CBSE Class 10 Answered
Explain the theorem:- Chords equidistant from the centre of a circle are equal in length?
Asked by | 27 Feb, 2009, 05:31: PM
Rememebr that the distance of a point from a line or line segment is always the perpendicular distance.
Consider that two chords AB and CD of a circle with center O.
Let the perpendicular from O meet AB at M and to CD at N.
Consider triangles OMB and OND ,
angle OMB = angle OND(90 degrees each)
So we see that both triangles are right triangles.
Now,
OB=OD(radii)
OM=ON(as it's given that the chords are equidistant from the center)
So,
triangle OMB congruent to triangle OND (RHS rule)
MB=ND( C.P.C.T.)
We know that the perpendicular from the centre of the circle bisects the chord.
So
MB=1/2(AB)
and
ND=1/2(CD)
Thus, MB=ND implies
1/2(AB)=1/2(CD)
i.e.
AB=CD
Hence proved.
Answered by | 28 Feb, 2009, 09:07: AM
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