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another geo problem

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another geo problem

by giatch » Fri Dec 26, 2008 3:52 pm
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

See attachment for figure.
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by crazykrans » Fri Dec 26, 2008 4:18 pm
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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Re: another geo problem

by kanha81 » Fri Apr 17, 2009 12:22 pm
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?
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by xunil56 » Fri Apr 17, 2009 1:22 pm
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)
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by arora007 » Fri Jun 04, 2010 4:54 am
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)
Xunil56 i can catch a small typo....

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.
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by gmatjedi » Sat Jun 05, 2010 4:40 am
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
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