In the circle shown in the picture, the radius is 6 and chord AB = 6. What is the area of the shaded region? (refer the image for other details)

karthikpandian19 wrote:In the circle shown in the picture, the radius is 6 and chord AB = 6. What is the area of the shaded region? (refer the image for other details) Area of the shaded region = Area of the sector - Area of the triangle OAB OAB is an equilateral triangle, as all sides are equal. So, each of the angles will be 60 degrees each. Area of sector = (1/2) * angle subtended at the center of circle * (pi)/180 * r² = (1/2) * 60 * (pi)/180 * 6² = 6(pi) Length of perpendicular from vertex O to the base AB = √(6)² - (3)² = √27 = 3√3 Area of triangle OAB = (1/2) * 6 * 3√3 = 9√3 Therefore, area of shaded region = 6(pi) - 9√3 The correct answer is E.

Anurag, Can you again explain me the Calculation for the area of the SECTOR?? Anurag@Gurome wrote:karthikpandian19 wrote:In the circle shown in the picture, the radius is 6 and chord AB = 6. What is the area of the shaded region? (refer the image for other details) Area of the shaded region = Area of the sector - Area of the triangle OAB OAB is an equilateral triangle, as all sides are equal. So, each of the angles will be 60 degrees each. Area of sector = (1/2) * angle subtended at the center of circle * (pi)/180 * r² = (1/2) * 60 * (pi)/180 * 6² = 6(pi) Length of perpendicular from vertex O to the base AB = √(6)² - (3)² = √27 = 3√3 Area of triangle OAB = (1/2) * 6 * 3√3 = 9√3 Therefore, area of shaded region = 6(pi) - 9√3 The correct answer is E.

Anurag, Can you again explain me the Calculation for the area of the SECTOR?? A circle has an angle of 360 degrees or 2(pi), and its area = (pi)r² Hence, a sector with an angle of A degrees, should have an area of [A/2(pi)] * (pi)r² = (A/2) * r²