Perimeter
This topic has expert replies
-
- Legendary Member
- Posts: 829
- Joined: Mon Jul 07, 2008 10:09 pm
- Location: INDIA
- Thanked: 84 times
- Followed by:3 members
-
- Legendary Member
- Posts: 1153
- Joined: Wed Jun 20, 2007 6:21 am
- Thanked: 146 times
- Followed by:2 members
We know that CD is parallel to the diameter.
Adjacent angles will be same. Therefore x=30
X+X = 60
Therefore the central angle would be = 120
arc CAE = 120/360 * 2pi5 = 10/3 pi
we have to find the length of EB & CB.
Triangles CAB & ABE are right triangle, and one of the angle is 30.
The diameter is the hypotenuese therefore
EB = 5 sqrt 3
CB = 5 sqrt 3
The total perimeter = 10/3 pi + 10sqrt3
The answer is D.
Adjacent angles will be same. Therefore x=30
X+X = 60
Therefore the central angle would be = 120
arc CAE = 120/360 * 2pi5 = 10/3 pi
we have to find the length of EB & CB.
Triangles CAB & ABE are right triangle, and one of the angle is 30.
The diameter is the hypotenuese therefore
EB = 5 sqrt 3
CB = 5 sqrt 3
The total perimeter = 10/3 pi + 10sqrt3
The answer is D.
shaded region's part of the circle can be found by
formula of the length of arc= @/360 * 2*pie*r
i.e x=30 (CD II AB)
thus @=x+x= 60
Lenght of the shaded arc= 60/360 *2 * pie * 5 = 10/6 *pie..(1)
Angle CBA = 30 , Angle ACB = 90, angle BAC=60
thus a right angle triangle will have sides in ratio : 1:sqroot 3 : 2
Thus,CB=5 sqroot3 , AB=10 and CA = 5
CB=EB=5sqroot3
Perimeter: 5sqroot3+5sqroot3+10/6 *pie (frm 1)
= 5/3 * pie + 10 sqroot3
IMO:B
formula of the length of arc= @/360 * 2*pie*r
i.e x=30 (CD II AB)
thus @=x+x= 60
Lenght of the shaded arc= 60/360 *2 * pie * 5 = 10/6 *pie..(1)
Angle CBA = 30 , Angle ACB = 90, angle BAC=60
thus a right angle triangle will have sides in ratio : 1:sqroot 3 : 2
Thus,CB=5 sqroot3 , AB=10 and CA = 5
CB=EB=5sqroot3
Perimeter: 5sqroot3+5sqroot3+10/6 *pie (frm 1)
= 5/3 * pie + 10 sqroot3
IMO:B
-
- Master | Next Rank: 500 Posts
- Posts: 222
- Joined: Fri Jan 04, 2008 8:10 pm
- Thanked: 15 times
shaded region's part of the circle can be found by
formula of the length of arc= @/360 * 2*pie*r
(@ is central angle)
i.e x=30 (CD II AB) (Inscribed angle)
thus @=x+x= 60 (Central angle would be 2 times the inscribed angle)
Lenght of the shaded arc= 60/360 *2 * pie * 5 = 10/6 *pie..(1)
Angle CBA = 30 , Angle ACB = 90, angle BAC=60
thus a right angle triangle will have sides in ratio : 1:sqroot 3 : 2
Thus,CB=5 sqroot3 , AB=10 and CA = 5
CB=EB=5sqroot3
Perimeter: 5sqroot3+5sqroot3+10/6 *pie (frm 1)
= 5/3 * pie + 10 sqroot3
IMO:B
formula of the length of arc= @/360 * 2*pie*r
(@ is central angle)
i.e x=30 (CD II AB) (Inscribed angle)
thus @=x+x= 60 (Central angle would be 2 times the inscribed angle)
Lenght of the shaded arc= 60/360 *2 * pie * 5 = 10/6 *pie..(1)
Angle CBA = 30 , Angle ACB = 90, angle BAC=60
thus a right angle triangle will have sides in ratio : 1:sqroot 3 : 2
Thus,CB=5 sqroot3 , AB=10 and CA = 5
CB=EB=5sqroot3
Perimeter: 5sqroot3+5sqroot3+10/6 *pie (frm 1)
= 5/3 * pie + 10 sqroot3
IMO:B
-
- Legendary Member
- Posts: 829
- Joined: Mon Jul 07, 2008 10:09 pm
- Location: INDIA
- Thanked: 84 times
- Followed by:3 members