The five sides of a regular pentagon are repr | vedclass

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Question

(d)

From triangle law, in given vector pentagon,

We have,

$B _2+ A _3= B _3 \quad \dots(i)$

$B _3+ A _4= B _4 \quad \dots(ii)$

$B _4+ A

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$- B _1$

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(d)

From triangle law, in given vector pentagon,

We have,

$B _2+ A _3= B _3 \quad \dots(i)$

$B _3+ A _4= B _4 \quad \dots(ii)$

$B _4+ A _5= B _5 \quad \dots(iii)$

$B _5+ A _1= B _1 \quad \dots(iv)$

Adding these equations, we have

$\Rightarrow\left( B _2+ B _3+ B _4+ B _5ight)$ $+\left( A _3+ A _4+ A _5+ A _1ight)$

$=\left( B _3+ B _1+ B _5+ B _1ight)$

$\Rightarrow B _2+ B _3+ B _4+ B _5+$ $\left(- A _2ight)=\left(- B _2ight) \quad \ldots( v )$

As from polygon law of vector addition,

$A _3+ A _4+ A _5+ A _1=- A _2$

$	ext { and }$ $B _3+ B _4+ B _5+ B _1=- B _2$

So, from Eq. $(v)$, we have

$B _2+ B _3+ B _4+ B _5= A _2- B _2=- B _1$

The five sides of a regular pentagon are represented by vectors $A _1, A _2, A _3, A _4$ 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

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