SChand CLASS 9 Chapter 13 Circles Exercise 13(B)

 Exercise 13 B

Question 1

Sol: (i) If Angle at a point=$360^{\circ}$

So $\begin{aligned} & 4 x-2+6 x+6+7 x-18=360^{\circ} \\ \Rightarrow & 17 x-14=360^{\circ} \\ \Rightarrow & 17 x=360^{\circ}+14 \\ \Rightarrow & 17 x=374^{\circ} \\ \Rightarrow & x=\frac{374}{17}=22^{\circ} \end{aligned}$

Now MQR = $7 x-18=7 \times 22-18^{\circ}$

$=154^{\circ}-18^{\circ}=136^{\circ}$

(ii) if $A D$ is the diameter of the circle with center $O$

So $m \angle A O D=180^{\circ}$

So $10 y=180^{\circ} \Rightarrow y=\frac{180^{\circ}}{10}=18^{\circ}$

NOW $ M \angle B O C=6 y=6 \times 18=108^{\circ}$

(iii) In the figure.

$\widehat{A C}=\widehat{B C}$

So $\angle A O C=\angle B O C$

So $8 y-8=6 y$

$\Rightarrow 8 y-6 y=8 \Rightarrow 2 y=8$

$\Rightarrow \quad y=4$

So $M \angle A O C=8 y-8=8 \times 4-8=32-8=240$

so $M \angle A O C=24^{\circ}$

(iv). In the given figure,

$O A=O B$

$\Rightarrow$ circles with centre $A$ and $B$ are equal and $\hat{C D}=\hat{E F}$

$45-6 x=9 x \Rightarrow 45=9 x+6 x$

$\Rightarrow 15 x=45 \Rightarrow x=\frac{45}{15}=3$

Now $m \angle E B F=9 x=9 \times 3=27^{\circ}$

if $m \angle C A D=45+6 x$, then

$45+6 x=9 x \Rightarrow 45=9 x-6 x$

$\Rightarrow 3 x=45 \Rightarrow x=\frac{45}{3}=15$

$m(\angle E B F)=9 x=9 \times 15=135^{\circ}$

(v)In the given figure, a circle with contre $O$ and $Q T$ and PS are diameters

$m P R=P Q+m Q R$

$=M S T+m Q R$

$(\because \angle P O Q=\angle S O T$ vertically opposite angles) $=55^{\circ}+100^{\circ}=155^{\circ}$

and $M P R T=m \angle P Q+M \angle Q R T$

$=55^{\circ}+180^{\circ}=235^{\circ}$

(vi) In the figure chord $A B=$ chord $C D$

$\begin{aligned}&\text { so } 4 y+y=y+68^{\circ} \\&5 y=y+68^{\circ} \Rightarrow 5 y-y=68^{\circ} \Rightarrow 4 y=68^{\circ} \\&\Rightarrow y=\frac{68^{\circ}}{4}=17^{\circ} \\&m \overbrace{A B}=4 y+y=5 y \\&=5 \times 17^{\circ}=85^{\circ}\end{aligned}$

Question 2

Sol: $\triangle A B C$ is inscribeb in a circle and $\angle P=\angle Q$

 so $A R=R R$

(IMAGE TO BE ADDED)

if equal chords subtend equal angles at the center 

So MPR =MQR 

Question 3

Sol: In the given figure.

A B=C D

$A C$ and $B D$ are joined

To prove: $A C=D B$

if $A B=D C$

so $m \hat{A B}=m \widehat{C}$

Subtracting MAD from both side

So $m \hat{A B}-\hat{A D}=m \overline{D C}-m \overline{A D}$

$\Rightarrow m A C=m D{D B}$

$\Rightarrow A C=D B$ Hence proved

Question 3

Sol: In the given figure,

$A B=C D$

$A C$ and $B D$ are Joinct

To prove: $A C=D B$

if $A B=D C$

so $M A B=m \bar{D}$

Substracting MAD form both sides 

So MAB - AD= MDC- MAD

=MAC = MDB 

= AC =DB Hence proved 

Question 4

Sol: (image to be added)

In the figure A C=D B

To prove: $A B=B C$

 $\begin{aligned} A C &=D B \\ \text { So } A C &=D B \end{aligned}$

Adding $M A D$ to both sides

$\hat{A C}+\widehat{D B}=\hat{A D}+\widehat{D B}$

$\widehat{C D}=\widehat{A B}$

$\Rightarrow C D=A B$

Hence $A B=C D$

Question 5

Sol: In the circle

$A B$ and $A C$ are two arcs

$X$ and $y$ are the mid points of Are $A B$ and are $A C$.

$x y$ is Joined which meet

$A B$ in $P$ and $A C$ in $Q$

To prove: $A P=A Q$

construction : Join $A X, A Y, B X$ and $B Y$

(image to be added)

proof: if $A B=A C$

So Arc A X B=arc A Y C

But X and Y are the mid point of $\widehat{A B}$ and $\widehat{A C}$

So $A X=X B$ and $A Y=4 C$

So $\angle X A Y=\angle X B A$ and $\angle Y A C=\angle Y C A$

So $\angle X A B$ or $\angle X A P=\angle Y A Q$

Now in $\triangle X A P$ and $\triangle Y A Q$

$A Y=A Y$

$\angle X A P=\angle Y A Q$

$\angle A X P=\angle A Y Q$

So $\triangle X A P \cong \triangle Y A Q$

So $A P=A Q$ Hence proved 

Question 6

Sol: In a circle with center $O$, chord $A B=B C=C D=D E$ $A D$ and $D E$ are joined

To prove: $A D=D E$

construction: Join AO,BO, CO, DO and EO

(image to be added)

Proof: if $A B=B C=C D=D E$

So $\hat{A B}+\hat{B C}+\hat{C D}=\hat{A D}$

similanly $B C+C D+D E=B E$

$\begin{aligned}&A B+B C+C D=B C+C D+D E \\&\Rightarrow A D=B E\end{aligned}$

So Center $\angle A O D=$ central $\angle B O E$

Now in $\triangle A O D$ and $\triangle B O E$,

$\begin{aligned}&O A=O B \\&O D=O E \\&\angle A O D=\angle B O E \\&\text { sO } \triangle A O D \cong \triangle B O E \\&\text { SO } A D=B E\end{aligned}$

Question 7

Sol: In a circle, arc APB = arc CQD AC and BD are joined 

To prove: AC = BD 

Construction : Join AD 

Proof: If Arc APB = Arc CQD 

So $\angle A D B=\angle C A D$

(Equal arcs subtends equal angles at the circumference )

But these are alternate angles

So AC||BD Hence proved  

Question 8

Sol: In the figure equilateral $\triangle A B C$ is inscribed in acircle $P$ and $s$ are midpoint ag ares $A B$ and AC respectively

To prove: $P Q=Q R=K S$

constructions: Join AP, BP, AS and CS

Proof : If p is the mid point of Arc and S is the mid point of Arc AC 

 (IMAGE TO BE ADDED)

So $A P=P B$ and $A S=S C$

So $\angle P A B=\angle P B A$ and $\angle C A S=\angle S A C$

But $A B=A C$

So $\angle P A B=\angle P B A=\angle C A S=\angle S A C$

Now in $\triangle A P Q$ and $\triangle A S R$

$\begin{aligned}&A P=A S \\&\angle P A Q=\angle S A R \\&\text { and } C A P Q=\angle A S R \\&\text { SO } \triangle P A Q \cong \triangle A S R \\&\text { so } P Q=R S \\&\text { NOW } \because \angle P A Q=\angle S A R\end{aligned}$

Adding $\angle Q A R$ to both sides

$\text { So } \angle M A+\angle Q A R=\angle Q A R+\angle S A R$

Now in $\triangle P A R$ and $\triangle S A Q$

$\begin{gathered}A P=A S \\\angle P A R=\angle S A Q \\\text { So } \angle A P Q=\angle A S \\\triangle P A R = \triangle S A Q \\\text { SO } P R=Q S\end{gathered}$

subtracting is and from (ii)

$\begin{aligned}&P R-P Q=Q S-R S \\&Q R=Q R \\&\text { SO } P Q=Q R=R S\end{aligned}$

Question 9

Sol: A regular hexagon ABCDEF Inscribe in a circle with center O join AO, BO , CO , DO ,EO and FO 

(IMAGE TO BE ADDED)

To prove: Each angle of hexagon = 60

Proof : If AB = BC = CD =DE= EF= FA 

So $\hat{A B}=\hat{B C}=\hat{C D}=\hat{D E}=E F=F F$

So each are will subtends angle at the center 

$=\frac{360^{\circ}}{6}=60^{\circ}$

Hence proved