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CAT Relation Between Arc and Chord - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Syllabus

Quick Facts

4 Questions around this concept.

Solve by difficulty

EASY

If two chords of congruent circles are equal then the corresponding minor arcs are equal. (True/False)

True

False

View Solution

Concepts Covered - 1

Relation Between Arc and Chord

Property 1: If two arcs of a circle (or of congruent circles) are congruent then the corresponding chords are equal.

Let's prove this.

We have given a circle C(O,r) in which, $\mathrm{\widehat{A B} \cong \widehat{C D}}$ .

And we need to prove chord AB = chord CD.

Case I When $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are minor arc.

Join OA, OB, OC and OD.

In triangle AOB and COD, we have.

OA = OC (Radii of a circle)

OB = OD (Radii of a circle)

∠ AOB = ∠ COD $\mathrm{\left [ \because \widehat{A B} \cong \widehat{C D} \Rightarrow \mathrm{m}(\widehat{A B})=\mathrm{m}(\widehat{C D}) \right ]}$

Therefore, ∆ AOB ≅ ∆ COD (by SAS rule)

This gives AB = CD (Corresponding parts of congruent triangles)

Case II When $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are major arc.

In this case, $\mathrm{\widehat{ BA}}$ and $\mathrm{\widehat{DC}}$ are minor arc

$\therefore \quad \widehat{A B} \cong \widehat{C D} \Rightarrow \widehat{B A} \cong \widehat{D C} \Rightarrow B A=D C$

$\Rightarrow \quad AB=CD$

Hence, in both cases, we have AB = CD.

Property 2: If two chords of a circle (or of congruent circles) are equal then their corresponding arcs (semicircular, minor or major) are congruent.

Let's prove this

Let given a circle C(O,r) in which chord AB = chord CD.

We need to prove $\mathrm{\widehat{A B} \cong \widehat{C D}}$ , where both $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are either semicircular, minor or major arcs.

Case I When AB and CD are diameters

In this case, $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are semicircles with the same radii.

So, $\mathrm{\widehat{A B} \cong \widehat{C D}}$

Thus, $\mathrm{A B=C D \Rightarrow \widehat{A B} \cong \widehat{C D}}$ .

Case II When chord AB = chord CD, where $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are minor arcs.

In triangle AOB and COD, we have.

AB = CD (given)

OA = OC (Radii of a circle)

OB = OD (Radii of a circle)

Therefore, ∆ AOB ≅ ∆ COD (by SAS rule)

This gives ∠ AOB = ∠ COD (Corresponding parts of congruent triangles)

Now,

$\begin{aligned} \angle A O B=\angle C O D & \Rightarrow \mathrm{m}(\widehat{A B})=\mathrm{m}(\widehat{C D}) \\ & \Rightarrow \widehat{A B} \cong \widehat{C D} \end{aligned}$

Case II When chord AB = chord CD, where $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are major arcs.

In this case, $\mathrm{\widehat{ BA}}$ and $\mathrm{\widehat{DC}}$ are minor arc

$\begin{aligned} \therefore \quad A B=C D & \Rightarrow B A=D C \\ & \Rightarrow \widehat{B A} \cong \widehat{D C} \\ & \Rightarrow \mathrm{m}(\widehat{B A})=\mathrm{m}(\widehat{D C}) \\ & \Rightarrow 360^{\circ}-\mathrm{m}(\widehat{A B})=360^{\circ}-\mathrm{m}(\widehat{D C}) \\ & \Rightarrow \mathrm{m}(\widehat{A B})=\mathrm{m}(\widehat{C D}) \\ & \Rightarrow \widehat{A B} \cong \widehat{C D} \end{aligned}$

Hence, in all the cases $A B=C D \Rightarrow \widehat{A B} \cong \widehat{C D}$ .

Property 2: If two chords of a circle (or of congruent circles) are equal then their corresponding arcs (semicircular, minor or major) are congruent.

Let's prove this

Let given a circle C(O,r) in which chord AB = chord CD.

We need to prove $\mathrm{\widehat{A B} \cong \widehat{C D}}$ , where both $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are either semicircular, minor or major arcs.

Case I When AB and CD are diameters

In this case, $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are semicircles with the same radii.

So, $\mathrm{\widehat{A B} \cong \widehat{C D}}$

Thus, $\mathrm{A B=C D \Rightarrow \widehat{A B} \cong \widehat{C D}}$ .

Case II When chord AB = chord CD, where $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are minor arcs.

In triangle AOB and COD, we have.

AB = CD (given)

OA = OC (Radii of a circle)

OB = OD (Radii of a circle)

Therefore, ∆ AOB ≅ ∆ COD (by SAS rule)

This gives ∠ AOB = ∠ COD (Corresponding parts of congruent triangles)

Now,

$\begin{aligned} \angle A O B=\angle C O D & \Rightarrow \mathrm{m}(\widehat{A B})=\mathrm{m}(\widehat{C D}) \\ & \Rightarrow \widehat{A B} \cong \widehat{C D} \end{aligned}$

Case II When chord AB = chord CD, where $\mathrm{\widehat{A B}}$ and $\mathrm{\widehat{CD}}$ are major arcs.

In this case, $\mathrm{\widehat{ BA}}$ and $\mathrm{\widehat{DC}}$ are minor arc

Hence, in all the cases $A B=C D \Rightarrow \widehat{A B} \cong \widehat{C D}$ .