class 9 maths ex 10.4 question 2
Answers!
Q.2 If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Sol.Given : AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.
To prove : (i) AP = PD (ii) PB = CP.
Construction : Draw OM⊥AB , ON⊥CD.
Join OP,
AM = MB =12AB [Perpendicular from centre bisects the chord]
CN = ND =12CD [Perpendicular from centre bisects the chord]
AM = ND and MB = CN [Since AB = CD (given)]
InΔsOMP and ONP , we have
OM = ON [Equal chords of a circle are equidistant from the centre]
∠OMP=∠ONP[Since Each = 90º]
OP = OP [Common]
By RHS' criterion of congruence,
ΔOMP≅ΔONP
⇒MP = PN ... (2) [C.P.C.T.]
Adding (1) and (2) , we have
AM MP = ND PN⇒AP = PD
Subtracting (2) from (1), we have
MB – MP = CN – PN⇒PB = CP
Hence (i) AP = PD and (ii) PB = CP
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Answer the question...
Q.2 If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Sol.
Sol.
Given : AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.
To prove : (i) AP = PD (ii) PB = CP.
Construction : Draw OM
To prove : (i) AP = PD (ii) PB = CP.
Construction : Draw OM
⊥
AB , ON
⊥
CD.
Join OP,
Join OP,
AM = MB =
1
2
AB [Perpendicular from centre bisects the chord]
CN = ND =
CN = ND =
1
2
CD [Perpendicular from centre bisects the chord]
AM = ND and MB = CN [Since AB = CD (given)]
In
AM = ND and MB = CN [Since AB = CD (given)]
In
Δs
OMP and ONP , we have
OM = ON [Equal chords of a circle are equidistant from the centre]
OM = ON [Equal chords of a circle are equidistant from the centre]
∠OMP=∠ONP
[Since Each = 90º]
OP = OP [Common]
By RHS' criterion of congruence,
OP = OP [Common]
By RHS' criterion of congruence,
ΔOMP≅ΔONP
⇒
MP = PN ... (2) [C.P.C.T.]
Adding (1) and (2) , we have
AM MP = ND PN
Adding (1) and (2) , we have
AM MP = ND PN
⇒
AP = PD
Subtracting (2) from (1), we have
MB – MP = CN – PN
Subtracting (2) from (1), we have
MB – MP = CN – PN
⇒
PB = CP
Hence (i) AP = PD and (ii) PB = CP
Hence (i) AP = PD and (ii) PB = CP
Comment