Question

Ncert solution for class 9th maths theorem 10.7 with proof

Answer 1

Theorem 10.7:

Chords equidistant from the centre of a circle are equal in length.

Given:

AB and CD are two chords of a circle C(O,r), OL perpendicular AB, OM perpendicular CD , OL= OM

To Prove: AB= CD

Construction: join OA & OC

Proof: WE KNOW THAT THE PERPENDICULAR FROM THE CENTRE OF A CIRCLE TO A CHORD BISECTS THE CHORD.

AL= 1/2AB & CM= 1/2CD........(1)

In right ∆OLA & ∆OMC

OA=OC (RADIUS)

OL= OM. (GIVEN)

Angle OLA= Angle OMC (EACH 90°)

∆OLA Congruent to ∆OMC

AL = CM

1/2AB = 1/2CD

[From eq 1]

AB = CD

Hence, AB=CD

==================================================================
Hope this will help you....

Answer 2

Statement : Chords equidistant from the centre of a circle are equal in length.

Given: OM and ON are perpendicular from the centre to the chords AB and CD and OM = ON.

To prove : chords AB = chord CD

Construction : Join OA and OC.

PROOF :

OM ⊥ AB $⇒ \frac{1}{2}$ AB = AM

ON ⊥ CD $⇒ \frac{1}{2}$ CD = CN

Consider, ΔAOM and ΔCON,

OA = OC . (radii of the circle)
OM = ON . (given)
∠OMA = ∠ONC = 90 ° . (given)

ΔAOM ≅ ∠CON . (RHS congruency)

AM = CN $⇒ \frac{1}{2}$ AB = $\frac{1}{2}$ CD $⇒$ AB = CD

The two chords are equal if they are equidistant from the centre.