Two circles touch internally sum of their area is 116pie m^2 and distance between their centres is 6 cm. find the area of circle enclosed.
Answers
Answered by
3
Let the radius of 2 circles be r1 and r2
Sum of area = 116 pi
pi(r1)2 + pi(r2)2 = 116*pi
i.e. (r1)2 + (r2)2 = 116
Also, r1 - r2 = 6
r1 = 6+r2
Substituting it back in the above equation
(6+r2)2 + (r2)2 = 116
2(r2)2 + 12r2 + 36 = 116
2(r2)2 + 12r2 -80 = 0
(r2)2 + 6r2 - 40 = 0
(r2+10) (r2 - 4) = 0
r2 = -10 or r2 = 4
Since, radius of a circle can't be negative, hence, r2 = 4cm and r1 = 10 cm
Sum of area = 116 pi
pi(r1)2 + pi(r2)2 = 116*pi
i.e. (r1)2 + (r2)2 = 116
Also, r1 - r2 = 6
r1 = 6+r2
Substituting it back in the above equation
(6+r2)2 + (r2)2 = 116
2(r2)2 + 12r2 + 36 = 116
2(r2)2 + 12r2 -80 = 0
(r2)2 + 6r2 - 40 = 0
(r2+10) (r2 - 4) = 0
r2 = -10 or r2 = 4
Since, radius of a circle can't be negative, hence, r2 = 4cm and r1 = 10 cm
Attachments:
Similar questions