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The following statement: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

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True

False

answer is A.

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Detailed Solution

There are two equal chords in the circle.
AB=CD
C:UsersadminDesktopjkjkhjh.PNG

Here,

AE = BE = DF = CF = \frac{x}{2} . . . . . . . . . (1)

By Pythagoras theorem in

Δ ABC

(AC)^{2} = (AB)^{2} + (BC)^{2} . . . . . . . . (2)

Again, in

Δ OEB

(OB)^{2} = (OE)^{2} + (EB)^{2}

Let

OB = r

(radius) and

EB = \frac{x}{2}

. Thus,

(r)^{2} = (OE)^{2} + {(\frac{x}{2})}^{2}

\Rightarrow O E^{2} = r^{2} - \frac{x^{2}}{4}

\Rightarrow O E^{2} = \sqrt{r^{2} - \frac{x^{2}}{4}} . . . . . . . . . (3)

Using Pythagoras theorem in

Δ OFC

(OC)^{2} = (OF)^{2} + (FC)^{2}

OB

= r

(radius) and,

FC = \frac{x}{2}

. Thus,

(r)^{2} = (OF)^{2} + {(\frac{x}{2})}^{2}

\Rightarrow O F^{2} = r^{2} - \frac{x^{2}}{4}

\Rightarrow O F^{2} = \sqrt{r^{2} - \frac{x^{2}}{4}} . . . . . . . . . (4)

From equation

(3) (4)

, we get,

OE = OF

Thus, equal chords of a circle are equidistant from the centre.
Option 1 is Correct.

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