question_answer

If the median drawn on the base of a triangle is half its base, then the triangle will be               [SSC (CGL) 2013]

A) obtuse angled    

B) equilateral

C) right angled

D) acute angled

Correct Answer: C

Solution :

Let ABC be a triangle and AD be the median

Then, \[AD=\frac{1}{2}BC\]

Now, by Apollonius theorem

\[A{{B}^{2}}+A{{C}^{2}}=2\,(A{{D}^{2}}+B{{D}^{2}})\]

Since, \[AD=BD=DC\]

\[=\frac{1}{2}BC\]

\[\therefore \] \[A{{B}^{2}}+A{{C}^{2}}=2\left[ {{\left( \frac{BC}{2} \right)}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}} \right]\]

\[\Rightarrow \]   \[A{{B}^{2}}+A{{C}^{2}}=2\,\,\left( \frac{B{{C}^{2}}}{4}+\frac{B{{C}^{2}}}{4} \right)\]

\[\Rightarrow \]   \[A{{B}^{2}}+A{{C}^{2}}=2\,\,\left( \frac{B{{C}^{2}}}{2} \right)\]

\[\Rightarrow \]   \[A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\]

Here, it follows Pythagoras theorem.

So, it is a right angled triangle.