Question

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

• A. $\frac{4}{3}\sqrt{11}$
• B. $\frac{2}{3}\sqrt{11}$
• C. $\frac{1}{3}\sqrt{11}$
• D. None of these
  

Answer 1

The correct option is A $\frac{4}{3}\sqrt{11}$

Given, $AD=3,CE=4$
Using Appaloneaus theorem for median $AD$
We have $c^2+b^2=2(\frac{a^2}{4}+9)$ ...(1)
Using Appaloneaus theorem for median $CE$
We have $b^2+a^2=2(\frac{c^2}{4}+10)$ ...(2)
Also, $a^2+c^2=b^2$
Adding (1) and (2)
$3b^2=2(\frac{b^2}{4}+25)\Rightarrow b^2=20$
Solving (1) and (2) we get,
$c=\frac{4}{\sqrt{3}}$ and $a=2\frac{4}{\sqrt{3}}$
Hence, area of triangle
$=\frac{1}{2}\left(\frac{4}{\sqrt{3}}\right)\left(2\sqrt{\frac{11}{3}}\right)=\frac{4}{3}\sqrt{11}$.