Question

Inscribe an equilateral triangle inside a circle of radius 5 cm and calculate the area outside the triangle and inside the circle?

Answer 1

`(25pi-(75sqrt(3))/4) "cm"^2 ~~ 46.06"cm"^2`

Explanation:

The roots of:

`z^3-1 = 0`

form the vertices of an equilateral triangle on the unit circle in the complex plane.

They are:

`1`, `omega` and `omega^2 = bar(omega)`

where:

`omega = cos((2pi)/3)+i sin((2pi)/3) = -1/2+sqrt(3)/2i`

is the primitive complex cube root of `1`

graph{((x-1)^2+y^2-0.003)((x+1/2)^2+(y-sqrt(3)/2)^2-0.003)((x+1/2)^2+(y+sqrt(3)/2)^2-0.003)(x^2+y^2-1) = 0 [-2.5, 2.5, -1.25, 1.25]}

The distance between `omega` and `bar(omega)` is `sqrt(3)`

The distance between `(omega+bar(omega))/2 = -1/2` and `1` is `3/2`.

So the area of the equilateral triangle is:

`1/2 xx "base" xx "height" = 1/2 * sqrt(3) * 3/2 = (3sqrt(3))/4`

Scaling up by a factor of `5`, the area of the triangle of radius `5"cm"` is:

`(5^2 * (3sqrt(3))/4)"cm"^2 = (75sqrt(3))/4 "cm"^2`

The area of a circle of radius `5"cm"` is given by the formula:

`A = pir^2 = pi(5 "cm")^2 = 25pi"cm"^2`

So the area outside the triangle and inside the circle is:

`(25pi-(75sqrt(3))/4) "cm"^2`

Answer 2

Area outside the triangle is `46.06` sq.unit and
area inside the triangle is `32.48` sq.unit

Explanation:

Radius of the circle is `r=5` and let `a` be the side of equilateral

triangle. Formula applied : `a/sqrt(3)= r or a= r*sqrt3` or

`a= 5*sqrt3 ~~ 8.66 (2dp)`. Area of the equilateral triangle is

`A_t=sqrt(3)/4*a^2=sqrt(3)/4*8.66^2~~32.48` sq.unit.

Area of the circle is `A_c=pi*r^2 orA_c=3.14*5^2` or

`A_c=3.14*25 ~~ 78.54` . Hence area outside the triangle

is `A_o=A_c-A_t= 78.54-32.48 ~~ 46.06` sq.unit [Ans]