Question

equal chord of a circle equal distance from the centre​

Answer 1

$\bf \large\red{\mid{\overline{\underline{Statement \: given} \red{:-}}}\mid}$

➺ Equal chords of a circle ( or congruent circles ) are equidistant from the centre (or centres).

$\\ \large \mathcal \color{purple} \underline{Given}{:-}$

AB and CD are two equal chords of a circle.

where, AB = CD and OL ⊥ AB and OM ⊥ CD.

⠀⠀

$\large \mathcal \color{purple} \underline{To \: \: prove}{:-}$

Chord AB and CD are equidistant from the centre O, i.e. OL = OM.

⠀⠀⠀

$\large \mathcal \color{purple} \underline{Construction}{:-}$

Join OA and OC at point O.

⠀⠀⠀

$\large \mathcal \color{purple} \underline{Proof}{:-}$

∵ The perpendicular from the centre of a chord bisects the chord. (Theorem 10.3)

⠀⠀

Therefore, ⠀⠀⠀⠀⠀⠀⠀

$\bf{OL ⊥ AB \implies \: AL = \dfrac{1}{2}AB \: \: — \: eq.(i)}$

$\bf{and, \: OM ⊥ CD \implies CM=\frac{1}{2}CD \: \: — eq.(ii)}$

$\\ \bf{But, \: \: \: \: \: AB = CD}$

$\bf{\implies \dfrac{1}{2}AB = \dfrac{1}{2}CD}$

$\bf{\implies AL = CM \: \: \: \: [ using \: eq.(i) \: and \: (ii) ] \: \: — eq. (iii) }$

Now, in right ∆s OAL and OCM, we have

⠀⠀⠀OA = OC ⠀⠀⠀( Radii of same circle )

⠀⠀⠀ AL = CM ⠀⠀ [ From eq. (iii) ]

and, ∠ALO = ∠ CMO ⠀⠀ ( Each 90° )

⠀⠀

⠀⠀⠀ So, ∆OAL ≅ ∆ OCM ⠀⠀( by R.H.S rule )

Thus, OL = OM⠀⠀( C.P.C.T. )

⠀⠀

Hence proved.

Answer 2

Step-by-step explanation:

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

\begin{gathered} \\ \large \mathcal \color{purple} \underline{Given}{:-}\end{gathered}

Given

:−

AB and CD are two equal chords of a circle.

where, AB = CD and OL ⊥ AB and OM ⊥ CD.

⠀⠀

\large \mathcal \color{purple} \underline{To \: \: prove}{:-}

Toprove

:−

Chord AB and CD are equidistant from the centre O, i.e. OL = OM.

⠀⠀⠀

\large \mathcal \color{purple} \underline{Construction}{:-}

Construction

:−

Join OA and OC at point O.

⠀⠀⠀

\large \mathcal \color{purple} \underline{Proof}{:-}

Proof

:−

∵ The perpendicular from the centre of a chord bisects the chord. (Theorem 10.3)

⠀⠀

Therefore, ⠀⠀⠀⠀⠀⠀⠀

\bf{OL ⊥ AB \implies \: AL = \dfrac{1}{2}AB \: \: — \: eq.(i)}OL⊥AB⟹AL=

2

1

AB—eq.(i)

\bf{and, \: OM ⊥ CD \implies CM=\frac{1}{2}CD \: \: — eq.(ii)}and,OM⊥CD⟹CM=

2

1

CD—eq.(ii)

\begin{gathered} \\ \bf{But, \: \: \: \: \: AB = CD}\end{gathered}

But,AB=CD

\bf{\implies \dfrac{1}{2}AB = \dfrac{1}{2}CD}⟹

2

1

AB=

2

1

CD

\bf{\implies AL = CM \: \: \: \: [ using \: eq.(i) \: and \: (ii) ] \: \: — eq. (iii) }⟹AL=CM[usingeq.(i)and(ii)]—eq.(iii)

\begin{gathered} \\ \end{gathered}

Now, in right ∆s OAL and OCM, we have

⠀⠀⠀OA = OC ⠀⠀⠀( Radii of same circle )

⠀⠀⠀ AL = CM ⠀⠀ [ From eq. (iii) ]

and, ∠ALO = ∠ CMO ⠀⠀ ( Each 90° )

⠀⠀

⠀⠀⠀ So, ∆OAL ≅ ∆ OCM ⠀⠀( by R.H.S rule )

Thus, OL = OM⠀⠀( C.P.C.T. )

⠀⠀

Hence proved.