Question

The medians of a right triangle which are drawn from the vertices of the acute angles are $5$ and $\sqrt{40}$. The value of the hypotenuse is:

• A. $10$
• B. $2\sqrt{40}$
• C. $\sqrt{13}$
• D. $2\sqrt{13}$
• E. none of these

Answer 1

Answer: (D)

$\left(2\sqrt{13}\right)$


From the given information it follows that for the right triangle $ABC$, we have
$[\frac{a}{2}{]}^2+b^2=25$and $a^2+[\frac{b}{2}{]}^2=40$.
$\frac{a^2+4b^2}{4}=25$ and $\frac{4a^2+b^2}{4}=40$
$a^2+4b^2=100$ and....(i) and $4a^2+b^2=160$...(ii)
Substitute value of $a^2$ of equation(i) in equation(ii), we get
$4(100-4b^2)+b^2=160$
$400-16b^2+b^2=160$
$15b^2=400-160$
$b^2=\frac{240}{15}$
$\therefore b^2=16$
Substitute $b^2$ in equation(i), we get
$a^2+4\left(16\right)=100$
$a^2+64=100$
$a^2=100-64$
$\therefore a^2=36$ and $b^2=16$.
Thus, $c^2=a^2+b^2$
$c^2=36+16$;
$c^2=52$
$c=\sqrt{52}$
$\therefore c=2\sqrt{13}$