Question

The length of the medians through acute angles of a right angled triangle are 3 and 4. Find the area of the triangle.

Answer 1

AD=3, CE=4
Using Appoloneous theorem for median AD, we have
$c^2+b^2=2(\frac{a^2}{4}+9)$..................(1)
Using Appoloneous theorem for median CE, we have
$b^2+a^2=2(\frac{c^2}{4}+16)$..................(2)
Also $a^2+c^2=b^2$
Adding (1) and (2),$3b^2=2(\frac{b^2}{4}+25)$ or $b^2=20$
Solving (1) and (2), we get $c=\frac{4}{\sqrt{3}}$ and $a=2\sqrt{\frac{11}{3}}$
Hence, Area of triangle $=\frac{1}{2}\left(\frac{4}{\sqrt{3}}\right)\left(2\sqrt{\frac{11}{3}}\right)=\frac{4}{3}\sqrt{11}$