Anonymous

# Area of a sector?

An equilateral triangle with a side of 2√3 is inscribed in a circle. What is the area of one of the sectors formed by the radii to the vertices of the triangle?

Update:

An equilateral triangle with a side of 2√3 is inscribed in a circle. What is the area of one of the sectors formed by the radii to the vertices of the triangle?

Ususally to find the area of sector, I use the central angle/360=area of sector/area of circle. I don't know the measure of the central angle or even how to find th measure of the central angle. I also don't know the length of the radii of the circle. I'm totally clueless.

Relevance
• Wal C
Lv 6

Well if you draw in the three radii from the vertices of the equilateral triangle you can quickly establish that because the triangle is rotationally symmetric then the size of each angle formed at the centre of the circle is 2π/3 radians

so half the chord length

= half the length of the side of the triangle

= √3

Now sinπ/3 (60°) = opposite / hypotenuse

So √3/2 = √3 / radius

Thus area of sector = ½ r²θ

= ½ 2² 2π/3

= 4π/3

Source(s): Me ;^))

Basically, we find the area of the circle, then the area of the triangle. We subtract them and divide the result by 3, since there are 3 sectors.

To find the area of the triangle, draw a line from one of its vertices through the center of the triangle. This line bisects the equilateral triangle into two right triangles. The hypotenuse is 2√3 and one leg is √3. So the remaining leg is √(12-3) = 3. The area of the right triangle is (1/2)√3(2√3) = 3, So the area of the equilateral triangle is 6.

To find the area of the circle, we first need to know its radius or diameter. The easiest way to do this is to note that the center of the triangle will also be the center of the circle. We will make a new triangle from one vertex, the center of the triangle and the intersection of two legs of the above right triangle. This triangle is a 30-60-90 triangle, and one leg is √3. So the other leg is 1 and the hypotenuse (which is a radius of the circle) is 2. So, we finally have the radius, and the area which is:

Circle area = Pi*R^2 = 4Pi

So the Area of Circle - Triangle is 4Pi - 6

The area of a sector is (4Pi-6)/3

• raj
Lv 7