Question

The five sides of a regular pentagon are represented by vectors $A_1,A_2,A_S,A_4$ and $A_5,$ in cyclic order as shown below

Corresponding vertices are represented by $B_1,B_2,B_3,B_4$ and $B_5$, drawn from the centre of the pentagon.
Then, $B_2+B_3+B_4+B_5$ is equal to

• A. $A_1$
• B. $-A_1$
• C. $B_1$
• D. $-B_1$

Answer 1

Answer: (D)

From triangle law, in given vector pentagon,

We have
$B_2+A_3=B_3...(i)$
$B_3+A_4=B_4...(ii)$
$B_4+A_5=B_5...(iii)$
$B_5+A_1=B_1...(iV)$
Adding these equations, we have
$\Rightarrow (B_2+B_3+B_4+B_5)$
$+(A_3+A_4+A_5+A_4)$
$=(B_3+B_4+B_5+B_4)$
$\Rightarrow (B_2+B_3+B_4,+B_5+$
$(-A_2)=(-B_2)...(v)$
As from polygon law of vector addition
$A_3+A_4+A_5+A_1=-A_2$
and $B_3+B_4+B_5+B_1=-B_2$
So, from Eq. (v), we have
$B_2+B_3+B_4+B_5$
$=A_2-B_2=-B_1$