22.) Q and R is the centre of two congruent circles intersecting each other at point C and D the line joining their centre intersects the circles in the point A and B such that A and B do not lie between Q and R . if CD=6cm and AB=12cm, determine the radius of either circles and the distance between the centre of two circles?
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For solutions, Refer to the attached picture.
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☛ A circle is the locus of a point which moves in a plane in such a way that its distance from a given fixed point is always constant.
☛ A line segment joining the centre and a point on the circle is called its radius.
☛ The perimeter of a circle is called its circumference.
Circumference = 2 π r
☛ A chord of a circle is a line segment joining any two points on the circle.
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Regrets for handwriting _/\_
_______________________
Let's see the related topics ;
☛ A circle is the locus of a point which moves in a plane in such a way that its distance from a given fixed point is always constant.
☛ A line segment joining the centre and a point on the circle is called its radius.
☛ The perimeter of a circle is called its circumference.
Circumference = 2 π r
☛ A chord of a circle is a line segment joining any two points on the circle.
_______________________
Thanks for the question !
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Heyy mate ❤✌✌❤
Here's your Answer. .
⤵️⤵️⤵️⤵️⤵️⤵️⤵️
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 = CD2 = 62 = 3 cm
And
O1P =O1O22 =AB − O1A − OB2 = 12 − r − r2 = 12 −2 r2= (6 − r) cm
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 .
✔✔✔
Here's your Answer. .
⤵️⤵️⤵️⤵️⤵️⤵️⤵️
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 = CD2 = 62 = 3 cm
And
O1P =O1O22 =AB − O1A − OB2 = 12 − r − r2 = 12 −2 r2= (6 − r) cm
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 .
✔✔✔
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