1) Find the area of sector MN of circle L The area of a circle is π*radius^2. The area of the entire circle, then, is 36π. The sector is 1/6 of the circle (60°/360°). Therefore, the area of the sector is 6π.
2) Find the area of triangle LMN The area of a triangle is 1/2*base*height. A line can be drawn bisecting &angl;L and bisecting side MN at a right angle. This gives us two equal 30-60-90 triangles. The hypotenuse of each is 6, making the base 3 and the height 3√3. The area of triangle LMN, then is 1/2*6*3√3 = 9√3.
(This could have been more easily determined if it was known that triangle LMN was equilateral)
3) Subtract the area of triangle LMN from the area of sector MN The area of the shaded region, then, is 6π - 9√3 = 3(2π - 3√3) ≈ 3.26
1 comment:
This problem is most easily done in three steps:
1) Find the area of sector MN of circle L
The area of a circle is π*radius^2. The area of the entire circle, then, is 36π. The sector is 1/6 of the circle (60°/360°). Therefore, the area of the sector is 6π.
2) Find the area of triangle LMN
The area of a triangle is 1/2*base*height. A line can be drawn bisecting &angl;L and bisecting side MN at a right angle. This gives us two equal 30-60-90 triangles. The hypotenuse of each is 6, making the base 3 and the height 3√3. The area of triangle LMN, then is 1/2*6*3√3 = 9√3.
(This could have been more easily determined if it was known that triangle LMN was equilateral)
3) Subtract the area of triangle LMN from the area of sector MN
The area of the shaded region, then, is
6π - 9√3 = 3(2π - 3√3) ≈ 3.26
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