ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

 Exercise 15.1

Question 1

Using the given information, find the value of x in each of the following figures :

Sol :

(i) ∠ADB and ∠ACB are in the same segment.

∠ADB = ∠ACB = 50°

Now in ∆ADB,

∠DAB + X + ∠ADB = 180°

(Angles of a triangles)

$\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$

$\Rightarrow x+92^{\circ}=180^{\circ}$

$\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$

(ii) ∠ABD=∠ACD

(Angles in the same segment)

But ∠ACD=32°

∴∠ABD=32°

Now in ΔABD

∠ABD+∠ADB+∠DAB=180°

⇒32°+45°+x=180°

⇒77°+x=180°

⇒x=180°-77°=103°

(iii) ∠BAD=∠BCD

(Angles in the same segement)

But ∠BAD=20°

$\therefore \angle B C D=20^{\circ}$

$\because \angle \mathrm{CEA}=90^{\circ}$

$\therefore \angle \mathrm{CED}=90^{\circ}$

Now in $\Delta$ CED,

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

$\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$

$\Rightarrow 110^{\circ}+x=180^{\circ}$

$\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$

(iv) In ΔABC

∠ABC+∠ACB+∠BAC=180°

(∵Sum of angles of a triangle)

$\Rightarrow 69^{\circ}+31^{\circ}+\angle \mathrm{BAC}=180^{\circ}$

$\Rightarrow \angle \mathrm{BAC}=180^{\circ}-100^{\circ}$

$\Rightarrow \angle \mathrm{BAC}=80^{\circ}$

Since $\angle \mathrm{BAC}$ and $\angle \mathrm{BAD}$ are in the same segment.

∴∠BAD=x=80°

(v) Given $\angle \mathrm{CPB}=120^{\circ}, \angle \mathrm{ACP}=70^{\circ}$

To find, $x^{\circ}$ i.e., $\angle \mathrm{PDB}$

$\because$ Reflex $\angle \mathrm{CPB}=\angle \mathrm{BPO}+\angle \mathrm{CPA}$

$\Rightarrow 120^{\circ}=\angle \mathrm{BPD}+\angle \mathrm{BPD}$

$(\because \mathrm{BPD}=\mathrm{CPA}$ are vertically opposite $\angle s)$

$\Rightarrow 2 \angle \mathrm{BPD}=120^{\circ}$

$\Rightarrow \angle \mathrm{BPD}=\frac{120^{\circ}}{2}=60^{\circ}$

Also $\angle \mathrm{ACP}$ and $\angle \mathrm{PBD}$ are in the same segment

$\therefore \angle \mathrm{PBD}=\angle \mathrm{ACP}=70^{\circ}$

Now, in $\Delta \mathrm{PBD}$

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

(Sum of all $\angle s$ in a triangle)

$\Rightarrow 70^{\circ}+x^{\circ}+60^{\circ}=180^{\circ}$

$\Rightarrow x=180^{\circ}-130^{\circ}$

$\Rightarrow x=50^{\circ}$

(vi) $\angle \mathrm{DAB}=\angle \mathrm{BCD}$

(∵ Angles in the same segment of the circle)

$\therefore \angle \mathrm{DAB}=25^{\circ}$ $\left(\because \angle \mathrm{BCD}=25^{\circ}\right.$ given $)$

In $\Delta \mathrm{DAP}$

$\mathrm{Ex} \cdot \angle \mathrm{CDA}=\angle \mathrm{DAP}+\angle \mathrm{DPA}$

$\Rightarrow x^{0}=\angle \mathrm{DAB}+\angle \mathrm{DPA}$

$\Rightarrow x^{\circ}=25^{\circ}+35^{\circ}$

$\Rightarrow x^{\circ}=60^{\circ}$

Question 2

If O is the centre of the circle, find the value of x in each of the following figures (using the given information):

Sol :

(i) ∠ACB = ∠ADB

(Angles in the same segment of a circle)

But ∠ADB = x°

$\therefore \angle \mathrm{ACB}=x^{\circ}$

Now in $\Delta \mathrm{ABC}$

$\angle \mathrm{CAB}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180^{\circ}$

$\Rightarrow 40^{\circ}+90^{\circ}+x^{\circ}=180^{\circ}$

(AC is the diameter)

$\Rightarrow 130^{\circ}+x^{\circ}=180^{\circ}$

$\Rightarrow x^{\circ}=180^{\circ}-130^{\circ}=50^{\circ}$

(ii) $\angle \mathrm{ACD}=\angle \mathrm{ABD}$

(Angles in the same segment)

$\therefore \angle \mathrm{ACD}=x^{\circ} \quad\left(\because \angle \mathrm{ABD}=x^{0}\right)$

Now in $\Delta \mathrm{OAC}$

$\mathrm{OA}=\mathrm{OC}$ [Radii of the same circle]

∴∠ACO=∠OAC

[opposite angles of equal sides]

∴x°=62°

(iii) ∠AOB+∠AOC+∠BOC=360° (Angles at a point)

$\Rightarrow \angle \mathrm{AOB}+80^{\circ}+130^{\circ}=360^{\circ}$

$\Rightarrow \angle \mathrm{AOB}+210^{\circ}=360^{\circ}$

$\Rightarrow \angle \mathrm{AOB}=360^{\circ}-210^{\circ}=150^{\circ}$

Now arc AB subtends $\angle \mathrm{AOB}$ at the centre $\angle \mathrm{ACB}$ at the remaining part of the circle.

$\therefore \angle \mathrm{AOB}=2 \angle \mathrm{ACB}$

$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 150^{\circ}=75^{\circ}$

(iv) $\angle \mathrm{ABC}+\angle \mathrm{CBD}=180^{\circ}$ (Linear pair)

$\Rightarrow \angle \mathrm{ABC}+75^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{ABC}=180^{\circ}-75^{\circ}=105^{\circ}$

Now arc AC subtends reflex $\angle \mathrm{AOC}$ at the centre and $\angle \mathrm{ABC}$ at the remaining part of the circle.

$\therefore$ Reflex $\angle A O C=2 \angle A B C$

$=2 \times 105^{\circ}=210^{\circ}$

(v) $\angle \mathrm{AOC}+\angle \mathrm{COB}=180^{\circ}$ (Linear pair)

$\Rightarrow  135^{\circ}+\angle \mathrm{COB}=180^{\circ}$

$\Rightarrow  \angle \mathrm{COB}=180^{\circ}-135^{\circ}=45^{\circ}$

Now arc BC subtends $\angle \mathrm{COB}$ at the centre and $\angle \mathrm{CDB}$ at the remaining part of the circle

$\therefore \angle \mathrm{COB}=2 \angle \mathrm{CDB}$

$\Rightarrow \angle \mathrm{CDB}=\frac{1}{2} \angle \mathrm{COB}$

$=\frac{1}{2} \times 45^{\circ}=\frac{45^{\circ}}{2}=22 \frac{1}{2}$

(vi) Arc AD subtends $\angle \mathrm{AOD}$ at the centre and $\angle \mathrm{ACD}$ at the remaining part of the circle

$\therefore \angle \mathrm{AOD}=2 \angle \mathrm{ACB}$

$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOD}=\frac{1}{2} \times 70^{\circ}=35^{\circ}$

$\because \angle \mathrm{CMO}=90^{\circ}$

$\therefore \angle \mathrm{AMC}=90^{\circ}$ $\left(\because \angle A M C+\angle C M O=180^{\circ}\right)$

Now in $\Delta \mathrm{ACM}$

$\angle \mathrm{ACM}+\angle \mathrm{AMC}+\angle \mathrm{CAM}=180^{\circ}$

$\Rightarrow 35^{\circ}+90^{\circ}+x^{\circ}=180^{\circ}$

$\Rightarrow 125^{\circ}+x^{\circ}=180^{\circ}$

$x^{\circ}=180^{\circ}-125^{\circ}=55^{\circ}$

Question 3

(a) In the figure (i) given below, AD || BC. If ∠ACB = 35°. Find the measurement of ∠DBC.

(b) In the figure (ii) given below, it is given that O is the centre of the circle and ∠AOC = 130°. Find∠ABC

Sol :

(a) Construction: Join AB

∠A = ∠C = 35° [∵ Alt angles]

Also $\angle B=\angle A=35^{\circ}$ ∵[Angles in same segment]

$\angle A B C=35^{\circ}$

(b) $\angle \mathrm{AOC}+$ reflex $\angle \mathrm{AOC}=360^{\circ}$

$\Rightarrow 130^{\circ}+$ reflex $\angle \mathrm{AOC}=360^{\circ}$

$\Rightarrow$ Reflex $\angle \mathrm{AOC}=360^{\circ}-130^{\circ}=230^{\circ}$

Now are AC subtends reflex $\angle \mathrm{AOC}$ at the centre and $\angle \mathrm{ABC}$ at the remaining part of the circle.

$\therefore$ Reflex $\angle \mathrm{AOC}=2 \angle \mathrm{ABC}$

$\Rightarrow \angle \mathrm{ABC}=\frac{1}{2}$ reflex $\angle \mathrm{AOC}$

$=\frac{1}{2} \times 230^{\circ}=115^{\circ}$

Question 4

(a) In the figure (i) given below, calculate the values of x and y.

(b) In the figure (ii) given below, O is the centre of the circle. Calculate the values of x and y.

Sol :

(a) ABCD is a cyclic quadrilateral

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

$\Rightarrow y+40^{\circ}+45^{\circ}=180^{\circ}$

$\Rightarrow y+85^{\circ}=180^{\circ}$

$\Rightarrow y=180^{\circ}-85^{\circ}=95^{\circ}$

$\angle \mathrm{ACB}=\angle \mathrm{ADB}$ (Angles in same segment)

$\Rightarrow x^{\circ}=40^{\circ}$

(b) Arc ADC subtends $\angle \mathrm{AOC}$ at the centre and $\angle \mathrm{ABC}$ at the remaining part of the circle

$\therefore \angle \mathrm{AOC}=2 \angle \mathrm{ABC}$

$\Rightarrow \angle \mathrm{ABC}=\frac{1}{2} \angle \mathrm{AOC}=\frac{1}{2} \times 120^{\circ}=60^{\circ}$

$\Rightarrow  x^{0}=60^{\circ}$

Again ABCD is a cyclic quadrilateral

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

$\Rightarrow 60^{\circ}+y^{\circ}=180^{\circ}$

$\Rightarrow y=180^{\circ}-60^{\circ}=120^{\circ}$

Question 5

(a) In the figure (i) given below, M, A, B, N are points on a circle having centre O. AN and MB cut at Y. If ∠NYB = 50° and ∠YNB = 20°, find ∠MAN and the reflex angle MON.

(b) In the figrue (ii) given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find

(i) ∠ACB

(ii) ∠OBC

(iii) ∠OAB

(iv) ∠CBA

Sol :

(a) ∠NYB = 50°, ∠YNB = 20°.

In $\Delta \mathrm{YNB}$

$\angle \mathrm{NYB}+\angle \mathrm{YNB}+\angle \mathrm{YBN}=180^{\circ}$

$\Rightarrow 50^{\circ}+20^{\circ}+\angle \mathrm{YBN}=180^{\circ}$

$\Rightarrow \angle \mathrm{YBN}+70^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{YBN}=180^{\circ}-70^{\circ}=110^{\circ}$

But $\angle \mathrm{MAN}=\angle \mathrm{YBN}$

(Angles in the same segment)

$\therefore \angle \mathrm{MAN}=110^{\circ}$

$\because$ Maior arc MN subtends reflex $\angle$ MON at the centre and $\angle \mathrm{MAN}$ at the remaining part of the circle

$\therefore$ Reflx $\angle \mathrm{MON}=2 \angle \mathrm{MAN}=2 \times 110^{\circ}=220^{\circ}$

(b) (i) $\angle \mathrm{AOB}+$ reflex $\angle \mathrm{AOB}=360^{\circ}$

(Angles at a point)

$\Rightarrow 140^{\circ}+$ reflex $\angle \mathrm{AOB}=360^{\circ}$

$\Rightarrow$ Reflex $\angle \mathrm{AOB}=360^{\circ}-140^{\circ}=220^{\circ}$

Now major arc $\mathrm{AB}$ subtends $\angle \mathrm{AOB}$ at the centre and $\angle \mathrm{ACB}$ is at the remaining part of the circle

$\therefore$ Reflex $\angle \mathrm{AOB}=2 \angle \mathrm{ACB}$

$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2}$ reflex $\angle \mathrm{AOB}$

$=\frac{1}{2} \times 220^{\circ}=110^{\circ}$

(ii) In quadrilateral , OACB,

$\angle \mathrm{OAC}+\angle \mathrm{ACB}+\angle \mathrm{AOB}+\angle \mathrm{OBC}=360^{\circ}$

$\Rightarrow 50^{\circ}+110^{\circ}+140^{\circ}+\angle \mathrm{OBC}=360^{\circ}$

$\Rightarrow 300^{\circ}+\angle \mathrm{OBC}=360^{\circ}$

$\Rightarrow \angle \mathrm{OBC}=360^{\circ}-300^{\circ}$

$\Rightarrow \angle \mathrm{OBC}=60^{\circ}$

(iii) In $\Delta$ OAB

OA=OB (∵ radii of the same circle)

But $\angle \mathrm{OAB}+\angle \mathrm{OBA}+\angle \mathrm{AOB}=180^{\circ}$

$\Rightarrow \angle \mathrm{OAB}+\angle \mathrm{OAB}+140^{\circ}=180^{\circ}$

$\Rightarrow 2 \angle \mathrm{OAB}=180^{\circ}-140^{\circ}=40^{\circ}$

$\therefore \angle \mathrm{OAB}=\frac{40^{\circ}}{2}=20^{\circ}$

(iv) But $\angle \mathrm{OBC}=60^{\circ}$

$\therefore \angle \mathrm{CBA}=\angle \mathrm{OBC}-\angle \mathrm{OBA}$

$=60^{\circ}-20^{\circ}=40^{\circ}$

Question 6

(a) In the figure (i) given below, O is the centre of the circle and ∠PBA = 42°. Calculate the value of ∠PQB

(b) In the figure (ii) given below, AB is a diameter of the circle whose centre is O. Given that ∠ECD = ∠EDC = 32°, calculate

(i) ∠CEF

(ii) ∠COF.

Sol :

(a) In ∆APB,

∠APB = 90° (Angle in a semi-circle)

But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle)

∠A + 90° + 42°= 180°

∠A + 132° = 180°

⇒ ∠A = 180° – 132° = 48°

But ∠A = ∠PQB

(Angles in the same segment of a circle)

$\therefore \angle \mathrm{PQB}=48^{\circ}$

(b) (i) In $\Delta \mathrm{EDC}$

Ext. $\angle \mathrm{CEF}=\angle \mathrm{ECD}+\angle \mathrm{EDC}$

$=32^{\circ}+32^{\circ}=64^{\circ}$

(Ext. angle of a triangle is equal to the sum of its interior opposite angles)

(ii) $\operatorname{arc} \mathrm{CF}$ subtends $\angle \mathrm{COF}$ at the centre and $\angle \mathrm{CDF}$ at the remaining part of the circle

$\therefore \angle \mathrm{COF}=2 \angle \mathrm{CDF}=2 \angle \mathrm{CDE}$

$=2 \times 32^{\circ}=64^{\circ}$

Question 7

(a) In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, ∠A = 35°, ∠Q = 25°. Find (i) ∠PRB (ii) ∠PBR (iii) ∠BPR.

(b) In the figure (ii) given below, it is given that ∠ABC = 40° and AD is a diameter of the circle. Calculate ∠DAC.

Sol :

(a) (i) ∠PRB = ∠BAP

(Angles in the same segment of the circle)

∴ ∠PRB = 35° (∵ ∠BAP = 35° given)

(ii) In ∆PRQ,

Ext. $\angle \mathrm{APR}=\angle \mathrm{PRQ}+\angle \mathrm{PQR}$

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

$=35^{\circ}+25^{\circ}=60^{\circ}$

But $\angle \mathrm{APB}=90^{\circ}$ (Angle in a semi circle)

$\therefore \angle B P R=\angle A P B-\angle A P R$

$=90^{\circ}-60^{\circ}=30^{\circ}$

(iii) $\angle \mathrm{APR}=\angle \mathrm{ABR}$

(Angles in the same segment of the circle)

$\Rightarrow 60^{\circ}=\angle \mathrm{ABR}$

In $\Delta P B Q$

Ext. $\angle \mathrm{PBR}=\angle \mathrm{Q}+\angle \mathrm{BPQ}$

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

(b) $\angle \mathrm{B}=\angle \mathrm{D}$

(Angles in the same segment)

$\therefore \angle D=40^{\circ}$

$\angle \mathrm{ACD}=90^{\circ}$ (Angle in the semi circle)

Now in $\triangle \mathrm{ADC}$

$\angle \mathrm{ACD}+\angle \mathrm{D}+\angle \mathrm{DAC}=180^{\circ}$

(Angle in a triangle)

$\Rightarrow 90^{\circ}+40^{\circ}+\angle \mathrm{DAC}=180^{\circ}$

$\Rightarrow  130^{\circ}+\angle \mathrm{DAC}=180^{\circ}$

$\Rightarrow \angle \mathrm{DAC}=180^{\circ}-130^{\circ}=50^{\circ}$

Question 8

(a) In the figure given below, P and Q are centres of two circles intersecting at B and C. ACD is a st. line. Calculate the numerical value of x.

(b) In the figure given below, O is the circumcentre of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate

(i)∠CAB

(ii)∠OAC

Sol :

Given that

(a) Arc AB subtends ∠APB at the centre

and ∠ACB at the remaining part of the circle

$\therefore \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{APB}=\frac{1}{2} \times 130^{\circ}=65^{\circ}$

But $\angle \mathrm{ACB}+\angle \mathrm{BCD}=180^{\circ}$

(Linear pair)

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

$\Rightarrow \angle B C D=180^{\circ}-65^{\circ}=115^{\circ}$

Major arc BD subtends reflex $\angle \mathrm{BQD}$ at the centre and $\angle \mathrm{BCD}$ at the remaining part of the circle

$\therefore$ Reflex $\angle \mathrm{BQD}=2 \angle \mathrm{BCD}$

$=2 \times 115^{\circ}=230^{\circ}$

But reflex angle $\mathrm{BOD}+x=360^{\circ}$

(Angles at a point)

$\therefore 230^{\circ}+x=360^{\circ}$

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

(b) Join OC

$\because \operatorname{In} \Delta \mathrm{ABC}, \mathrm{AC}=\mathrm{BC}$

$\therefore \quad \angle \mathrm{A}=\angle \mathrm{B}$

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

$\Rightarrow \angle A+\angle A+56^{\circ}=180^{\circ}$

$\Rightarrow 2 \angle A=180^{\circ}-56^{\circ}=1.24^{\circ}$

$\therefore \angle \mathrm{A}=\frac{124}{2}=62^{\circ}$ or $\angle \mathrm{CAB}=62^{\circ}$

$\because \mathrm{OC}$ is the radius of the circle

$\therefore \mathrm{OC}$ bisects $\angle \mathrm{ACB}$

$\therefore \angle \mathrm{OCA}=\frac{1}{2} \angle \mathrm{ACB}=\frac{1}{2} \times 56^{\circ}=28^{\circ}$

Now in $\Delta$ OAC

$\mathrm{OA}=\mathrm{OC}$ (radii of the same circle)

$\therefore \angle \mathrm{OAC}=\angle \mathrm{OCA}=28^{\circ}$

Question 9

(a) In the figure (i) given below, chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

(b) In the figure (ii) given below, C is a point on the minor arc AB of the circle with centre O. Given ∠ACB = p°, ∠AOB = q°, express q in terms of p. Calculate p if OACB is a parallelogram.

Sol :

(a) ∠CBE = ∠CAE

(Angle in the same segment of a circle)

⇒ ∠CAE = 65°

∠AEC = 90° (Angle in a semi circle)

Now in ∆AEC

∠AEC + ∠CAE + ∠ACE = 180° (Angle of a triangle)

⇒ 90°+ 65° +∠ACE = 180°

⇒ 155° + ∠ACE = 180°

⇒ ∠ACE = 180° – 155° – 25°

∵AC || ED (given)

∴∠ACE = ∠DEC (alternate angles)

∴∠DEC = 25°

(b) Major arc AB subtends reflex $\angle \mathrm{AOB}$ at the centre and $\angle \mathrm{ACB}$ at the remaining part of the circle.

$\therefore$ reflex $\angle \mathrm{AOB}=2 \angle \mathrm{ACB}=2 p \quad \ldots(i)$

But reflex $\angle \mathrm{AOB}+q=360^{\circ}$

$\Rightarrow \quad$ Reflex $\angle \mathrm{AOB}=360^{\circ}-q$ ...(ii)

From (i) and (ii)

$2 p=360^{\circ}-q$

$q=360^{\circ}-2 p=2\left(180^{\circ}-p\right)$

If OACB is a parallelogram, then

p=q 

$\Rightarrow p=360^{\circ}-2 p$

$\Rightarrow 3 p=360^{\circ}$

$\Rightarrow p=\frac{360^{\circ}}{3}=120^{\circ}$

$\therefore \quad q=p=120^{\circ}$

Question 10

(a) In the figure (i) given below, straight lines AB and CD pass through the centre O of a circle. If ∠OCE = 40° and ∠AOD = 75°, find the number of degrees in :

(i) ∠CDE

(ii) ∠OBE.

(b) In the figure (ii) given below, I is the incentre of ∆ABC. AI produced meets the circumcircle of ∆ABC at D. Given that ∠ABC = 55° and ∠ACB = 65°, calculate

(i) ∠BCD

(ii) ∠CBD

(iii) ∠DCI

(iv) ∠BIC.

Sol :

(a) (i) ∠CED = 90° (Angle in semi-circle)

In ∆CED

∠CED + ∠CDE + ∠DCE = 180°

⇒ 90° +∠CDE + 40° = 180°

⇒ 130° + ∠CDE = 180°

⇒ ∠CDE = 180° – 130° = 50°

(ii) In $\Delta \mathrm{OBD}$,

Ext. $\angle \mathrm{AOD}=\angle \mathrm{OBE}+\angle \mathrm{ODB}$

$\Rightarrow  \angle \mathrm{AOD}=\angle \mathrm{OBE}+\angle \mathrm{CDE}$

$\Rightarrow 75^{\circ}=\angle \mathrm{OBE}+50^{\circ}$

$\Rightarrow \angle \mathrm{OBE}=75^{\circ}-50^{\circ}=25^{\circ}$

(b) Join BI and CI

In $\Delta \mathrm{ABC}$

$\angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180^{\circ}$

(Angles of a triangle)

$\Rightarrow \angle B A C+55^{\circ}+65^{\circ}=180^{\circ}$

$\Rightarrow \angle B A C=180^{\circ}-120^{\circ}=60^{\circ}$

$\because I$ is incentre

$\therefore$ I lies on the bisectors of angle of the $\Delta \mathrm{ABC}$

$\therefore \angle B A D=\angle C A D=\frac{60^{\circ}}{2}=30^{\circ}$

But $\angle \mathrm{BCD}=\angle \mathrm{BAD}=30^{\circ}$

(Angles in the same segment)

Similarly $\angle \mathrm{CBD}=\angle \mathrm{CAD}=30^{\circ}$

and $\angle \mathrm{IBC}=\frac{55^{\circ}}{2}=27 \frac{1}{2}^{\circ}$

and $\angle \mathrm{ICB}=\frac{65^{\circ}}{2}=32 \frac{1}{2}^{\circ}$

$\therefore \quad \angle \mathrm{BIC}=180^{\circ}-\left(27 \frac{1}{2}^{\circ}+32 \frac{1}{2} \circ\right)$

$=180^{\circ}-60^{\circ}=120^{\circ}$

Now $\angle \mathrm{ICD}=\angle \mathrm{ICB}+\angle \mathrm{BCD}$

$=32 \frac{1}{2}^{\circ}+30^{\circ}=62 \frac{1}{2} 0$

Question 11

O is the circumcentre of the triangle ABC and D is mid-point of the base BC. Prove that ∠BOD = ∠A.

Sol :

In the given figure, O is the centre of circumcentre of ∆ABC.

D is mid-point of BC. BO, CO and OD are joined.

To prove : 

∠BOD=∠A

Proof : Arc BC subtends ∠BOC on the centre and ∠A on the remaining part of the circle.

$\therefore \angle \mathrm{BOC}=2 \angle \mathrm{A}$

In $\Delta \mathrm{OBD}$ and $\triangle \mathrm{BCO}$

$\mathrm{OD}=\mathrm{OD}$ (common)

$\mathrm{BD}=\mathrm{CD}$ (D is mid-point of BC)

$\mathrm{OB}=\mathrm{OC}$ (Radii of the same circle)

$\therefore \Delta \mathrm{OBD} \cong \Delta \mathrm{BCO}$ (SSS axiom)

$\therefore \angle \mathrm{BOD}=\angle \mathrm{COD}$ (c.p.c.t)

$\therefore \angle \mathrm{BOD}=\frac{1}{2} \angle \mathrm{BOC}$...(i)

But $2 \angle \mathrm{A}=\angle \mathrm{BOC}$

$\Rightarrow \angle \mathrm{A}=\frac{1}{2} \angle \mathrm{BOC}$...(ii)

From (i) and (ii)

∴∠BOD=∠A

Hence proved

Question 12

In the given figure, AB and CD are equal chords. AD and BC intersect at E. Prove that AE = CE and BE = DE.

Sol :

In the given figure, AB and CD are two equal chords

AD and BC intersect each other at E.

To prove : AE = CE and BE = DE

Proof:

In ∆AEB and ∆CED

AB = CD (given)

∠A = ∠C (angles in the same segment)

∠B = ∠D (angles in the same segment)

∴ ∆AEB ≅ ∆CED (ASA axiom)

∴ AE = CE and BE = DE (c.p.c.t.)

Question 13

(a) In the figure (i) given below, AB is a diameter of a circle with centre O. AC and BD are perpendiculars on a line PQ. BD meets the circle at E. Prove that AC = ED.

(b) In the figure (ii) given below, O is the centre of a circle. Chord CD is parallel to the diameter AB. If ∠ABC = 25°, calculate ∠CED.

Sol :

(a) Given: AB is the diameter of a circle with centre O.

AC and BD are perpendiculars on a line PQ,

such that BD meets the circle at E.

To prove : AC=ED

Construction : Join AE

Proof : ∠AEB=90°

(angle in a semi-circle)

But $E D \perp P Q$ (given)

$\therefore  A E \| P Q$

But BD and AC are perpendicular to PQ

$\therefore$ AEDC is a rectangle.

Hence AC=ED

(Opposite sides of a rectangle) Q.E.D.

(b) AB||CD and O is the centre of the circle where AB is diameter, ∠ABC=25°. Join OC and OD

Arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle

$\therefore   \angle \mathrm{AOC}=2 \angle \mathrm{ABC}=2 \times 25^{\circ}=50^{\circ}$

But $\angle \mathrm{OCD}=\angle \mathrm{AOC} \quad$ (alternate angles)

$=50^{\circ}$

But $\angle \mathrm{ODC}=\angle \mathrm{OCD}=50^{\circ}$

$(\because$ in $\Delta \mathrm{OCD}, \mathrm{OC}=\mathrm{OD})$

In $\Delta \mathrm{OCD}$

$\angle \mathrm{COD}+\angle \mathrm{OCD}+\angle \mathrm{ODC}=180^{\circ}$

(Angles of a triangle)

$\Rightarrow \angle C O D+50^{\circ}+50^{\circ}=180^{\circ}$

$\Rightarrow \angle C O D+100^{\circ}=180^{\circ}$

$\therefore \angle C O D=180^{\circ}-100^{\circ}=80^{\circ}$

Now arc CD subtends ∠COD at the centre and ∠CED at the remaining point of the circle

$\therefore \angle \mathrm{COD}=2 \angle \mathrm{CED}$

$\Rightarrow 80^{\circ}=2 \angle \mathrm{CED}$

$\Rightarrow \angle \mathrm{CED}=80^{\circ} \times \frac{1}{2}=40^{\circ}$

Question 14

In the adjoining figure, O is the centre of the given circle and OABC is a parallelogram. BC is produced to meet the circle at D.

Prove that ∠ABC = 2 ∠OAD.

Sol :

Given: In the figure,

OABC is a || gm and O is the centre of the circle.

BC is produced to meet the circle at D.

To Prove : ∠ABC = 2∠OAD.

Construction: Join AD.

Proof : Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle.

$\therefore \angle \mathrm{AOC}=2 \angle \mathrm{ADC}$

But $\angle \mathrm{OAD}=\angle \mathrm{ADC}$ (alternate angles)

$\because \angle \mathrm{AOC}=2 \angle \mathrm{OAD}$

But $\angle \mathrm{ABC}=\angle \mathrm{AOC}$

(opposite angles of a ||gm)

$\therefore \angle \mathrm{ABC}=2 \angle \mathrm{OAD}$

Hence proved

Question 15

(a) In the figure (i) given below, P is the point of intersection of the chords BC and AQ such that AB = AP. Prove that CP = CQ.

(b) In the figure (i) given below, AB = AC = CD, ∠ADC = 38°. Calculate :

(i) ∠ABC (ii) ∠BEC.

Sol :

(a) Given: Two chords AQ and BC intersect each other at P

inside the circle. AB and CQ are joined and AB = AP.

To Prove : CP = CQ

Construction : Join AC.

Proof: In ∆ABP and ∆CQP

∴ ∠B = ∠Q

(Angles in the same segment)

$\angle \mathrm{BAP}=\angle \mathrm{PCQ}$

(Angles in the same segment)

$\angle \mathrm{BPA}=\angle \mathrm{CPQ}$

(Vertically opposite angles)

$\therefore \quad \Delta \mathrm{ABP} \sim \Delta \mathrm{CQP}$

(AAA axioms of similarity)

$\therefore \quad \frac{\mathrm{AB}}{\mathrm{CQ}}=\frac{\mathrm{AP}}{\mathrm{CP}}$

But AB=AP (given)

∴CQ=CP

(b) $\mathrm{AB}=\mathrm{AC}=\mathrm{CD}, \angle \mathrm{ADC}=38^{\circ}$

In $\triangle \mathrm{ACD}, \mathrm{AC}=\mathrm{CD}$  (given)

$\therefore \angle \mathrm{CAD}=\angle \mathrm{ADC}=38^{\circ}$

and ext. $\angle \mathrm{ACB}=\angle \mathrm{CAD}+\angle \mathrm{ADC}$

$=38^{\circ}+38^{\circ}=76^{\circ}$

In $\Delta \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}$

$\therefore \angle \mathrm{ABC}=\angle \mathrm{ACB}=76^{\circ}$

and $\angle \mathrm{BAC}=180^{\circ}-(\angle \mathrm{ABC}+\angle \mathrm{ACB})$

$=180^{\circ}-\left(76^{\circ}+76^{\circ}\right)=180^{\circ}-152^{\circ}=28^{\circ}$

But $\angle \mathrm{BEC}=\angle \mathrm{BAC}$

(Angles in the same segment)

$\therefore \angle \mathrm{BEC}=28^{\circ}$

Question 16

(a) In the figure (i) given below, CP bisects ∠ACB. Prove that DP bisects ∠ADB.

(b) In the figure (ii) given below, BD bisects ∠ABC. Prove that $\frac{A B}{B D}=\frac{B E}{B O}$

Sol :

(a)Given: In the figure, CP is the bisector of

∠ACB meeting the circle at P.

PD is joined

Prove : DP bisects $\angle \mathrm{ADB}$

Proof $: \because \angle \mathrm{ACB}$ and $\angle \mathrm{ADB}$ are in the same segment of the circle

$\therefore \angle \mathrm{ACB}=\angle \mathrm{ADB}$

Similarly, $\angle \mathrm{ACP}$ and $\angle \mathrm{ADP}$ are in the same segment

$\therefore \angle \mathrm{ACP}=\angle \mathrm{ADP}$

But $\angle \mathrm{ACP}=\frac{1}{2} \angle \mathrm{ACB}$

$(\because \mathrm{CP}$ is the angle bisector)

and $\angle \mathrm{ACB}=\angle \mathrm{ADB}$ (proved)

$\therefore \angle \mathrm{ADP}=\frac{1}{2} \angle \mathrm{ADB}$

Hence, DP bisects $\angle \mathrm{ADB}$

Hence proved.

(b) Given : In the figure, $\Delta \mathrm{ABC}$ is inscribed in a circle. BD bisects $\angle \mathrm{ABC}$.

To Prove : $\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BE}}{\mathrm{BC}}$

Construction : Join CD.

Proof : $\operatorname{In} \Delta \mathrm{ABE}$ and $\Delta \mathrm{BCD}$

$\angle \mathrm{A}=\angle \mathrm{D}$

(Angles in the same segment) 

$\angle \mathrm{ABE}=\angle \mathrm{BDC}$

$(\because \mathrm{BD}$ is the bisector of $\angle \mathrm{ABC})$

$\therefore \Delta \dot{A B E} \sim \Delta B A C$ (AA axiom)

$\therefore \frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BE}}{\mathrm{BC}}$

Hence proved

Question 17

(a) In the figure (ii) given below, chords AB and CD of a circle intersect at E.

(i) Prove that triangles ADE and CBE are similar.

(ii) Given DC =12 cm, DE = 4 cm and AE = 16 cm, calculate the length of BE.

(b) In the figure (ii) given below, AB and CD are two intersecting chords of a circle. Name two triangles which are similar. Hence, calculate CP given that AP = 6cm, PB = 4 cm, and CD = 14 cm (PC > PD).

Sol :

(a) Given: Two chords AB and CD intersect each other

at E inside the circle.

To Prove :

(i) $\Delta \mathrm{ADE} \sim \Delta \mathrm{CBE}$

(ii) If $\mathrm{DC}=12 \mathrm{~cm}, \mathrm{DE}=4 \mathrm{~cm}$ and

$\mathrm{AE}=16 \mathrm{~cm},$ calculate the length of $\mathrm{BE}$

Proof: $(i)$ In $\triangle \mathrm{ADE}$ and $\Delta \mathrm{CBE}$ $\angle \mathrm{D}=\angle \mathrm{B}$

{Angles in the same segment}

$\angle \mathrm{A}=\angle \mathrm{C}\{$ Angles in the same segment $\}$

$\therefore \Delta \mathrm{ADE} \sim \Delta \mathrm{CBE}$

(AA axiom of similarity)

(ii) DC=12 cm, DE=4 cm

∴EC=12-4=8 cm

AE=16 cm

∵Chords AB and CD intersects each other at E

∴AE×EB=CE×ED

$\Rightarrow 16 \times \mathrm{EB}=8 \times 4$

$\Rightarrow  \mathrm{EB}=\frac{8 \times 4}{16}=2 \mathrm{~cm}$

(b) Now in ΔAPD and ΔCPB,

$\angle \mathrm{DAB}=\angle \mathrm{DCB}$

$\angle \mathrm{CDA}=\angle \mathrm{CBA}$

{Angles in the same segment of the circle}

$\therefore \Delta \mathrm{APD} \sim \Delta \mathrm{CPB}$

(AA axiom of similarity)

$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{CP}}{\mathrm{PB}}$

$\Rightarrow A P \times P B=C P \times P D$

$\Rightarrow 6 \times 4=x(14-x)$

$\Rightarrow 24=14 x-x^{2}$

$\Rightarrow x^{2}-14 x+24=0$

$\Rightarrow x^{2}-12 x-2 x+24=0$

$\Rightarrow x(x-12)-2(x-12)=0$

$\Rightarrow(x-12)(x-2)=0$

Either x-12=0

then x=12

or x-2=0

then x=2

∴CP=12 cm or 2 cm

But CP>PD (given)

∴CP=12 cm

Question 18

In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE. (2008)

Sol :

In the figure, AE and BC intersect each other at D.

AB is joined.

$\angle \mathrm{CDE}=90^{\circ}, \mathrm{AB}=5 \mathrm{~cm}$

$B D=4 \mathrm{~cm}$

and $\mathrm{CD}=9 \mathrm{~cm}$

In right $\Delta \mathrm{ADB}$, $A B^{2}=A D^{2}+B D^{2}$

(Pythagoras Theorem)

$\Rightarrow(5)^{2}=A D^{2}+(4)^{2}$

$\Rightarrow 25=A D^{2}+16$

$\Rightarrow A D^{2}=25-16=9=(3)^{2}$

$\therefore A D=3 \mathrm{~cm}$

$\because$ Chords AE and BC of the circle intersect each other at $\mathrm{D}$ inside the circle

$\therefore A D \times D E=B D \times D C$

$\Rightarrow 3 \times \mathrm{DE}=4 \times 9$

$\Rightarrow \mathrm{DE}=\frac{4 \times 9}{3}=\frac{36}{3}$

$\Rightarrow \mathrm{DE}=12 \mathrm{~cm}$

Question 19

(a) In the figure (i) given below, PR is a diameter of the circle, PQ = 7 cm, QR = 6 cm and RS = 2 cm. Calculate the perimeter of the cyclic quadrilateral PQRS.

(b) In the figure (ii) given below, the diagonals of a cyclic quadrilateral ABCD intersect in P and the area of the triangle APB is 24 cm². If AB = 8 cm and CD = 5 cm, calculate the area of ∆DPC.

Sol :

(a) PR is the diameter of the circle

PQ = 7 cm, QR = 6 cm, RS = 2 cm.

In $\Delta \mathrm{PQR}$

$\angle \mathrm{Q}=90^{\circ}$

(Angle in a semi-circle)

$\therefore \quad P R^{2}=P Q^{2}+Q R^{2}=(7)^{2}+(6)^{2}$

=49+36=85

Again in ΔPSR

∠S=90°

(Angle in a semi-circle)

$\therefore \quad P R^{2}=P S^{2}+R S^{2}$

$\Rightarrow \quad 85=P S^{2}+(2)^{2} \Rightarrow 85=P S^{2}+4$

$\Rightarrow P S^{2}=85-4=81=(9)^{2}$

$\therefore \quad P S=9 \mathrm{~cm}$

Now the perimeter of PQRS,

=7+6+2+9=24 cm

(b) In ΔAPB and ΔDPC

$\angle \mathrm{APB}=\angle \mathrm{DPC}$

(Vertically opposite angles)

$\angle \mathrm{ABP}=\angle \mathrm{DCP}$

(Angles in the same segment)

$\therefore \Delta \mathrm{APB} \sim \Delta \mathrm{DPC}$

(AA axiom of similarity)

$\therefore \frac{\text { area of } \Delta \mathrm{APB}}{\text { area of } \Delta \mathrm{DPC}}=\frac{\mathrm{AB}^{2}}{\mathrm{CD}^{2}}$

$\Rightarrow   \frac{24}{\text { area of } \Delta \mathrm{DPC}}=\frac{(8)^{2}}{(5)^{2}}$

$\Rightarrow \frac{24}{\text { area of } \Delta \mathrm{DPC}}=\frac{64}{25}$

$\Rightarrow$ area of $\Delta \mathrm{DPC}$

$=\frac{24 \times 25}{64}=\frac{75}{8}=9 \frac{3}{8} \mathrm{~cm}^{2}$

Question 20

(a) In the figure (i) given below, QPX is the bisector of ∠YXZ of the triangle XYZ. Prove that XY : XQ = XP : XZ,

(b) In the figure (ii) given below, chords BA and DC of a circle meet at P. Prove that:

(i) ∠PAD = ∠PCB

(ii) PA. PB = PC . PD.

 Sol :

(a) Given: ∆XYZ is inscribed in a circle.

Bisector of ∠YXZ meets the circle at Q.

QY is joined.

To Prove : XY : XQ = XP : XZ

Proof : In ΔXYQ and ΔXPZ

∠Q=∠Z

∠YXQ=∠PXZ

(∵XQ is the bisector of ∠YXZ)

∴∆XYZ～∆XPZ

∴$\frac{X Y}{X P}=\frac{X Q}{X Z} \Rightarrow \frac{X Y}{X Q}=\frac{X P}{X Z}$

$\Rightarrow X Y: X Q=X P: X Z$  [Q.E.D]

(b) Given : Two chords BA and DC meet each other at P outside the circle. AD and BC are joined.

To prove :

(i) ∠PAD=∠PCB

(ii) PA.PB=PC.PD

Proof : ∠PAD+∠DAB=∠PCB+∠BCD

(each 180°)

But ∠DAB=∠BCD

(angle in the same segment)

∴∠PAD=∠PCB

Now in ΔPBC and ΔPAD

∠PCB=∠PAD (proved)

∠P=∠P (common)

∴ΔPBC～ΔPAD

(AA axiom of similarity)

∴ $\frac{P C}{P A}=\frac{P B}{P D}$

⇒PA.PB=PC.PD (Q.E.D)