Exercise 9.1 (Revised) - Chapter 10 - Circles - Ncert Solutions class 9 - Maths

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Chapter 9: Circles - NCERT Solutions for Class 9 Maths | Expert Answers & PDF

Ex 9.1 Question 1.

Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer.

I Part: Two circles are said to be congruent if and only if one of them can be superposed on the other so as to cover it exactly.

Let $\mathrm{C}(\mathrm{O}, r)$ and $\mathrm{C}\left(\mathrm{O}^{\prime}, s\right)$ be two circles. Let us imagine that the circle $\mathrm{C}\left(\mathrm{O}^{\prime}, s\right)$ is superposed on $\mathrm{C}(\mathrm{O}, r$ ) so that $O^{\prime}$ coincide with $O$. Then it can easily be seen that $C\left(O^{\prime}, s\right)$ will cover $C(O, r)$ completely if and only if $r=s$.

Hence we can say that two circles are congruent, if and only if they have equal radii.

II Part: Given: In a circle $(\mathrm{O}, r), \mathrm{AB}$ and $\mathrm{CD}$ are two equal chords, subtend $\angle \mathrm{AOB}$ and $\angle \mathrm{COB}$ at the centre.

To Prove: $\angle \mathrm{AOB}=\angle \mathrm{COD}$

Proof: In $\triangle \mathrm{AOB}$ and $\triangle C O D$,
$A B=C D[$ Given $]$
$A O=C O[$ Radii of the same circle]
$\mathrm{BO}=\mathrm{DO}$ [Radii of the same circle]
$\therefore \triangle \mathrm{AOB} \cong \triangle$ COD [By SSS axiom]
$\Rightarrow \angle \mathrm{AOB}=\angle \mathrm{COD}[\mathrm{By} \quad \mathrm{CPCT}]$

Hence Proved.

Ex 9.1 Question 2.

Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer.

Given: In a circle $(\mathrm{O}, r), \mathrm{AB}$ and $\mathrm{CD}$ subtend two angles at the centre such that $\angle \mathrm{AOB}=\angle \mathrm{COD}$

To Prove: $A B=C D$

Proof: : In $\triangle A O B$ and $\triangle C O D$,
$\mathrm{AO}=\mathrm{CO}$ [Radii of the same circle]
$\mathrm{BO}=\mathrm{DO}$ [Radii of the same circle]
$\angle \mathrm{AOB}=\angle \mathrm{COD}$ [Given]
$\therefore \triangle \mathrm{AOB} \cong \triangle C O D$ [By SAS axiom]
$\Rightarrow \mathrm{AB}=\mathrm{CD}[\mathrm{By}$ CPCT]

Hence proved.