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(D) From the polygon law of vector addition,for a closed polygon,the sum of vectors taken in cyclic order is zero.For the outer sides of the pentagon:

Vedclass · Accepted Answer

$-B_1$

Vedclass · Answer

(D) From the polygon law of vector addition,for a closed polygon,the sum of vectors taken in cyclic order is zero.For the outer sides of the pentagon: $A_1 + A_2 + A_3 + A_4 + A_5 = 0$.Therefore,$A_3 + A_4 + A_5 + A_1 = -A_2$.From the triangle law of vector addition applied to the triangles formed by the center and the sides:$B_1 + A_1 = B_2$$B_2 + A_2 = B_3$$B_3 + A_3 = B_4$$B_4 + A_4 = B_5$$B_5 + A_5 = B_1$Summing these equations:$(B_1 + B_2 + B_3 + B_4 + B_5) + (A_1 + A_2 + A_3 + A_4 + A_5) = (B_2 + B_3 + B_4 + B_5 + B_1)$Since the sum of $A_i$ is zero,we confirm the consistency.Alternatively,consider the sum $S = B_2 + B_3 + B_4 + B_5$. From the geometry,$B_2 + A_3 = B_3$,$B_3 + A_4 = B_4$,$B_4 + A_5 = B_5$,and $B_5 + A_1 = B_1$.Adding these: $B_2 + (A_3 + A_4 + A_5 + A_1) = B_1$.Since $A_1 + A_2 + A_3 + A_4 + A_5 = 0$,we have $A_3 + A_4 + A_5 + A_1 = -A_2$.Substituting this: $B_2 - A_2 = B_1$.From the triangle formed by the center and side $A_2$,we have $B_2 + A_2 = B_3$ is incorrect based on the diagram orientation; rather $B_1 + A_1 = B_2$,$B_2 + A_2 = B_3$,etc. Actually,$B_1 + A_1 = B_2 \implies A_1 = B_2 - B_1$. Summing $B_2 + B_3 + B_4 + B_5 = (B_1 + A_1) + (B_2 + A_2) + (B_3 + A_3) + (B_4 + A_4) - (A_1 + A_2 + A_3 + A_4) = B_1 + B_2 + B_3 + B_4 + B_5 - (A_1 + A_2 + A_3 + A_4)$.Given the symmetry and the closed loop $B_1 + A_1 + (-B_2) = 0$,the correct result is $-B_1$.

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

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