A) 1 unitB) 2 unitsC) 3 unitsD) 4 units If 4 root 3 sq. unit, .pdf

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A) 1 unit B) 2 units C) 3 units D) 4 units If 4 root 3 sq. unit, then the radius of the circle is:- A) 1 unit B) 2 units C) 3 units D) 4 units Ohm is the cube root of unity) is z + wz (where z,wz and z lies on a circle centered at the origin and if the area of the triangle whose vertices are Solution Without loss of generality assume that ` ``z = r + 0i` . Since we have that `w = (e^(2pi i))^(1/3)` ( using `e^(2pi i) = 1 + 0i = 1` ) then `w = e^((2pi)/3i)` In cartesian coordinates, `w = cos((2pi)/3) + sin((2pi)/3)i = -1/2 + sqrt(3)/2i` so that `z + wz = r/2 + sqrt(3)/2 ri` Now, the points O (origin), `z`, `z+wz`, `wz` form a rhombus with sides length `r` . The triangle of interest as area equal to half that of this rhombus. The area of the rhombus is equal to `(r/2) times (sqrt(3)/2r)` (the rectangle in the middle) `+` `2 times (1/2) times (r/2) times (sqrt(3)/2 r)` (the two identical triangles at each end) which in total is `2 times (r/2) times (sqrt(3)/2 r) = sqrt(3)/2r^2` . The triangle of interest then has area `sqrt(3)/4 r^2` . Given that this area is equal to `4sqrt(3)` squared units then this implies that `r = 4`. Answer is D) the radius of the circle is length 4 units.

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