In the above circle, the radius is 6 and the chord AB=6. What is the area of the shaded region?
2pi -3root3
4pi -4root3
4pi -9root3
6pi -6root3
6pi-9root3
Ans: 6pi-9root3
Area of the shaded region = Area of the arc - Area of the triangle
Area of the arc = Area of the circle * angle of the arc/2pi
Area of the triangle = Area of equilateral triangle formed by the radii and chord
= pi/6 * 6^2 (since radius=6) - 1/2 * 6 *6root3/2
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another geo problem
Source: Beat The GMAT — Problem Solving |
giatch wrote:In the above circle, the radius is 6 and the chord AB=6. What is the area of the shaded region?
[A] 2pi -3root3
4pi -4root3
[C] 4pi -9root3
[D] 6pi -6root3
[E] 6pi-9root3
See attachment for figure.
Revisiting this problem. Can anyone show a different perspective on this problem. Brent, Ian, Staurt, Cramya?
Want to Beat GMAT.
Always do what you're afraid to do. Whoooop GMAT
Always do what you're afraid to do. Whoooop GMAT
Area of the sector - Area of the triangle is the best way to get the answer.
Area of the sector = angle of the radius of the circle / 360 * (area of the circle)
area of the triangle can be calculated by drawing a triangle from the center to A and another line from center to B. From this we know the angle is 60 deg each because it forms an equilateral triangle with equal sides.
Thus,
Area of the Sector = 60/360 * (2pi) (6*6) = 6pi
Area of the triangle = 9 sqrt(3)
Ans: 6pi-9sqrt(3)
Area of the sector = angle of the radius of the circle / 360 * (area of the circle)
area of the triangle can be calculated by drawing a triangle from the center to A and another line from center to B. From this we know the angle is 60 deg each because it forms an equilateral triangle with equal sides.
Thus,
Area of the Sector = 60/360 * (2pi) (6*6) = 6pi
Area of the triangle = 9 sqrt(3)
Ans: 6pi-9sqrt(3)
Xunil56 i can catch a small typo....xunil56 wrote:Area of the sector - Area of the triangle is the best way to get the answer.
Area of the sector = angle of the radius of the circle / 360 * (area of the circle)
area of the triangle can be calculated by drawing a triangle from the center to A and another line from center to B. From this we know the angle is 60 deg each because it forms an equilateral triangle with equal sides.
Thus,
Area of the Sector = 60/360 * (2pi) (6*6) = 6pi
Area of the triangle = 9 sqrt(3)
Ans: 6pi-9sqrt(3)
instead of
Area of the Sector = 60/360 * (2pi) (6*6) = 6pi
it should be
Area of the Sector = 60/360 * (pi) (6*6) = 6pi
remember area of sector is (m/360)* pi * r^2.
step 1 -calculate area of sector
as radius as 6 and chord ab is 6, then equilateral triangle with angle 60
area=60/360 *pi*6^2=6pi
step 2-calculate area of triangle
drop perpendicular from b to radius ao and create 30/60/90 triangle
hypotenuse=radius=6
therefore ht =3rt3
area=1/2* 6* 3rt3=9rt3
step 3 subtract area of triangle from sector
=6pi-9rt3
as radius as 6 and chord ab is 6, then equilateral triangle with angle 60
area=60/360 *pi*6^2=6pi
step 2-calculate area of triangle
drop perpendicular from b to radius ao and create 30/60/90 triangle
hypotenuse=radius=6
therefore ht =3rt3
area=1/2* 6* 3rt3=9rt3
step 3 subtract area of triangle from sector
=6pi-9rt3
