Question

There is an equilateral triangle with each side 4 and a circle with the centre on one of the vertices of that triangle. The arc of that circle divides the triangle into two parts of equal area. How long is the radius of the circle?

Answer 1

Hint: Find the area of the equilateral triangle with the given side. Take its half as the arc of the circle divides it into two equal halves. Now find the area of the sector and equate them together to get the radius of the circle.
Complete step-by-step answer:
We have been given an equilateral triangle, whose side is 4. Let us take \[a\] as the side of the given equilateral triangle. Thus, \[a = 4\].
Now it is said that there is a circle whose centre is the vertex of the equilateral triangle. Thus we can draw its figure as,

Now it is said that the arc of the circle divides the triangle into two equal areas. Thus we need to find the radius of the circle. First let us find the area of the equilateral triangle.
We know the area of the equilateral triangle as,
Area of the equilateral triangle= \[\dfrac{{\sqrt 3 }}{4}{a^2}\].
Now let us substitute \[a = 4\] in the above equation.
Thus, area of triangle =\[\dfrac{{\sqrt 3 }}{4}{a^2} = \dfrac{{\sqrt 3 }}{4}{4^2}\]
\[\dfrac{{\sqrt 3 }}{4}{4^2} = \dfrac{{\sqrt 3 }}{4} \times 4 \times 4 = \sqrt 3 \times 4 = 4\sqrt 3 \]
As the arc of the circle divides the triangle into two equal areas,
The area of the triangle =\[\dfrac{{4\sqrt 3 }}{2} = 2\sqrt 3 \]
Hence the area of the equilateral triangle =\[4\sqrt 3 \].
Let us consider the radius of the circle as \[r\].
We know that in an equilateral triangle all the sides are the same and the angles in an equilateral triangle are also the same.
i.e. \[\theta = 60^\circ \]
We can use the formula of the area of the sector to get the radius of the circle.
We know that area of sector=\[\dfrac{\theta }{{360}} \times \pi {r^2}\].
We know, \[\theta = 60^\circ \]thus substitute this in its place and get the area of the sector.
\[\dfrac{\theta }{{360}} \times \pi {r^2} = \dfrac{{60}}{{360}} \times \pi {r^2} = \dfrac{{\pi {r^2}}}{6}\]
Hence are of the sector=\[\dfrac{{\pi {r^2}}}{6}\]
Thus we can equate them together with the area of the triangle to the area of the sector we got. As the sector of the circle divides its area of triangle into two equal halves, the area of half of the triangle is equal to the area of the sector. You can understand this from the figure.
Are of the trainable = are of the sector
\[2\sqrt 3 = \dfrac{{\pi {r^2}}}{6}\]
Now apply cross multiplication property and simplify it.
\[\begin{gathered}
  {r^2} = \dfrac{{2\sqrt 3 \times 6}}{\pi } = \dfrac{{12\sqrt 3 }}{\pi } \ \\
  r = \sqrt {\dfrac{{12\sqrt 3 }}{\pi }} \ \\
\end{gathered} \]
Hence we got the radius of the circle as, \[r = \sqrt {\dfrac{{12\sqrt 3 }}{\pi }} \].

Note: Don’t forget to divide the area of the triangle obtained as the arc of the circle divides it into two equal halves. You should know the formula for area of sector. From the figure it is clear about the sector of the circle with the inbuilt portion of the triangle.