Question

In $\Delta XYZ$, $YZ=2$ units, $XZ=3$ units and medians $XP,YQ$ are such that $XP$ is perpendicular to $YQ$. Then the area of $\Delta XYZ$ is

• A. $\frac{\sqrt{14}}{5}\text{sq.units.}$
• B. $\frac{2\sqrt{14}}{5}\text{sq.units.}$
• C. $\sqrt{\frac{8}{15}}\text{sq.units.}$
• D. $\frac{\sqrt{2}}{5}\text{sq.units.}$

Answer 1

The correct option is B $\frac{2\sqrt{14}}{5}\text{sq.units.}$

Let the length of median $YQ$ be ${}^{\prime }3a^{\prime }$ and $XP$ be ${}^{\prime }3b^{\prime }$ .
Centroid divides the median $2:1$ from the vertex. Therefore,
In $\Delta XGQ$
$a^2+4b^2=\frac{9}{4}-\left(1\right)$
and similarly in $\Delta PGY$
$4a^2+b^2=1-\left(2\right)$
on solving equation $\left(1\right)$ and $\left(2\right)$
we get,
$b=\sqrt{\frac{8}{15}}a=\frac{1}{2}\times \sqrt{\frac{7}{15}}$
After drawing all the three medians the original triangle is divided into six triangles that are all of the same area.
Area $a_1=a_2=a_3=a_4=a_5=a_6$

Now according to our question -
$\text{Area}\Delta XGY=\text{Area}\Delta XGZ=\text{Area}\Delta YGZ=\frac{1}{3}\text{Area}\Delta XYZ$
Therefore,
Area of $\Delta XYZ$= $3\times \frac{1}{2}\times 2a\times 2b=2\frac{\sqrt{14}}{5}\text{sq.units.}$