Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $l$ is the median from the vertex $A$ to the side $BC$, then?

• A. $4l^2=a^2+4bc\cos A$
• B. $4l^2=2b^2+2c^2-a^2$
• C. $4l^2=b^2+c^2+2bc\cos A$
• D. $4l^2={(2s-a)}^2-4bc{\sin }^2\frac{A}{2}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $4l^2=a^2+4bc\cos A$
B $4l^2=2b^2+2c^2-a^2$
C $4l^2={(2s-a)}^2-4bc{\sin }^2\frac{A}{2}$
D $4l^2=b^2+c^2+2bc\cos A$

In $△ABC,$

Formula for median through $A$ to $BC$ is $l=\frac{1}{2}\left(\sqrt{2b^2+2c^2-a^2}\right)$

Squaring on both sides

$l^2=\frac{1}{4}(2b^2+2c^2-a^2)⟹4l^2=2b^2+2c^2-a^2$

$4l^2=2b^2+2c^2-a^2=2(b^2+c^2-a^2)+a^2=a^2+2\left(2bc\cos A\right)=a^2+4bc\cos A$ $(\because \cos A=\frac{b^2+c^2-a^2}{2bc})$

$4l^2=a^2+\left(2bc\cos A\right)+2bc\cos A=a^2+(b^2+c^2-a^2)+2bc\cos A=b^2+c^2+2bc\cos A$

$4l^2=b^2+c^2+2bc\cos A=b^2+c^2+2bc-2bc+2bc\cos A=(b+c{)}^2-2bc(1-\cos A)=(b+c{)}^2-4bc{\sin }^2\frac{A}{2}=(2s-a{)}^2-4bc{\sin }^2\frac{A}{2}$