- Messages
- 398
- Reaction score
- 685
- Points
- 103
O is the center of the sector OAB. CD is another arc, with center O and radius r cm. DB = 2cm. If the area of ABDC is one-third the area of the sector OCD find the value of r ?
-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Pure Math (As) , Help :(
- Thread starter GorgeousEyes
- Start date
- Messages
- 213
- Reaction score
- 9
- Points
- 28
Radius of CD=r and so radius of OB=r-2
area of ABDC=1/3 area of OCD
1/2(r square x Theta-(r-2)square x theta)= 1/3 r Sq x Theta
1/2(r(sq) theta -(r-2) sq Theta)=1/3 r(sq) Theta
2r(sq)=3(( r(sq)-(r-2)sq)) simplifying it
r(sq) -6r+ 6=0
solve for r using quadratic formula you get
r=3 +sq.rt(3)
area of ABDC=1/3 area of OCD
1/2(r square x Theta-(r-2)square x theta)= 1/3 r Sq x Theta
1/2(r(sq) theta -(r-2) sq Theta)=1/3 r(sq) Theta
2r(sq)=3(( r(sq)-(r-2)sq)) simplifying it
r(sq) -6r+ 6=0
solve for r using quadratic formula you get
r=3 +sq.rt(3)