Q1 of 15 Page 86

Prove that chords of the same length in a circle are at the same distance from the centre.


D is the centre of the circle. EF and GH are two chords of same length.


DI and DJ are two perpendiculars drawn on GH and EF respectively.


We have to prove that, DI = DJ.


In ΔDJF and ΔDIG we have,


DF = DG [radius of the same circle]


∠DJF = ∠DIG = 90° [∵DI and DJ are perpendiculars on GH and EF]


JF = IG [∵Perpendicular drawn from the centre bisects the chords]


∴∆DJF≅∆DIG [RHS congruency]


∴ DI = DJ [similar sides of congruent triangle]


∴ The chords are at same distance from the centre.


More from this chapter

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4

Prove that the angle made by two equal chords drawn from a point on the circle is bisected by the diameter through that point.

 5

Draw a square and a circle through all four vertices. Draw diameters parallel to the sides of the square and draw a polygon joining the end points of these diameters and the vertices of the square.


Prove that this polygon is a regular octagon.

 2

Two chords intersect at a point on a circle and the diameter through this point bisects the angle between the chords. Prove that the chords have the same length.

 3

In the picture on the right, the angles between the radii and the chords are equal.


Prove that the chords are of the same length.