Q. If `z` lies on a circle centered at the origin and if the area of the triangle whose vertices are `z, omegaz`and `z+omegaz` (where `omega` is the cube root of unity) is `4sqrt3` sq. unit, then the radius of the circle is:-
A) 1 unit
B) 2 units
C) 3 units
D) 4 units
1 Answer | Add Yours
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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