Question

Choose the correct option, if $BL$ and $CM$ are the median of a triangle $ABC$, right angled at $A$, then which one is true
$
  {\text{A}}{\text{.4}}\left( {B{L^2} + C{M^2}} \right) = 5B{C^2} \\
  {\text{B}}{\text{.4}}\left( {B{L^2} + C{M^2}} \right) = 3B{C^2} \\
  {\text{C}}{\text{.2}}\left( {B{L^2} + C{M^2}} \right) = 5B{C^2} \\
  {\text{D}}{\text{.2}}\left( {B{L^2} + C{M^2}} \right) = 3B{C^2} \\
 $

Answer 1

Hint: Here, at first we have to write what is given to us in the question and the most important hint is to remember the basic properties of a right angles triangle and medians. Usage of Pythagoras theorem is required and we can find the desired answer using this theorem alone.

Complete step by step answer:

Given- ${\text{BL and CM}}$ are median on ${\text{AC and AB}}$ respectively.
In $\Delta {\text{ABC}}$, by using Pythagoras theorem
${\text{B}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}$…. (1)
In $\Delta {\text{ABL}}$, by using Pythagoras theorem
$
  {\text{B}}{{\text{L}}^2}{\text{ = A}}{{\text{L}}^2}{\text{ + A}}{{\text{B}}^2} \\
  {\text{B}}{{\text{L}}^2}{\text{ = }}\dfrac{{{\text{A}}{{\text{C}}^2}}}{4} + {\text{A}}{{\text{B}}^2} \\
 $
Since ${\text{BL}}$ is the median on ${\text{AC}}$
Then ${\text{AL = }}\dfrac{{{\text{AC}}}}{2}$
Solving the above equation, we get
$4{\text{B}}{{\text{L}}^2} = {\text{A}}{{\text{C}}^2} + 4{\text{A}}{{\text{B}}^2}$ …. (2)
In $\Delta {\text{AMC}}$, by using Pythagoras theorem
$
  {\text{C}}{{\text{M}}^2}{\text{ = A}}{{\text{C}}^2}{\text{ + A}}{{\text{M}}^2} \\
  {\text{C}}{{\text{M}}^2}{\text{ = A}}{{\text{C}}^2}{\text{ + }}\dfrac{{{\text{A}}{{\text{B}}^2}}}{4} \\
 $
Since ${\text{CM}}$ is median on ${\text{AB}}$
Then ${\text{AM = }}\dfrac{{{\text{AB}}}}{2}$
Solving the equation, we get
${\text{4C}}{{\text{M}}^2}{\text{ = 4A}}{{\text{C}}^2}{\text{ + A}}{{\text{B}}^2}$ …. (3)
Adding equations (2) and (3), we get
${\text{4B}}{{\text{L}}^2}{\text{ + 4C}}{{\text{M}}^2}{\text{ = A}}{{\text{C}}^2}{\text{ + 4A}}{{\text{B}}^2}{\text{ + 4A}}{{\text{C}}^2}{\text{ + A}}{{\text{B}}^2}$
Solving it further
$
  {\text{4B}}{{\text{L}}^2}{\text{ + 4C}}{{\text{M}}^2}{\text{ = 5A}}{{\text{C}}^2}{\text{ + 5A}}{{\text{B}}^2} \\
  4\left( {{\text{B}}{{\text{L}}^2}{\text{ + C}}{{\text{M}}^2}} \right) = 5\left( {{\text{A}}{{\text{C}}^2}{\text{ + A}}{{\text{B}}^2}} \right) \\
    \\
 $
$4\left( {{\text{B}}{{\text{L}}^2}{\text{ + C}}{{\text{M}}^2}} \right) = 5{\text{B}}{{\text{C}}^2}$…. (Using (1))

So, the correct answer is “Option A”.

Note: As the medians are given in the question we can use the Pythagoras theorem to solve it as well as we add the two equations where we find median values using the theorem and then check the options given and select the right answer.
Pythagoras Theorem applies to a right angled triangle and is as follows:
${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Base}}} \right)^2} + {\left( {{\text{Perpendicular}}} \right)^2}$ .