S Chand Class 10 CHAPTER 14 Circle Exercise 14 A

  Exercise 14 A

Question 1

Ans: (i) $A B$ is diameter of the circle

So $\angle A C B=90^{\circ}$

So $\angle B A C+\angle A B C=90^{\circ}$

$\Rightarrow 46^{\circ}+\angle A D C=90^{\circ}$

$\Rightarrow \angle A B C=90^{\circ}-46^{\circ}=44^{\circ}$

 

(ii) $A C$ is diameter of the circle

So $\angle A D C=\angle A B C=90^{\circ}$

So $\angle B A C+\angle B C A=90^{\circ}$.

$\begin{aligned}\Rightarrow & \angle B C A=90^{\circ}-70^{\circ}=20^{\circ} \\\text { and } & \angle D A C+\angle D C A=90^{\circ} \\\Rightarrow & \text { 40 }+\angle D C A=90^{\circ} \\\Rightarrow & \angle D C A=90^{\circ}-40^{\circ}=50^{\circ} \\& \text { So } \angle B C D=\angle B C A+\angle A C D=20^{\circ}+50^{\circ}=70^{\circ}\end{aligned}$

(iii) Arc $A B$ subtends $\angle A O B$ at the centre and $\angle A C B$ at 

He remaining part of the circle So $\angle A O B=2 \angle A C B$

$\Rightarrow \angle x=2 \angle y \Rightarrow 2 \angle y=120^{\circ}$

$\Rightarrow \quad \angle y=\frac{120^{\circ}}{2}=60^{\circ}$

So $\quad \angle y=60^{\circ}$

(iv) Bc is diameter of the circle 

So $\angle B A C=90^{\circ}$

$\Rightarrow \angle B A D+\angle O A C=90^{\circ}$

$\Rightarrow \angle B A D+65^{\circ}=90^{\circ}$

$\Rightarrow \angle B A D=90^{\circ}-65^{\circ}=25^{\circ}$

But $\angle B C D=\angle B A D$

Question 2

Ans: In the figure $O$ is the centre af the circle $\angle O A B$ $=30^{\circ}$ and $\angle O C B=40^{\circ}$ Join $O B$

(IMAGE TO BE ADDED)

In $\triangle O A B, O A=O B$

So $\angle O B A=\angle O A B=30^{\circ}$

similarly in $\triangle O B C, O B=O C$

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

So $\angle A B C=\angle O B A+\angle O B C=30^{\circ}+40^{\circ}=70^{\circ}$

Now are $A B$ subtends $\angle A O C$ at the center

and $\angle A B C$ at the remaining part of the circle So $\angle A O C=2 \angle A B C=2 \times 70^{\circ}=140^{\circ}$.

So $\angle A O C=140$

Question 3

Ans: In the figure $O$ is the center and $\mathrm{BOC}$ is the diameter of the circle $\angle A O B=80^{\circ}$

 $A C$ and $A B$ are joined

(IMAGE TO BE ADDED)

Now arc AB subtends 

$\angle A O B$ at the center and $\angle A C B$ at the remaining part of the circle 

So $\angle A O B=2 \angle A C B$

$\Rightarrow 80^{\circ}=2 \angle A C B \Rightarrow \angle A C B=\frac{80^{\circ}}{2}=40^{\circ}$

But in $\triangle O A C \cdot O A=O C$

So $\angle O A C=\angle O C A=\angle A C B=40^{\circ}$

But $\angle B A C=90^{\circ}$

$\begin{aligned}&\Rightarrow \angle O A C+\angle O A B=90^{\circ} \\&\Rightarrow 40^{\circ}+\angle O A B=90^{\circ} \\&\Rightarrow \angle O A B=90^{\circ}-40^{\circ}=50^{\circ}\end{aligned}$

Hence $\angle O A B=50^{\circ}$ and $\angle O A C=40^{\circ}$

Question 4

Ans: In the figure 0 is the center of the circle $\angle A O B=140^{\circ}, \angle O A C=50^{\circ}$

Join $A B$

(IMAGE TO BE ADDED)

So  Reflex $\angle A O B=360^{\circ}-140^{\circ}=220^{\circ}$

Now major arc AB subtends 

Ref $\angle A O B$ at the center $\angle A C B$ at the

Remaining part of the circle

(i) So ref $\angle A O B=2 \angle A C B$

$=220^{\circ}=2 \angle A C B \Rightarrow \angle A C B=\frac{1}{2} \times 2200$

$\Rightarrow \angle A C B=110^{\circ}$

(ii)In quad. OACB

$\angle B O A+\angle O A C+\angle A C B+\angle O B C=360^{\circ}$

$=140^{\circ}+50^{\circ}+110^{\circ}+\angle O B C=360^{\circ}$

$\Rightarrow 300^{\circ}+\angle O B C=360^{\circ}$

$\Rightarrow \angle O B C=360^{\circ}-300^{\circ}=60$

(iii)

$\begin{aligned} & \angle O A B+\angle O B A+\angle A O B=180^{\circ} \\ \Rightarrow & \angle O A B+\angle O B A+140^{\circ}=180^{\circ} \\ \Rightarrow & \angle \angle O A B=180^{\circ}-140^{\circ}=40^{\circ} \\ \text { So } \angle O A B=\frac{40}{2}=20^{\circ} \end{aligned}$

(iv) In $\triangle A B C$

$\angle B A C+\angle C B A+\angle A C B=180^{\circ}$

$\begin{aligned}&\Rightarrow 30^{\circ}+\angle C B A+110^{\circ}=180^{\circ} \\&\Rightarrow 140^{\circ}+\angle C B A=180^{\circ} \\&\Rightarrow \angle C B A=180^{\circ}-140^{\circ}=40^{\circ}\end{aligned}$

Question 5

Ans: (i) In the figure $\angle A D C=50^{\circ} \angle B D C=40^{\circ}$

AB is the diameter of the circle 

If $\angle B D A=90^{\circ}$

So $\angle C O A=\angle B D A-\angle B D C=90^{\circ}-40^{\circ}=50^{\circ}$

(ii)  $\angle B C A=90^{\circ}$

So $\angle B A C+\angle A B C=90^{\circ}$

$\Rightarrow \angle D A C+50^{\circ}=90^{\circ}$

$\Rightarrow \angle B A C=90^{\circ}-50^{\circ}$

$\Rightarrow \angle B A C=40$

Hence (i) $\angle C D A=50^{\circ}$ (ii) BAC = $40^{\circ}$ and (iii) $\angle B C A=90^{\circ}$

Question 6

Ans: In the figure AC is the diameter of the circle and $\angle P C D=75^{\circ}$

ii $\begin{aligned} & \angle A B C=90^{\circ} \\ \text { ii } & \angle B C D+\angle B A D=180^{\circ} \\ \Rightarrow & 75^{\circ}+\angle B A D=180^{\circ} \\ \Rightarrow & \angle B A D=180^{\circ}-75^{\circ}=105^{\circ} \\ & \text { But } \angle E A F=\angle B A D=105^{\circ} \\ & \text { SO } \angle E A F=105^{\circ} \end{aligned}$

Question 7

Ans: A quadrilateral ABCD which is inscribed in a circle with center O. CD is produced to E $\angle A D E=$70^{\circ}$\text { and } \angle O B A=45^{\circ}$

Join OA, OB OCI AC

$A B C D$ is a cyclic. quadrilateral

So Ext. $\angle A D E=$ interior opposite $\angle A B C$

So $\angle A B C=70^{\circ}$

(IMAGE TO BE ADDED)

Arc AC subtends $\angle A O C$ at the center and $\angle A B C$ on the

remaining part of the circle So $\angle A O C=2 \angle A B C$

$=2 \times 70^{\circ}=140^{\circ}$

Now in $\triangle O A C, O A=O C$

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

But $\angle O C A+\angle O A C+\angle A O C=180^{\circ}$

$\begin{aligned}&\Rightarrow \angle O C A+\angle O C A+140^{\circ}=180^{\circ} \\&\Rightarrow 2 \angle O C A=180^{\circ}-140^{\circ}=40^{\circ}\end{aligned}$

So $\angle O C A=\frac{40^{\circ}}{2}=20^{\circ}$

and $\angle O A C=20^{\circ}$

Now $\angle B A C=\angle B A O+\angle O A C$

$=45^{\circ}+20^{\circ}=65^{\circ}$

Hence (i) $\angle B A C=65^{\circ}$ and (ii $\angle O C A=20^{\circ}$

Question 8

Ans:  In the figure ABCD is a cyclic quadrilateral in Which O is the center of the circle

$\angle A O C=120^{\circ}$

But $\angle A O C+ Ref  \cdot \angle A O C=360^{\circ}$

(IMAGE TO BE ADDED)

So Ref $\angle A O C=360^{\circ}-\angle A O C$

$=360^{\circ}-120^{\circ}=240^{\circ}$

Now arc ADC subtends $\angle A O C$ at the center and  \angle A B C at the remaining part of the circle 

So 

$\begin{aligned} & \angle A O C=2 \angle A B C \\ \Rightarrow & 120^{\circ}=2 x \end{aligned}$

$\begin{aligned} \Rightarrow & x=\frac{120^{\circ}}{2}=60^{\circ} \\ & \text { But } \angle B+\angle D=180^{\circ} \\ \Rightarrow & x+y=180^{\circ} \\ \Rightarrow & 60^{\circ}+y=180^{\circ} \\ \Rightarrow & y=180^{\circ}-60^{\circ} \\ \Rightarrow & y=120^{\circ} \end{aligned}$

Hence $x=60^{\circ}$ and $y=120^{\circ}$

Question 9

Ans: In the figure, $A B \| C D$

0 is the center of the circle and $\angle A D C=25^{\circ}$

 Join $O A$ and $O B$

(IMAGE TO BE ADDED)

if $A B \| C D$

So $\angle B A D=\angle A D C=25^{\circ}$

In $\triangle O R D \mid O A=O D$

So $\angle O A D=\angle A D O$ or $\angle A D C=25^{\circ}$

So $\angle O A B=\angle O A D+\angle B A D=25^{\circ}+25^{\circ}=50^{\circ}$

But in $\triangle O A B$

$O A=O B$

So $\angle O A B=\angle O B A=50^{\circ}$

But $\angle O A B+\angle O B A+\angle A O B=180^{\circ}$

$\begin{aligned}&\Rightarrow 50^{\circ}+50^{\circ}+\angle A O B=180^{\circ} \\&\Rightarrow \angle A O B+100^{\circ}=180^{\circ} \\&\Rightarrow \angle A O B=180^{\circ}-100^{\circ}=80^{\circ}\end{aligned}$

Now arc AB subtends  $\angle A O B$ at the centre and $\angle A E D$ at the remaining part of the circle 

So $\begin{aligned} & \angle A O B=2 \angle A E B \\ \Rightarrow & \angle A E B=\frac{1}{2}\left(A O B=\frac{1}{2} \times 80^{\circ}=40^{\circ} \right.\\ & \text { Hence } \angle A E B=40^{\circ} \end{aligned}$

Question 10

Ans: In the figure $A B C D$ is a cycric quadrilatoral In which $A D ||D C$

$\angle B C D=100^{\circ} \text { and } \angle B A C=40^{\circ}$

(IMAGE TO BE ADDED)

if $A B \| D C$ and $A C$ is its transversal

So $\angle B A C=\angle A C D=40^{\circ}$

But $\angle B C D=100^{\circ}$

So

$\angle B C A=\angle B C D-\angle A C D=100^{\circ}-40^{\circ}=60^{\circ}$

if $A B C D$ is a cyclic quadrilateral

So $\angle B A D+\angle B C D=180^{\circ}$

$\begin{aligned}&\Rightarrow \angle B A D+100^{\circ}=180^{\circ} \\

&\Rightarrow \angle B A D=180^{\circ}-100^{\circ}=80^{\circ} \\&\Rightarrow \angle B A C+\angle C A D=80^{\circ} \\&\Rightarrow 40^{\circ}+\angle C A D=80^{\circ} \\&\Rightarrow \angle C A D=80^{\circ}-40^{\circ}=40^{\circ}\end{aligned}$

But $\angle C B D=\angle C A D$

$=40^{\circ}$

Hence ii $\angle C A D=40^{\circ}$ (ii) $\angle C B D=40^{\circ}$

(iii) $\angle B C A=60^{\circ}$

Question 11

Ans: ABCD is a parallelogram and a circle pass through A and D, Intersect AB at E and DC at F

EF is joined 

$\angle B E F=80^{\circ}$

If ADFE is a cyclic quadrilateral 

So Ext. $\angle BE F=$ int $\cdot$ OPP. $\angle A B F$

So $\angle A D F=80^{\circ}$ or $\angle A D C=80^{\circ}$

But In ||gm ABCD 

$\angle A D C=\angle A B C$

So $\angle A B C=80^{\circ}$

Question 12

Ans: In the figure , AD is the diameter of the circle and ABCD is a cyclic quadrilateral $\angle B C D=125^{\circ}$

Join BD 

(IMAGE TO BE ADDED)

 

(i) If ABCD is a cyclic quadrilateral 

So  $\angle B A D+\angle B C D=180^{\circ}$

$\angle B A D+125^{\circ}=180^{\circ}$

$\angle 13 A D=180^{\circ}-125^{\circ}=55^{\circ}$

(ii) Now in $\triangle A B D$.

$\angle A B D=90^{\circ}$

So $\angle B A D+\angle A D B=90^{\circ}$

$\Rightarrow 55^{\circ}+\angle A O B=90^{\circ}$

$\Rightarrow \angle A D B=90^{\circ}-55^{\circ}=35^{\circ}$

So $\angle A D B=35^{\circ}$

Hence  $\angle B A D$ or $\angle D A B=55^{\circ}$ and  $\angle A D B=35^{\circ}$

Question 13

Ans: In the figure, 

ABCD is a cyclic trapezium in which AD||BC and $\angle B=70^{\circ}$

(i) If AD||BC 

So $\angle A B C+\angle B A D=180^{\circ}$

$70^{\circ}+\angle B A D=180^{\circ}$

$\Rightarrow \angle B A D=180^{\circ}-70^{\circ}$

SO $\angle B A D=110^{\circ}$

(ii) In Cyclic Trapezium 

$\angle B A D+\angle B C D=180^{\circ}$

$110^{\circ}+\angle B C D=180^{\circ}$

$\Rightarrow \angle B C D=180^{\circ}-110^{\circ}$

$\Rightarrow \angle B C D=180^{\circ}-110^{\circ}=70^{\circ}$

Hence $\angle B A D=110^{\circ}, \angle B C D=70^{\circ}$

Question 14

Ans: In the Figure $A B$ is the diameter as the circle $A P Q$ and $R B Q$ are straight lines $\angle A=35^{\circ}$ and $\angle Q=25^{\circ}$

(IMAGE TO BE ADDED)

(i) $\angle P R B$ and $\angle P A B$ are in the same segment So $\angle P R B=\angle P A B=35^{\circ}$

(ii) In $\triangle P B Q$

Ext. $\angle P B R=\angle B P Q+\angle Q$

$=90^{\circ}+25^{\circ}=115^{\circ}$

(iii) In $\triangle B R P$

$\angle B R P+\angle P B R+\angle BP R=180^{\circ}$

$\Rightarrow 35^{\circ}+115^{\circ}+\angle B P R=180^{\circ}$

$\Rightarrow 150^{\circ}+\angle B P R=180^{\circ}$

$\Rightarrow \angle B P R=180^{\circ}-150^{\circ}=30^{\circ}$

Question 15

Ans: In the figure O is the center of the circle and $\angle A O C=130^{\circ}$

So Reflex $\angle A O C=360^{\circ}-130^{\circ}=230^{\circ}$

Now major  arc AC subtends reflex $\angle A O C$ at the center and $\angle A B C$ at the remaining pant of the circle

So $\operatorname{Reflex} \angle A O C=2 \angle A B C$

$\Rightarrow 230^{\circ}=2 \angle A B C \Rightarrow \angle A B C=\frac{230^{\circ}}{2}=115^{\circ}$

Question 16

Ans: In the figures AOB is the diameter of the circle  $\angle B C D=140^{\circ}$

If ABCD is a cyclic quadrilateral 

So $\angle B C D+\angle B A D=180^{\circ}$

$140^{\circ}+\angle B A D=180^{\circ}$

$\Rightarrow \angle B A D=180^{\circ}-140^{\circ}$

$\angle B A B=40^{\circ}$

Now in $\triangle A O B$

$\angle A D B=90^{\circ}$

So $\angle BA D+\angle D B A=90^{\circ}$

$\Rightarrow 40^{\circ}+\angle D B A=90^{\circ}$

$\Rightarrow \angle D B A=90^{\circ}-40^{\circ}=50^{\circ}$

Question 17

Ans: In the figure ABCD is a cyclic Quadrilateral in Which AB is produced to D 

$\angle C B D=65^{\circ}$

If ABCD is a cyclic quadrilateral 

So Ext. $\angle C B D=$ interior opposite $\angle A E C$

$\Rightarrow \angle A E C=\angle C B D=65^{\circ}$

IF Arc $A B C$ subtends $\angle A O C$ at the center and LAEC at the remaining part of the circle So $\angle A O C=2 \angle A E C=2 \times 65^{\circ}=130^{\circ}$

But $\angle A O C+$ reflex $\angle A O C=360^{\circ}$

$\Rightarrow 130^{\circ}+x^{\circ}=360^{\circ} \Rightarrow x^{\circ}=360^{\circ}-130^{\circ}=230^{\circ}$

Question 18

Ans: In the figure, 

Two diagonals of a cyclic quadrilateral ABCD intersect Each other at P inside the circle.

AB= 8cm, CD = 5cm and Area $(\triangle A P B)=24 \mathrm{CB}^{2}$

If diagonals AC and BD intersect each other at P 

So, AP. PC = BP. PD 

$\begin{aligned}\Rightarrow \frac{A P}{B P}=\frac{D P}{C P} \Rightarrow \frac{A P}{D P}=\frac{B P}{C P} & \\\text { and } \angle A P B=& \angle C P D\end{aligned}$

So $\triangle A P B \backsim \triangle C P D$

So $\frac{\text { area } \triangle A P B}{\text { area } \triangle C P B}=\frac{A B^{2}}{C D^{2}}$

$\Rightarrow \frac{24 \mathrm{~cm}^{2}}{\text { area } \mathrm{△CPD}}=\frac{(8)^{2}}{(5)^{2}}$

$\Rightarrow \frac{24}{\text { area } \triangle C P D}=\frac{64}{25}$

Area △CPD = $\frac{24 \times 25}{64}=\frac{3 \times 25}{8}$

$=\frac{75}{8}=9 \frac{3}{8} \mathrm{~cm}^{2}$

Question 19

Ans: In the figure, 

(i) O is the center of the circle and $\angle B A D=30^{\circ}$

If ABCD is a cyclic quadrilateral 

So $\angle B A D+\angle B C D=180^{\circ}$

$\Rightarrow 30^{\circ}+p=180^{\circ}$

$\Rightarrow p=180^{\circ}-30^{\circ}=150^{\circ}$

Arc BCD subtends $\angle B O D$ at the center and $\angle B A D$ at the remaining part of the circle 

So  $\angle D O D=2 \angle B A D$

$q=2 \times 30^{\circ}=60^{\circ}$

if $\angle B A D$ and $\angle B E D$ are in the same segment

So

$\begin{aligned}& \angle B E D=\angle B A D \\\Rightarrow & r=300\end{aligned}$

Hence $p=150^{\circ}, q=60^{\circ}$  and $r=30^{\circ}$

Question 20

Ans: In the figure ABCD is a cyclic quadrilateral in which its diagonals AC and BD intersect each other at P 

$\angle B A D=65^{\circ}, \angle A B D=70^{\circ}$ $\angle B D C=45^{\circ}$

if $A B C D$ is a cyclic quadrilateral

So $\angle B A D+\angle B C D=180^{\circ}$

$\begin{aligned}\Rightarrow & 65^{\circ}+\angle B C D=180^{\circ} \\\Rightarrow & \angle BC D=180^{\circ}-65^{\circ}=115^{\circ}\end{aligned}$

(ii) In $\triangle A B D$,

$\angle B A D+\angle A B D+\angle A D B=180^{\circ}$

$\Rightarrow 65^{\circ}+70^{\circ}+\angle A D B=180^{\circ}$

$\Rightarrow 135^{\circ}+\angle A D B=180^{\circ}$

$\Rightarrow \angle A D B=1.80^{\circ}-135^{\circ}=45^{\circ}$

$\text { if } \angle A D C=\angle A D B+\angle B D C=45^{\circ}+45^{\circ}=90^{\circ}$

So $A D C$ is a semicircle

Hence is the diameter of the circle .

Question 21

Ans:  In the figure $A O B$ is the diameter as the circle with center O

$\angle E C D=\angle E D C=32^{\circ}$

(IMAGE TO BE ADDED)

(i) In $\triangle C DE$

Ext $\angle C E F=\angle E C D+\angle E D C=32^{\circ}+32^{\circ}=64^{\circ}$

(ii) Arc CF subtends $\angle C O F$ at the center and $\angle C D F$ or $\angle C F F$ at the remaining part of the circle

So $\angle C O F=2 \angle C D F=2 \times 32^{\circ}=64^{\circ}$

Question 22

Ans: In the figure ABCD is a cyclic quadrilateral 

AC and BD are joined 

$\angle D A C=27^{\circ}, \angle D B A=50^{\circ}, \angle A D B=33^{\circ}$ $\angle A D B=\angle A C B$

So $\angle A C B=33^{\circ}$

similarly $\angle A B D=\angle A C D$

So $\angle A C B=50^{\circ}$

So $\angle D C B=\angle A C B+\angle A C D$

$=33^{\circ}+50^{\circ}=83^{\circ}$

In cyclic quad. ABCD 

$\begin{aligned} \angle D C B+\angle B A D &=180^{\circ} \\ \Rightarrow 83^{\circ}+\angle B A D &=180^{\circ} \end{aligned}$

$\begin{aligned} \Rightarrow & \angle B A D=180^{\circ}-83^{\circ}=97^{\circ} \\ \Rightarrow & \angle D A C+\angle O A C=97^{\circ} \\ \Rightarrow & \angle B A C+27^{\circ}=97^{\circ} \\ \Rightarrow & \angle D A C=97^{\circ}-27^{\circ}=70^{\circ} \\ & or \angle C A B=70^{\circ} \\ & \angle D B C=\angle D A C \end{aligned}$

=$27^{\circ}$

Hence (i)  $\angle D B C=27^{\circ}$ (ii) $\angle D C B=83^{\circ}$  $\angle C A B=70^{\circ}$

Question 23

Ans: In the figure $A B C D E$ is a pentagon inscribed in the circle with centre 0

$A B=B C=C D$ and $\angle A B C=132^{\circ}$

Join $B E$ and $C E$

(IMAGE TO BE ADDED)

(i) In cyclic quad. $A E C B$

$\angle A E C+\angle A B C=180^{\circ}$

(IMAGE TO BE ADDED)

$\Rightarrow \angle A E C+132^{\circ}=180^{\circ}$

$\Rightarrow A E C=180^{\circ}-132^{\circ}=48^{\circ}$

if $A B=B C$

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

$=\frac{48^{\circ}}{2}=24$

(ii) if $A B=B C=C D$

So $\quad \angle A E B=\angle B E C=\angle C E D=24^{\circ}$

So $\angle A E D=\angle A E B+\angle A E C+\angle C E D$

$=24^{\circ}+24^{\circ}+24^{\circ}=72^{\circ}$

(iii) Arc CD subtends $\angle C O D$ at the center and 

$\angle C E D$ at the remaining part of the circle

 So $\angle C O D=2 \angle C E D=2 \times \angle 4^{\circ}=48^{\circ}$