Question

Find the length of the median of one the legs of an isosceles triangle with sides the lengths $18\,,\,18\,and\,6$?

Answer 1

Hint:In this question we have to find the length of the median falling on the side of length $6cm$. We know that in isosceles triangles the median is the perpendicular bisector of the third side. Therefore, we can use Pythagoras theorem to find its length.

Complete step by step answer:

In the above question, it is given that the given triangle is an isosceles triangle and AC is the median.
Therefore, C is the mid-point of length BD and $BC = DC$.
Now, we will prove the triangle ABC and triangle ADC are congruent.
In $\vartriangle ABC\,\,and\,\,\vartriangle ADC$
$AB = AD\,\,$(given)
$BC = CD$ (C is the mid-point)
 $AC = AC$ (common)
Therefore, $\vartriangle ABC\,\,and\,\,\vartriangle ADC$ are congruent.
Hence,$\angle BCA = \angle DCA$
We also know that
$\angle BCA + DCA = {180^ \circ }$
Now using the above result,
$\angle BCA + \angle BCA = {180^ \circ }$
$ \Rightarrow 2\angle BCA = {180^ \circ }$
Now, divide both sides by $2$
$ \Rightarrow \angle BCA = {90^ \circ }$
Therefore, we can use Pythagoras theorem in triangle ABC.
Now, as we know that C is the mid-point of side BD.
Therefore, $BC = \dfrac{{1BD}}{2}$
Now, substituting the value of $BD = 6cm$.
$ \Rightarrow BC = \dfrac{1}{2} \times 6$
$ \Rightarrow BC = 3\,cm$
Now, using Pythagoras theorem in triangle ABC.
$A{B^2} = A{C^2} + B{C^2}$
Now substituting the value of AB and BC from the given question.
${\left( {18} \right)^2} = A{C^2} + {\left( 3 \right)^2}$
$ \Rightarrow A{C^2} = {\left( {18} \right)^2} - {\left( 3 \right)^2}$
$ \Rightarrow A{C^2} = 324 - 9$
Now, taking square root both sides
$ \Rightarrow AC = \sqrt {315} $
$ \Rightarrow AC = 17.75\,cm$
Therefore, the required value of median is $17.75\,cm$.

Note:The property we have proved in this question i.e. about perpendicular bisectors is a well known property. So, you have to keep in mind that property always while doing the questions of isosceles triangles. Also, the angles opposite to equal sides are also equal in an isosceles triangle.