Question

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

Answer 1

Hint: Start by drawing a neat diagram followed by using the theorem that the midpoint of the hypotenuse of a right-angled triangle is equidistant from its all three vertice.

Complete step-by-step answer:
First, we will draw a neat diagram of the situation given in the question.

According to the question, ABC is a right-angled triangle, and we know that the midpoint of the centre of the hypotenuse of a right-angled triangle is equidistant from its all three vertices and the median of a triangle divides the side into two equal halves so we can conclude that:
AD = BD = BC
And it is given in the question that AD = 6. Therefore, we can say that AD = BD = CD = 6.
So, using the above result, we can say that the length of the hypotenuse is BD + CD = 12.
Now, we know the Pythagoras theorem states that the sum of squares of the perpendicular and the base of a right-angled triangle is equal to the square of the hypotenuse of the triangle. So, using this, we get
${{\left( \text{perpendicular} \right)}^{\text{2}}}\text{+}{{\left( \text{base} \right)}^{\text{2}}}\text{=}{{\left( \text{hypotenuse} \right)}^{2}}$
$\Rightarrow {{\left( AB \right)}^{\text{2}}}\text{+}{{\left( AC \right)}^{\text{2}}}\text{=}{{\left( BC \right)}^{2}}$
$\Rightarrow {{\left( 10 \right)}^{\text{2}}}\text{+}{{\left( AC \right)}^{\text{2}}}\text{=}{{\left( 12 \right)}^{2}}$
$\Rightarrow {{\left( AC \right)}^{\text{2}}}\text{=144-100}$
$\Rightarrow AC\text{=}\sqrt{44}$
Therefore, the length of the side AC of the triangle is $\sqrt{44}$ , which can be written as $2\sqrt{11}$ .
Note: Remember that for a right-angled triangle, the circumcentre is the centre of the hypotenuse, and the radius of the circumcircle is equal to half the length of the hypotenuse.