Question

12.
A particle moves in a plane from A to E along the shown path. It is given that AB=BC=CD =DE 10 metre.
Then the magnitude of net displacement of particle is:
108°
108-
C
108°
(A) 10 m
(B) 15 m
(C) 5 m
(D) 20 m​

Answer 1

Answer:

Explanation:

Given figure represents a regular Pentagon. In this figure That particle start from A and continue to B, C,D then E. All the sides are equal to 10 m.

A to B = 10 m

B to C = 10m

C to D = 10 m

D to E = 10 m

E to A = 10 m

So magnitude of AE = 10 metre.

Answer 2

Answer:10 m

Explanation:

ABCDE is a regular pentagon

If we take O as a point which is equidistant from B, C and D, then angle OEA= angle OAE = 54°

ΔOEA is congruent to ΔOBC

AE = BC = 10 m