Perpendicular From Centre of Circle Bisects the Chord and Vice Versa


 
 
Concept Explanation
 

Perpendicular From Centre of Circle Bisects the Chord and Vice Versa

Theorem 1   The perpendicular  from the centre of a circle to a chord bisects the chord.

Given      A chord AB of a circle C (O,r) and perpendicular OD to the chord AB.

TO PROVE   AD = DB

CONSTRUCTION   Join OA and OB.

PROOF    In triangles AOD and BOD, we have

                   dpi{120} large OA = OB = r       [ Radii of the same circle ]

                 Large OD=OD                  [ Common ]

and,       dpi{120} large angle ODA = angle ODB       [ Each equal to large 90^{circ}]

So, by RHS - criterion of congruence, we have

                dpi{120} large Delta ADO cong Delta BDO

large Rightarrow                dpi{120} large AD =BD

Hence Proved

Theorem 2: A line drawn from the centre of a circle to the mid point of the chord is perpendicular to the chord

Given      A chord AB of a circle C (O,r) and a line OD which bisects the chord AB.

TO PROVE   OD is perpendicular to AB

CONSTRUCTION   Join OA and OB.

PROOF    In triangles AOD and BOD, we have

                   dpi{120} large OA = OB = r       [ Radii of the same circle ]

                 Large OD=OD                  [ Common ]

and,       dpi{120} large AD =BD                      [ Given ]

So, by SSS - criterion of congruence, we have

                dpi{120} large Delta ADO cong Delta BDO

large Rightarrow   dpi{120} large angle ODA = angle ODB      

 But they are linear pairs so

 dpi{120} large angle ODA = angle ODBlarge 90^{circ}

     Hence Proved

 

Illustration: If the length of a chord of a circle is 24 cm and is at a distance of 9 cm from the centre of the circle, Find the radius of the circle .

The length of a chord of a circle  AB = 24 cm

Perpendicular distance from the centre OD =  9 cm

As the perpendicular bisects the chord AD = 12 cm

large OA^2 = OD^2+ AD^2

large OA^2 = 9^2+ 12^2 = 81+144= 225= 15^2

OA = 15 cm

 

 

Sample Questions
(More Questions for each concept available in Login)
Question : 1

In the figure given below, large OP=ON=4 units and OM = 3 units.

The length of the chord is ____________

Right Option : A
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Question : 2

In the figure, OM perp PN and PN=7;units. What is the length of PM ?

        

 

Right Option : B
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Explanation
Question : 3

In the figure given below, large OP=ON=4 units and OM = 3 units.

The perpendicular distance of chord PN from the centre O is    ____________

Right Option : D
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Explanation
 
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