $\frac{OC}{OM}=\frac{2}{1}\Rightarrow OM=\frac{1}{2}$

$ar△BOC=\frac{1}{2}\times 1\times 1=\frac{1}{2}$ sq. units

$ar△AOB=ar△AOC=ar△BOC=\frac{1}{2}$ sq. units [$\because$ O is centroid]

$ar△ABC=ar△AOB+ar△BOC+ar△AOC$

$=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$

$=\frac{3}{2}=1.5$ sq. units