Question

In a right triangle the lengths of the medians of the acute angles are $9$ cm and $13$ cm. Find the length of the hypotenuse of the triangle.

Answer 1

$E$ is the mid point. $BC$ so that $AE=9$ units. $D$ is the midpoint on $AD$ so that $BD=13$ units.

As $BC=2EC$ and $AC=2DC$ and $DCE$

1) a corner of both it follows that $ABC$ and $DEC$ are similar and $AB=2DE$ we therefore only need to find length of $DE$

Let $CE=x$ $CD=y$

Pythagoras give,

${BD}^2={CD}^2+{BC}^2$

$\Rightarrow {\left(13\right)}^2=y^2+{\left(2x\right)}^2$

$y^2+{4x}^2=169⟶\left(1\right)$

${\left(AE\right)}^2={\left(AC\right)}^2+{\left(CE\right)}^2$

$9^2={\left(2y\right)}^2+x^2$

$x^2+{4y}^2=81⟶\left(2\right)$

If we add $\left(1\right)$ and $\left(2\right)$ we get

${5x}^2+{5y}^2=250$

$\Rightarrow x^2+y^2=50$

${DE}^2=x^2+y^2=50=2\times 25$

$DE=5\sqrt{2}.AB=2DE=10\sqrt{2}$ or $Ca$ $14,14$