Question

In $△ABC,m\angle A={90}^{\circ }$ and $\overline{AD}$ is its median. If $AD=6,AB=10$ then find $AC$.

Answer 1

The median on the hypotenuse of a right triangle divides the triangle into two isosceles triangles that is the median equals one-half the hypotenuse.

Also the median equals one-half the hypotenuse.

As $AD=6$

$CB=12$

Given:$\angle A={90}^{\circ }$ and $AB=10$

Also $AD=6$

Therefore according to Pythagoras Theorem,

$(AB{)}^2+(CA{)}^2=(BC{)}^2$

$(10{)}^2+(CA{)}^2=(12{)}^2$

$(CA{)}^2=144-100=44$

$CA=\sqrt{44}$

$CA=2\sqrt{11}$

Hence $AC=11units.$