O1 AND O2 ARE THE CENTRES OF TWO CONGRUENT CIRCLES INTERSECTING EACH OTHER AT POINTS C AND D. THE LINE JOINING THEIR CENTRES INTERSECTS THE CIRCLES AT A AND B SUCH THAT AB > O1O2. IF CD = 6 CM AND AB = 12 CM FIND THE RADIUS OF THE EITHER CIRCLE.
Answer :
As circle are congruent , So they both have same radius . and AB > O1 O2 , SO we form our diagram , As :
Here
O1C = O1D = O2C = O2C = O1A = O2B = r
And
From O1C = O1D = O2C = O2C ,
We get Quadrilateral O1CO2D is a rhombus .
And we know diagonal of rhombus are perpendicular bisector , So
CD O1O2 And CD bisect O1O2 , So
CP =
And
O1P =
Now triangle O1PC , we apply Pythagoras theorem , and get
O1C2 = O1P2 + CP2 , Substitute values , we get
r2 = ( 6 - r )2 + 32
r2 = 36 + r2 - 12r + 9
12r = 45
r = 3.75
So,
Radius of either circle = 3.75 cm ( Ans )
As circle are congruent , So they both have same radius . and AB > O1 O2 , SO we form our diagram , As :
Here
O1C = O1D = O2C = O2C = O1A = O2B = r
And
From O1C = O1D = O2C = O2C ,
We get Quadrilateral O1CO2D is a rhombus .
And we know diagonal of rhombus are perpendicular bisector , So
CD O1O2 And CD bisect O1O2 , So
CP =
And
O1P =
Now triangle O1PC , we apply Pythagoras theorem , and get
O1C2 = O1P2 + CP2 , Substitute values , we get
r2 = ( 6 - r )2 + 32
r2 = 36 + r2 - 12r + 9
12r = 45
r = 3.75
So,
Radius of either circle = 3.75 cm ( Ans )