Rs Aggarwal 2021 2022 Solutions for Class 9 Maths Chapter 12 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among Class 9 students for Maths Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 9 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 436:

#### Answer:

Let *AB* be the chord of the given circle with centre *O* and a radius of 10 cm.

Then *AB* =16 cm and *OB* = 10 cm

From *O*, draw *OM* perpendicular to *AB*.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ *BM* = $\left(\frac{16}{2}\right)\mathrm{cm}=8\mathrm{cm}$

In the right Δ*OMB*, we have:

*OB ^{2}^{ }= OM^{2} + MB^{2} * (Pythagoras theorem)

⇒ 10

^{2}=

*OM*

^{2}+ 8

^{2}

⇒ 100 =

*OM*

^{2}+ 64

⇒

*OM*

^{2}= (100 - 64) = 36

⇒ $OM=\sqrt{36}\mathrm{cm}=6\mathrm{cm}$

Hence, the distance of the chord from the centre is 6 cm.

#### Page No 436:

#### Answer:

Let* AB* be the chord of the given circle with centre *O* and a radius of 5 cm.

From *O*, draw *OM* perpendicular to *AB*.

Then *OM *= 3 cm and *OB* = 5 cm

From the right Δ*OMB*, we have:

*OB ^{2}^{ }= OM^{2} + MB^{2} * (Pythagoras theorem)

⇒ 5

^{2}= 3

^{2}+

*MB*

^{2}⇒ 25 = 9 +

*MB*

^{2}⇒

*MB*= (25

^{2}*−*9) = 16

⇒ $MB=\sqrt{16}\mathrm{cm}=4\mathrm{cm}$

Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:

*AB = 2 × MB*= (2 × 4) cm = 8 cm

Hence, the required length of the chord is 8 cm.

#### Page No 436:

#### Answer:

Let *AB* be the chord of the given circle with centre *O*. The perpendicular distance from the centre of the circle to the chord is 8 cm.

Join *OB.*

Then *OM* = 8 cm and *AB =* 30 cm

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ $MB=\left(\frac{AB}{2}\right)=\left(\frac{30}{2}\right)\mathrm{cm}=15\mathrm{cm}$

From the right Δ*OMB*, we have:

*OB ^{2}^{ }= OM^{2} + MB^{2}*

⇒

*OB*= 8

^{2}^{2}+ 15

^{2}

⇒

*OB*

^{2}= 64 + 225

⇒

*OB*= 289

^{2}⇒ $OB=\sqrt{289}\mathrm{cm}=17\mathrm{cm}$

Hence, the required length of the radius is 17 cm.

#### Page No 436:

#### Answer:

We have:

(i)

Let *AB* and *CD* be two chords of a circle such that *AB* is parallel to *CD* on the same side of the circle.

Given:* AB* = 8 cm, *CD* = 6 cm and *OB = OD *= 5 cm

Join* **OL* and *OM*.

The perpendicular from the centre of a circle to a chord bisects the chord.

∴ $LB\mathit{=}\frac{\mathit{A}\mathit{B}}{\mathit{2}}=\left(\frac{8}{2}\right)=4\mathrm{cm}$

Now, in right angled Δ*BLO*, we have:

*OB ^{2} = LB^{2} + LO^{2}*

⇒

*LO*−

^{2}^{ }= OB^{2}*LB*

^{2}⇒

*LO*

^{2}

^{ }= 5

^{2}− 4

^{2}

⇒

*LO*= 25 − 16 = 9

^{2}^{ }∴

*LO*= 3 cm

Similarly, $MD\mathit{=}\frac{\mathit{C}\mathit{D}}{\mathit{2}}=\left(\frac{6}{2}\right)=3\mathrm{cm}$

In right angled Δ

*DMO*, we have:

*OD*

^{2}= MD^{2}+ MO^{2}⇒

*MO*−

^{2}= OD^{2}*MD*

^{2}⇒

*MO*= 5

^{2}^{2}− 3

^{2}

⇒

*MO*= 25 − 9 = 16

^{2}⇒

*MO*= 4 cm

∴ Distance between the chords = (

*MO*−

*LO)*= (4 − 3) cm = 1 cm

(ii)

Let

*AB*and

*CD*be two chords of a circle such that

*AB*is parallel to

*CD*and they are on the opposite sides of the centre.

Given:

*AB*= 8 cm and

*CD*= 6 cm

Draw

*OL*⊥

*AB*and

*OM*⊥

*CD*.

Join

*OA*and

*OC*.

*OA = OC =*5 cm (Radii of a circle)

The perpendicular from the centre of a circle to a chord bisects the chord.

∴ $AL\mathit{=}\frac{\mathit{A}\mathit{B}}{\mathit{2}}=\left(\frac{8}{2}\right)=4\mathrm{cm}$

Now, in right angled ΔOLA, we have:

*OA*

^{2}= AL^{2}+ LO^{2}⇒

*LO*

^{2}^{ }= OA^{2}− AL^{2}⇒

*LO*= 5

^{2}^{2}

*−*4

^{2}

⇒

*LO*= 25

^{2}*−*16 = 9

∴

*LO*= 3 cm

Similarly, $CM\mathit{=}\frac{\mathit{C}\mathit{D}}{\mathit{2}}=\left(\frac{6}{2}\right)=3\mathrm{cm}$

In right angled Δ

*CMO*, we have:

*OC*

^{2}= CM^{2}+ MO^{2}⇒

*MO*

^{2}= OC^{2}− CM^{2}⇒

*MO*

^{2}*= 5*

^{ }^{2}

*−*3

^{2}

⇒

*MO*

^{2}^{ }= 25

*−*9 = 16

∴

*MO*= 4 cm

Hence, distance between the chords = (

*MO + LO*) = (4 + 3) cm = 7 cm

#### Page No 436:

#### Answer:

Let *AB *and *CD* be two chords of a circle such that *AB* is parallel to *CD* and they are on the opposite sides of the centre.

Given: *AB *= 30 cm and *CD* = 16 cm

Draw *OL *⊥* AB* and *OM *⊥* CD*.

Join *OA* and *OC*.

*OA = OC* = 17 cm (Radii of a circle)

The perpendicular from the centre of a circle to a chord bisects the chord.

∴

Now, in right angled Δ*OLA*, we have:

*OA ^{2} = AL^{2} + LO^{2}*

⇒

*LO*−

^{2}^{ }= OA^{2}*AL*

^{2}⇒

*LO*= 17

^{2}^{2}− 15

^{2}

⇒

*LO*289 − 225 = 64

^{2}=∴

*LO*= 8 cm

Similarly, $CM\mathit{=}\left(\frac{\mathit{C}\mathit{D}}{\mathit{2}}\right)=\left(\frac{16}{2}\right)=8\mathrm{cm}$

In right angled Δ

*CMO*

*,*we have:

⇒

*OC*

^{2}*= CM*

^{2}+*MO*

^{2}

⇒

*MO*−

^{2}= OC^{2}*CM*

^{2}⇒

*MO*= 17

^{2}^{2}− 8

^{2}

⇒

*MO*= 289 − 64 = 225

^{2}∴

*MO*= 15 cm

Hence, distance between the chords = (

*LO + MO*) = (8 + 15) cm = 23 cm

#### Page No 436:

#### Answer:

*CD* is the diameter of the circle with centre *O* and is perpendicular to chord *AB*.

Join *OA*.

Given: *AB* = 12 cm and *CE* = 3 cm

Let *OA = OC *= *r* cm (Radii of a circle)

Then *OE* = (*r* - 3) cm

Since the perpendicular from the centre of the circle to a chord bisects the chord, we have:

$AE\mathit{=}\left(\frac{\mathit{A}\mathit{B}}{\mathit{2}}\right)=\left(\frac{12}{2}\right)\mathrm{cm}=6\mathrm{cm}$

Now, in right angled Δ*OEA*, we have:

⇒ *OA*^{2} = *OE ^{2} + AE^{2}*

⇒

*r*

^{2}= (

*r*− 3)

^{2}+ 6

^{2}

⇒

*r*

^{2}=

*r*

^{2}− 6

*r*+ 9 + 36

⇒

*r*

^{2}−

*r*

^{2}+ 6

*r*= 45

⇒ 6

*r*= 45

$\Rightarrow r=\left(\frac{45}{6}\right)\mathrm{cm}=7.5\mathrm{cm}$

⇒

*r*= 7.5 cm

Hence, the required radius of the circle is 7.5 cm.

#### Page No 436:

#### Answer:

*AB* is the diameter of the circle with centre *O,* which bisects the chord* CD* at point *E*.

Given: *CE = ED = *8 cm and *EB *= 4 cm

Join *OC*.

Let *OC = OB* = *r* cm (Radii of a circle)

Then* OE *= (*r* − 4) cm

Now, in right angled Δ*OEC*, we have:

*OC ^{2} = OE^{2} + EC^{2} * (Pythagoras theorem)

⇒

*r*

^{2}= (

*r*− 4)

^{2}+ 8

^{2}

⇒

*r*

^{2}=

*r*

^{2}− 8

*r*+ 16 + 64

⇒

*r*

^{2}−

*r*

^{2}+ 8

*r*= 80

⇒ 8

*r*= 80

$\Rightarrow r=\left(\frac{80}{8}\right)\mathrm{cm}=10\mathrm{cm}$

⇒

*r*= 10 cm

Hence, the required radius of the circle is 10 cm.

#### Page No 437:

#### Answer:

Given: *BC* is a diameter of a circle with centre *O *and *OD** *⊥* AB. *

**To prove:** *AC* parallel to *OD* and *AC = 2 × OD*

**Construction:** Join *AC*.

**Proof:**

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Here, *OD *⊥* AB*

*D* is the mid point of *AB.*

i.e., *AD = BD*

Also, *O* is the mid point of *BC*.

i.e., *OC = OB*

Now, in Δ*ABC**, *we have:

*D* is the mid point of *AB* and *O* is the mid point of *BC.*

According to the mid point theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.

$\mathrm{i}.\mathrm{e}.,OD\parallel AC\mathrm{and}OD\mathit{=}\frac{\mathit{1}}{\mathit{2}}AC$

∴ *AC = *2* × OD*

Hence, proved.

#### Page No 437:

#### Answer:

Given:*O* is the centre of a circle in which chords *AB* and *CD* intersect at *P* such that* PO* bisects ∠*BPD**.*

**To prove:** *AB = CD*

**Construction: **Draw *OE *⊥* AB *and *OF *⊥* CD*

**Proof:** In Δ*OEP** *and Δ*OFP**, *we have:

*∠OEP = ∠OFP * (90° each)

*OP = OP * (Common)

∠*OPE** = ∠OPF * (∵ OP bisects ∠BPD )

Thus, Δ*OEP** *≅ Δ*OFP* (AAS criterion)

⇒ *OE = OF *

Thus, chords *AB* and *CD* are equidistant from the centre *O*.

⇒ *AB = CD *(∵ Chords equidistant from the centre are equal)

∴ *AB = CD*

#### Page No 437:

#### Answer:

Given: *AB* and *CD* are two parallel chords of a circle with centre *O*.

*POQ *is a diameter which is perpendicular to *AB*.

**To prove: ***PF *⊥ *CD* and *CF = FD*

**Proof: **

*AB || CD* and *POQ *is a diameter.

∠*PEB* = 90° (Given)

*∠PFD = ∠PEB * (∵ *AB || CD,* Corresponding angles)

Thus, *PF *⊥* CD *

∴ *OF *⊥* CD*

We know that the perpendicular from the centre to a chord bisects the chord.

i.e., *CF = FD*

Hence,* POQ* is perpendicular to *CD* and bisects it.

#### Page No 437:

#### Answer:

**Given**: Two distinct circles

**To prove: **Two distinct circles cannot intersect each other in more than two points.

**Proof:** Suppose that two distinct circles intersect each other in more than two points.

∴ These points are non-collinear points.

Three non-collinear points determine one and only one circle.

∴ There should be only one circle.

This contradicts the given, which shows that our assumption is wrong.

Hence, two distinct circles cannot intersect each other in more than two points.

#### Page No 437:

#### Answer:

Given: *OA* = 10 cm, *O'A* = 8 cm and *AB* = 12 cm

$AD\mathit{=}\left(\frac{\mathrm{AB}}{2}\right)=\left(\frac{12}{2}\right)=6\mathrm{cm}$

Now, in right angled Δ*ADO*, we have:

*OA*^{2} = *AD*^{2} + *OD*^{2}

⇒ *OD ^{2} = OA^{2} - AD^{2}*

= 10

^{2}- 6

^{2}

= 100 - 36 = 64

∴

*OD*= 8 cm

Similarly, in right angled Δ

*ADO*', we have:

*O'A*

^{2}=

*AD*

^{2}+

*O'D*

^{2}

^{ }

⇒

*O'D*

^{2}

^{ }=

*O'A*

^{2}-

*AD*

^{2}

^{ }

= 8

^{2}- 6

^{2}

= 64 - 36

= 28

⇒ $O\mathit{\text{'}}D=\sqrt{28}=2\sqrt{7}$ cm

Thus,

*OO'*= (

*OD + O'D*)

= $\left(8+2\sqrt{7}\right)\mathrm{cm}$

Hence, the distance between their centres is $\left(8+2\sqrt{7}\right)\mathrm{cm}$.

#### Page No 437:

#### Answer:

**Given:** Two equal circles intersect at point *P* and *Q*.

A straight line passes through *P* and meets the circle at points *A* and *B.*

**To prove:** *QA = QB*

**Construction:** Join *PQ*.

**Proof: **

Two circles will be congruent if and only if they have equal radii.

Here, *PQ* is the common chord to both the circles.

Thus, their corresponding arcs are equal (if two chords of a circle are equal, then their corresponding arcs are congruent).

So, arc *PCQ *= arc *PDQ*

∴ ∠*QAP** = ∠QBP *(Congruent arcs have the same degree in measure)

Hence, *QA = QB* (In isosceles triangle, base angles are equal)

#### Page No 437:

#### Answer:

**Given: ***AB* and *CD* are two chords of a circle with centre *O*. Diameter* POQ* bisects them at points *L* and *M*.

**To prove:** *AB *||* CD*

**Proof: ***AB* and *CD* are two chords of a circle with centre *O.* Diameter *POQ* bisects them at *L *and *M*.

Then *OL *⊥* AB*

Also, *OM *⊥* CD*

∴ ∠ *ALM = ∠ LMD* = 90^{o}

Since alternate angles are equal, we have:

*AB|| CD *

#### Page No 437:

#### Answer:

Two circles with centres *A* and* B *of respective radii 5 cm and 3 cm touch each other internally.

The perpendicular bisector of *AB* meets the bigger circle at *P *and *Q.*

Join *AP*.

Let *PQ* intersect *AB *at point* L*.

Here, *AP *= 5 cm

Then* AB *= (5 - 3) cm = 2 cm

Since *PQ *is the perpendicular bisector of *AB, *we have:

$AL\mathit{=}\left(\frac{\mathit{A}\mathit{B}}{\mathit{2}}\right)=\left(\frac{2}{2}\right)=1\mathrm{cm}$

Now, in right angled Δ*PLA*, we have:

*AP ^{2} = AL^{2} + PL^{2}*

⇒

*PL*

^{2}= AP^{2}- AL^{2}= 5

^{2}- 1

^{2}

= 25 - 1 = 24

⇒ $\mathrm{PL}=\sqrt{24}=2\sqrt{6}\mathrm{cm}$

Thus

*PQ = 2 × PL*

= $\left(2\times 2\sqrt{6}\right)=4\sqrt{6}\mathrm{cm}$

Hence, the required length of

*PQ*is $4\sqrt{6}\mathrm{cm}$.

#### Page No 438:

#### Answer:

We have:

OB = OC, ∠BOC = ∠BCO = *y*

External ∠OBA = ∠BOC + ∠BCO = (2*y*)

Again, OA = OB, ∠OAB = ∠OBA = (2*y*)

External ∠AOD = ∠OAC + ∠ACO

Or *x* = ∠OAB + ∠BCO

Or *x* = (2*y*) + *y* = 3*y*

Hence, *x* = 3*y*

#### Page No 438:

#### Answer:

Let *AC* =* a*.

Since, *AB *= 2*AC*, ∴ *AB* = 2*a*.

From centre *O*, perpendicular is drawn to the chords *AB* and *AC* at points *M* and *N*, respectively.

It is given that *OM* = *p* and* ON* = *q.*

We know that perpendicular drawn from the centre to the chord, bisects the chord.

∴* AM* = *MB* = *a * ...(1)

and *AN* = *NC* = $\frac{a}{2}$* * ...(2)

In ∆*OAN*,

(*AN*)^{2} + (*NO*)^{2} = (*OA*)^{2} (Pythagoras theorem)

$\Rightarrow {\left(\frac{a}{2}\right)}^{2}+{\left(q\right)}^{2}={\left(r\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{2}}{4}+{q}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{2}+4{q}^{2}}{4}={r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}+4{q}^{2}=4{r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=4{r}^{2}-4{q}^{2}....\left(3\right)$

In ∆*OAM*,

(*AM*)^{2} + (*MO*)^{2} = (*OA*)^{2} (Pythagoras theorem)

$\Rightarrow {\left(a\right)}^{2}+{\left(p\right)}^{2}={\left(r\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}={r}^{2}-{p}^{2}....\left(4\right)$

From eq. (3) and (4),

$4{r}^{2}-4{q}^{2}={r}^{2}-{p}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4{r}^{2}-{r}^{2}+{p}^{2}=4{q}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 3{r}^{2}+{p}^{2}=4{q}^{2}$

Hence, 4*q*^{2} = *p*^{2} + 3*r*^{2}.

#### Page No 438:

#### Answer:

**Given:** AB and AC are chords of the circle with centre O. AB = AC, OP ⊥ AB and OQ ⊥ AC

**To prove:** PB = QC

**Proof:**

AB = AC (Given)

⇒ $\frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{AC}$

The perpendicular from the centre of a circle to a chord bisects the chord.

∴ MB = NC ...(i)

Also, OM = ON (Equal chords of a circle are equidistant from the centre)

and OP = OQ (Radii)

⇒ OP - OM = OQ - ON

∴ PM = QN ...(ii)

Now, in ΔMPB and ΔNQC, we have:

MB = NC [From (i)]

∠PMB = ∠QNC [90° each]

PM = QN [From (ii)]

i.e., ΔMPB ≅ ΔNQC (SAS criterion)

∴ PB = QC (CPCT)

#### Page No 438:

#### Answer:

**Given:*** BC* is a diameter of a circle with centre *O. AB* and* CD* are two chords such that *AB *||* CD.*

**TO prove:** *AB = CD*

**Construction:** Draw *OL *⊥* AB *and *OM *⊥* CD.*

**Proof:**

In Δ*OLB* and Δ*OMC*, we have:

*∠OLB = ∠OMC* [90° each]

*∠OBL = ∠OCD* [Alternate angles as *AB || CD*]

*OB = OC * [Radii of a circle]

∴ Δ*OLB* ≅ Δ*OMC** * (AAS criterion)

Thus, *OL = OM * (CPCT)

We know that chords equidistant from the centre are equal.

Hence, *AB = CD*

#### Page No 438:

#### Answer:

Let Δ*ABC** *be an equilateral triangle of side 9 cm.

Let *AD* be one of its median.

Then, *AD *⊥* BC* (Δ*ABC* is an equilateral triangle)

Also, $BD\mathit{=}\left({\displaystyle \frac{BC}{2}}\right)=\left(\frac{9}{2}\right)=4.5\mathrm{cm}$

In right angled ΔADB, we have:

*AB ^{2} = AD^{2} + BD^{2}*

⇒

*AD*

^{2}= AB^{2}- BD^{2}$\Rightarrow AD=\sqrt{A{B}^{2}-B{D}^{2}}$

$=\sqrt{{\left(9\right)}^{2}-{\left(\frac{9}{2}\right)}^{2}}\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

$=\frac{9\sqrt{3}}{2}\mathrm{cm}$

In the equilateral triangle, the centroid and circumcentre coincide and

*AG*:

*GD*= 2 : 1.

Now, radius = $AG\mathit{=}\frac{\mathit{2}}{\mathit{3}}AD$

$\Rightarrow AG=\left(\frac{2}{3}\times \frac{9\sqrt{3}}{2}\right)=3\sqrt{3}\mathrm{cm}$

∴ The radius of the circle is $3\sqrt{3}\mathrm{cm}$.

#### Page No 438:

#### Answer:

**Given:** *AB *and* AC* are two equal chords of a circle with centre *O.*

**To prove:** *∠OAB = ∠OAC*

**Construction: **Join *OA, OB* and *OC*.

**Proof:**

In Δ*OAB* and Δ*OAC*, we have:

*AB = AC * (Given)

*OA = OA* (Common)

*OB = OC *(Radii of a circle)

∴ Δ *OAB *≅* Δ OAC *(By SSS congruency rule)

⇒ *∠OAB = ∠OAC * (CPCT)

Hence, point *O *lies on the bisector of *∠BAC.*

#### Page No 438:

#### Answer:

**Given: ***OPQR* is a square. A circle with centre *O* cuts the square at *X* and *Y.*

**To prove: ***QX = QY*

**Construction:** Join *OX* and *OY*.

**Proof:**

In Δ*OXP* and Δ*OYR*, we have:

*∠OPX = ∠ORY * (90° each)

*OX = OY* (Radii of a circle)

*OP = OR * (Sides of a square)

∴ Δ*OXP* ≅ *ΔOYR *(BY RHS congruency rule)

⇒* PX = RY* (By CPCT)

⇒ *PQ - PX* = *QR - RY* (*PQ* and *QR* are sides of a square)

⇒ *QX = QY*

Hence, proved.

#### Page No 439:

#### Answer:

Given: Two circles with centres *O* and *O*' intersect at two points *A* and *B*.

Draw a line *PQ* parallel to *OO*' through *B*, *OX *perpendicular to *PQ*, *O*'*Y* perpendicular to *PQ*, join all.

We know that perpendicular drawn from the centre to the chord, bisects the chord.

∴ *PX* = *XB* and *YQ* = *BY*

∴ *PX* + *YQ* = *XB* + *BY*

On adding *XB* + *BY* on both sides, we get

*PX* + *YQ* + *XB* + *BY* = 2(*XB* + *BY*)

⇒ *PQ* = 2(*XY*)

⇒ *PQ* = 2(*OO*')

Hence, *PQ* = 2*OO*'.

#### Page No 456:

#### Answer:

(i) Join *BO.*

In Δ*BOC*, we have:

*OC = OB* (Radii of a circle)

⇒ ∠*OBC** = *∠*OCB*

∠*OBC* = 30° ...(i)

In Δ*BOA*, we have:

*OB = OA* (Radii of a circle)

⇒∠*OBA* = ∠*OAB** * [∵ ∠*OAB* = 40°]

⇒∠*OBA* = 40° ...(ii)

Now, we have:

∠*ABC* = ∠*OBC* + ∠*OBA*

= 30° + 40° [From (i) and (ii)]

∴ ∠*ABC** *= 70°

The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

i.e., ∠*AOC** *= 2∠*ABC*

= (2 × 70°) = 140°

(ii)

Here, ∠*BOC* = {360° - (90° + 110°)}

= (360° - 200°) = 160°

We know that ∠*BOC* = 2∠*BAC*

$\Rightarrow \angle BAC\mathit{=}\frac{\mathit{\angle}\mathit{B}\mathit{O}\mathit{C}}{\mathit{2}}=\left(\frac{160\xb0}{2}\right)=80\xb0$

Hence, ∠*BAC* = 80°

#### Page No 456:

#### Answer:

(i)

The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

Thus, ∠*AOB** *= 2∠*OCA*

$\Rightarrow \angle OCA=\left(\frac{\angle AOB}{2}\right)=\left(\frac{70\xb0}{2}\right)=35\xb0$

(ii)

*OA = OC* (Radii of a circle)

∠*OAC** = ∠OCA* [Base angles of an isosceles triangle are equal]

= 35°

#### Page No 457:

#### Answer:

From the given diagram, we have:

∠*ACB** = ∠PCB*

∠*BPC** *= (180° - 110°) = 70° (Linear pair)

Considering ΔPCB, we have:

*∠PCB *+* ∠BPC *+* ∠PBC *= 180° (Angle sum property)

⇒ ∠*PCB* + 70° + 25° = 180°

⇒ ∠*PCB** *= (180° – 95°) = 85°

⇒ ∠*ACB* = ∠*PCB* = 85°

We know that the angles in the same segment of a circle are equal.

∴ ∠*ADB* = ∠*ACB* = 85°

#### Page No 457:

#### Answer:

It is clear that *BD* is the diameter of the circle.

Also, we know that the angle in a semicircle is a right angle.

i.e., ∠*BAD* = 90°

Now, considering the Δ*BAD*, we have:

∠*ADB* + ∠*BAD* + ∠*ABD* = 180° (Angle sum property of a triangle)

⇒ ∠*ADB* + 90° + 35° = 180°

⇒ ∠*ADB* = (180° - 125°) = 55°

Angles in the same segment of a circle are equal.

Hence, ∠*ACB* = ∠*ADB* = 55°

#### Page No 457:

#### Answer:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

∠*AOB** *= 2∠*ACB*

= 2 × 50° [Given]

∠*AOB* = 100° ...(i)

Let us consider the triangle Δ*OAB**.*

*OA = OB *(Radii of a circle)

Thus, ∠*OAB* = ∠*OBA*

In Δ*OAB*, we have:

∠*AOB* + ∠*OAB** *+ ∠*OBA* = 180°

⇒ 100° + ∠*OAB* + ∠*OAB* = 180°

⇒ 100° + 2∠*OAB* = 180°

⇒ 2∠*OAB* = 180° – 100° = 80°

⇒ ∠*OAB* = 40°

Hence, ∠*OAB** = *40°

#### Page No 457:

#### Answer:

(i)

We know that the angles in the same segment of a circle are equal.

*i.e., ∠ABD = ∠ACD *= 54°

(ii)

We know that the angles in the same segment of a circle are equal.

i.e., *∠BAD = ∠BCD* = 43°

(iii)

In Δ*ABD**, *we have:

*∠BAD *+* ∠ADB *+* ∠DBA =* 180° (Angle sum property of a triangle)

⇒ 43° + ∠*ADB* + 54° = 180°

⇒ ∠*ADB* = (180° – 97°) = 83°

⇒ ∠*BDA* = 83°

#### Page No 457:

#### Answer:

Angles in the same segment of a circle are equal.

i.e., *∠CAD = ∠CBD *= 60°

We know that an angle in a semicircle is a right angle.

i.e., *∠ADC* = 90°

In Δ*ADC**, *we have:

*∠ACD *+* ∠ADC *+* ∠CAD *= 180° (Angle sum property of a triangle)

⇒ ∠*ACD* + 90° + 60° = 180°

⇒∠*ACD* = 180° – (90° + 60°) = (180° – 150°) = 30°

⇒∠*CDE* = ∠*ACD* = 30° (Alternate angles as *AC* parallel to *DE*)

Hence, ∠*CDE* = 30°

#### Page No 457:

#### Answer:

*∠BCD = ∠ABC *= 25° (Alternate angles)

Join *CO* and *DO*.

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by an arc at any point on the circumference.

Thus, *∠BOD** = 2∠BCD*

⇒∠*BOD** *= 2 × 25° = 50°

Similarly, ∠*AOC** = 2∠ABC*

⇒* ∠AOC *= 2 × 25° = 50°

*AB* is a straight line passing through the centre.

i.e., *∠AOC + ∠COD + ∠BOD *= 180°

⇒ 50° + ∠*COD* + 50° = 180°

⇒ ∠*COD** *= (180° – 100°) = 80°

$\Rightarrow \angle CED=\frac{1}{2}\angle COD\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \angle CED=\left(\frac{1}{2}\times 80\xb0\right)=40\xb0$

∴ *∠CED *= 40°

#### Page No 458:

#### Answer:

(i)

∠*CED* = 90° (Angle in a semi circle)

In Δ*CED*, we have:

*∠CED +∠EDC + ∠DCE *= 180° (Angle sum property of a triangle)

⇒ 90° + 40° + ∠*DCE* = 180°

⇒ ∠*DCE* = (180° – 130°) = 50° ...(i)

∴ *∠DCE =* 50°

(ii)

As *∠AOC* and *∠BOC* are linear pair, we have:

*∠BOC =* (180° – 80°) = 100° ...(ii)

In Δ *BOC,* we have:

*∠OBC *+* ∠OCB *+* ∠BOC *= 180° (Angle sum property of a triangle)

⇒ *∠ABC + ∠DCE + ∠BOC* = 180° [∵ *∠OBC = ∠ABC* and *∠OCB = ∠DCE*]

⇒ ∠*ABC** *= 180° – (∠*BOC* + ∠*DCE*)

⇒ ∠*ABC* = 180° – (100° + 50°) [From (i) and (ii)]

⇒ ∠*ABC* = (180° - 150°) = 30°

#### Page No 458:

#### Answer:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

*∠AOB = 2∠ACB*

= 2∠*DCB** * [∵∠*ACB** = ∠DCB*]

∴ $\angle DCB=\frac{1}{2}\angle AOB\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \angle DCB=\left(\frac{1}{2}\times 40\xb0\right)=20\xb0$

Considering Δ*DBC**,* we have:

*∠BDC *+* ∠DCB *+* ∠DBC* = 180°

⇒ 100° + 20° + ∠*DBC* = 180°

⇒ ∠*DBC** = *(180° – 120°) = 60°

⇒ *∠OBC = ∠DBC* = 60°

Hence, ∠*OBC* = 60°

#### Page No 458:

#### Answer:

*OA = OB* (Radii of a circle)

Thus, *∠OBA = ∠OAB =* 25°

Join OB.

Now in Δ*OAB*, we have:

∠*OAB** *+* ∠OBA *+* ∠AOB *= 180° (Angle sum property of a triangle)

$\Rightarrow $25° + 25° + ∠*AOB* = 180°

$\Rightarrow $50° + ∠*AOB** =* 180°

$\Rightarrow $∠*AOB** *= (180° – 50°) = 130°

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

i.e., *∠AOB = 2∠ACB*

$\Rightarrow $$\angle ACB\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\angle}AOB=\left(\frac{1}{2}\times 130\xb0\right)=65\xb0$

Here,*∠ACB = ∠ECB*

∴ *∠ECB* = 65° ...(i)

Considering the right angled Δ*BEC*, we have:

∠*EBC** *+* ∠BEC + ∠ECB* = 180° (Angle sum property of a triangle)

$\Rightarrow $∠*EBC* + 90° + 65° = 180° [From(i)]

$\Rightarrow $∠*EBC* = (180° – 155°) = 25°

Hence, ∠*EBC** *= 25°

#### Page No 458:

#### Answer:

**(i)**

*OB = OC* (Radii of a circle)

⇒ *∠OBC = ∠OCB =* 55°

Considering Δ

*BOC*, we have:

*∠BOC*+

*∠OCB*+

*∠OBC*= 180° (Angle sum property of a triangle)

⇒∠

*BOC*+ 55° + 55° = 180°

⇒∠

*BOC*= (180° - 110°) = 70°

**(ii)**

*OA = OB*(Radii of a circle)

⇒ ∠

*OBA*

*= ∠OAB*= 20°

Considering Δ

*AOB*, we have:

*∠AOB*+

*∠OAB*+

*∠OBA*= 180° (Angle sum property of a triangle)

⇒∠

*AOB*+ 20° + 20° = 180°

⇒∠

*AOB*= (180° - 40°) = 140°

∴

*∠AOC = ∠AOB - ∠BOC*

= (140° - 70°)

= 70°

Hence,

*∠AOC*= 70°

#### Page No 458:

#### Answer:

In the given figure, *OD* is parallel to *BC*.

∴ ∠*BCO* = ∠*COD *(Alternate interior angles)

⇒ $\angle COD=30\xb0$ ...(1)

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc *CD* subtends ∠*COD* at the centre and ∠*CBD* at *B* on the circle.

∴ ∠*COD* = 2∠*CBD
⇒ $\angle CBD=\frac{30\xb0}{2}=15\xb0$ * (from (1))

∴ $y=15\xb0$ ...(2)

Also, arc

*AD*subtends ∠

*AOD*at the centre and ∠

*ABD*at

*B*on the circle.

∴ ∠

*AOD*= 2∠

*ABD*

⇒ $\angle ABD=\frac{90\xb0}{2}=45\xb0$...(3)

⇒ $\angle ABD=\frac{90\xb0}{2}=45\xb0$

In ∆

*ABE*,

*x*+

*y*+ ∠

*ABD*+ ∠

*AEB*= 180

^{∘}(Sum of the angles of a triangle)

*⇒ x*+ 15

^{∘}+ 45

^{∘}+ 90

^{∘}= 180

^{∘}(from (2) and (3))

*⇒ x*= 180

^{∘}− (90

^{∘}+ 15

^{∘}+ 45

^{∘})

*⇒ x*= 180

^{∘}− 150

^{∘}

*⇒ x*= 30

^{∘}

Hence,

*x*= 30

^{∘}and

*y*= 15

^{∘}.

#### Page No 459:

#### Answer:

In the given figure, *BD *= *OD *and *CD *⊥ *AB*.

Join *AC* and *OC*.

In ∆*ODE* and ∆*DBE*,

∠*DOE* = ∠*DBE * (given)

∠*DEO * = ∠*DEB = *90^{∘}

*OD* = *DB *(given)

∴ By AAS conguence rule, ∆*ODE* ≌ ∆*BDE*,

Thus, *OE* = *EB * ...(1)

Now, in ∆*COE* and ∆*CBE*,

*CE* = *CE * (common)

∠*CEO * = ∠*CEB = *90^{∘}

*OE* = *EB *(from (1))

∴ By SAS conguence rule, ∆*COE* ≌ ∆*CBE*,

Thus, *CO* = *CB * ...(2)

Also, *CO* = *OB* = *OA* (radius of the circle) ...(3)

From (2) and (3),

*CO* = *CB *= *OB*

∴ ∆*COB* is equilateral triangle.

∴ ∠*COB* = 60^{∘} ...(4)

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc *CB *subtends ∠*COB* at the centre and ∠*CAB* at *A* on the circle.

∴ ∠*COB* = 2∠*CAB
⇒ $\angle CAB=\frac{60\xb0}{2}=30\xb0$ * (from (4))

Hence, ∠

*CAB*= 30

^{∘}.

#### Page No 459:

#### Answer:

Here, *PQ* is the diameter and the angle in a semicircle is a right angle.

i.e., ∠*PRQ* = 90°

In Δ*PRQ*, we have:

*∠QPR *+* ∠PRQ *+* ∠PQR *= 180° (Angle sum property of a triangle)

⇒ ∠*QPR* + 90° + 65° = 180°

⇒∠*QPR** *= (180° – 155°) = 25°

In Δ*PQM**, PQ* is the diameter.

∴∠*PMQ* = 90°

In Δ*PQM*, we have:

*∠QPM *+* ∠PMQ *+* ∠PQM *= 180° (Angle sum property of a triangle)

⇒∠*QPM** *+ 90° + 50° = 180°

⇒ ∠*QPM* = (180° – 140°) = 40°

Now, in quadrilateral *PQRS*, we have:

*∠QPS *+* ∠SRQ* = 180° (Opposite angles of a cyclic quadrilateral)

⇒*∠QPR** *+* ∠RPS *+* ∠PRQ *+* ∠PRS* = 180°

⇒ 25° + 40° + 90° + ∠*PRS* = 180°

⇒ ∠*PRS* = 180° – 155° = 25°

∴ ∠*PRS** *= 25°

Thus, ∠*QPR* = 25°; ∠*QPM* = 40°; ∠*PRS** *= 25°

#### Page No 459:

#### Answer:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc *AEB *subtends ∠*APB* at the centre and ∠*ACB* at *C* on the circle.

∴ ∠*APB* = 2∠*ACB
⇒ $\angle ACB=\frac{150\xb0}{2}=75\xb0$ * ...(1)

Since

*ACD*is a straight line, ∠

*ACB +*∠

*BCD*= 180

^{∘}

⇒ ∠

*BCD*= 180

^{∘}− 75

^{∘}

⇒ ∠

*BCD*= 105

^{∘}

*...(2)*

Also, arc

*BFD*subtends reflex ∠

*BQD*at the centre and ∠

*BCD*at

*C*on the circle.

∴ reflex ∠

*BQD*= 2∠

*BCD*

⇒ $\mathrm{reflex}\angle BQD=2\left(105\xb0\right)=210\xb0$...(3)

⇒ $\mathrm{reflex}\angle BQD=2\left(105\xb0\right)=210\xb0$

Now,

reflex ∠

*BQD*+ ∠

*BQD =*360

^{∘}

*⇒*210

^{∘}+

*x =*360

^{∘}

*⇒*

*x =*360

^{∘}−

*210*

^{∘}

*⇒*

*x =*150

^{∘}

Hence,

*x =*150

^{∘}.

#### Page No 459:

#### Answer:

Join *OB* and *OC.*

*∠BOC = 2∠BAC* (As angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference)

= 2 × 30° [∵ ∠*BAC* = 30°]

= 60° ...(i)

Consider Δ*BOC*, we have:

*OB = OC * [Radii of a circle]

⇒* ∠OBC = ∠OCB * ...(ii)

In Δ*BOC*, we have:

*∠BOC *+* ∠OBC *+ *∠OCB* = 180 (Angle sum property of a triangle)

⇒ 60° +* ∠OCB *+* ∠OCB* = 180° [From (i) and (ii)]

⇒ 2∠*OCB* = (180° - 60°) = 120°

⇒ ∠*OCB** *= 60° ...(ii)

Thus we have:

*∠OBC = ∠OCB = ∠BOC = *60°

Hence, Δ*BOC* is an equilateral triangle.

i.e., *OB = OC = BC*

∴ *BC *is the radius of the circumcircle.

#### Page No 459:

#### Answer:

Join AD

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc *AXC *subtends ∠*AOC* at the centre and ∠*ADC* at *D* on the circle.

∴ ∠*AOC* = 2∠*ADC
⇒ $\angle ADC=\frac{1}{2}\left(\angle AOC\right)$ * ...(1)

Also, arc

*DYB*subtends ∠

*DOB*at the centre and ∠

*DAB*at

*A*on the circle.

∴ ∠

*DOB*= 2∠

*DAB*

⇒ $\angle DAB=\frac{1}{2}\left(\angle DOB\right)$...(2)

⇒ $\angle DAB=\frac{1}{2}\left(\angle DOB\right)$

Now, in ∆ADE,

∠

*AEC*= ∠

*ADC +*∠

*DAB*(Exterior angle)

⇒ ∠

*AEC*= $\frac{1}{2}\left(\angle AOC+\angle DOB\right)$ (from (1) and (2))

Hence, ∠

*AEC*= $\frac{1}{2}$(angle subtended by arc

*CXA*at the centre + angle subtended by arc

*DYB*at the centre).

#### Page No 482:

#### Answer:

(i) ∠BDC = ∠BAC = 40° (Angles in the same segment)

In ΔBCD, we have:

∠BCD + ∠DBC + ∠BDC = 180° (Angle sum property of a triangle)

⇒ ∠BCD + 60° + 40° = 180°

⇒ ∠BCD = (180° - 100°) = 80°

(ii) ∠CAD = ∠CBD (Angles in the same segment)

= 60°

#### Page No 482:

#### Answer:

In cyclic quadrilateral* PQRS,* we have:

*∠PSR *+* ∠PQR* = 180°

⇒ 150° + ∠*PQR** *= 180°

⇒ ∠*PQR* = (180° – 150°) = 30°

∴ ∠*PQR* = 30° ...(i)

Also, ∠*PRQ* = 90° (Angle in a semicircle) ...(ii)

Now, in Δ*PRQ*, we have:

*∠PQR *+* ∠PRQ *+* ∠RPQ* = 180°

⇒ 30° + 90° + ∠*RPQ* = 180° [From(i) and (ii)]

⇒ ∠*RPQ* = 180° – 120° = 60°

∴ ∠*RPQ* = 60°

#### Page No 482:

#### Answer:

Reflex ∠*AOC* + ∠*AOC = *360^{∘}

*⇒ *Reflex ∠*AOC + *130^{∘} + *x = *360^{∘}

*⇒ *Reflex ∠*AOC = *360^{∘} − 130^{∘}

*⇒ *Reflex ∠*AOC = *230^{∘}

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc *AC *subtends reflex ∠*AOC* at the centre and ∠*ABC* at *B* on the circle.

∴ ∠*AOC* = 2∠*ABC
⇒ $\angle ABC=\frac{230\xb0}{2}=115\xb0$ * ...(1)

Since

*ABP*is a straight line, ∠

*ABC +*∠

*PBC*= 180

^{∘}

⇒ ∠

*PBC*= 180

^{∘}− 115

^{∘}

⇒ ∠

*PBC*= 65

^{∘}

*...(2)*

Hence, ∠

*PBC*= 65

^{∘}.

#### Page No 482:

#### Answer:

Given: *ABCD* is a cyclic quadrilateral.

Then *∠ABC + ∠ADC* = 180°

⇒ 92° + ∠*ADC* = 180°

⇒ ∠*ADC* = (180° – 92°) = 88°

Again, *AE* parallel to *CD.*

Thus, *∠EAD = ∠ADC* = 88° (Alternate angles)

We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

∴ *∠BCD = ∠DAF*

⇒ *∠BCD = ∠EAD + ∠EAF*

= 88° + 20° = 108°

Hence, ∠*BCD* = 108°

#### Page No 482:

#### Answer:

*BD = DC*

⇒ *∠BCD = ∠CBD* = 30°

In Δ*BCD**,* we have:

*∠BCD *+* ∠CBD *+* ∠CDB *= 180° (Angle sum property of a triangle)

⇒ 30° + 30° + ∠*CDB* = 180°

⇒ *∠CDB *= (180° – 60°) = 120°

The opposite angles of a cyclic quadrilateral are supplementary.

Thus, *∠CDB *+* ∠BAC* = 180°

⇒ 120° + ∠*BAC* = 180°

⇒ ∠*BAC* = (180° – 120°) = 60°

∴ ∠*BAC** *= 60°

#### Page No 482:

#### Answer:

We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.

Thus, *∠AOC = 2∠ADC*

⇒ 100° = 2*∠ADC*

∴ *∠ADC =* 50°

The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.

Thus, *∠ADC +∠ABC* = 180°

⇒ 50° + *∠ABC *= 180°

⇒ *∠ABC* = (180° – 50°) = 130°

∴ *∠ADC* = 50° and* ∠ABC* = 130°

#### Page No 483:

#### Answer:

(i)

Given: Δ*ABC* is an equilateral triangle.

i.e., each of its angle = 60°

⇒ *∠BAC = ∠ABC = ∠ACB *= 60°

Angles in the same segment of a circle are equal.

i.e., *∠BDC = ∠BAC* = 60°

∴ *∠BDC *= 60°

(ii)

The opposite angles of a cyclic quadrilateral are supplementary.

Then in cyclic quadrilateral *ABEC, *we have:

*∠BAC + ∠BEC* = 180°

⇒ 60° + *∠BEC* = 180°

⇒ *∠BEC* = (180° – 60°) = 120°

∴ *∠BDC* = 60° and *∠BEC* = 120°

#### Page No 483:

#### Answer:

Given: *ABCD* is a cyclic quadrilateral.

∴ *∠DAB *+* ∠DCB *= 180° ( Opposite angles of a cyclic quadrilateral are supplementary)

⇒ ∠*DAB* + 100° = 180°

⇒ *∠DAB* = (180° – 100°) = 80°

Now, in Δ*ABD**,* we have:

⇒ *∠DAB + ∠ABD + ∠ADB *= 180°

⇒ 80° + 50° + *∠ADB *= 180°

⇒ *∠ADB* = (180° – 130°) = 50°

Hence, *∠ADB *= 50°

#### Page No 483:

#### Answer:

*O* is the centre of the circle and ∠*BOD* = 150°.

Thus, reflex angle ∠*BOD* = (360° – 150°) = 210°

Now, $x=\frac{1}{2}\left(\mathrm{reflex}\angle BOD\right)=\left(\frac{1}{2}\times 210\xb0\right)=105\xb0$

∴ *x* = 105°

Again, *x* + *y* = 180° (Opposite angles of a cyclic quadrilateral)

⇒ 105° + *y* = 180°

⇒ *y* = (108° - 105°)= 75°

∴ *y* = 75°

Hence, *x* = 105° and *y* = 75°

#### Page No 483:

#### Answer:

*O* is the centre of the circle and ∠*DAB* = 50°.

*OA = OB* (Radii of a circle)

⇒ *∠OBA = ∠OAB *= 50°

In Δ*OAB*, we have:

*∠OAB *+* ∠OBA *+* ∠AOB* = 180°

⇒ 50° + 50° +*∠AOB* = 180°

⇒ *∠AOB* = (180° – 100°) = 80°

Since *AOD* is a straight line, we have:

∴ *x* = 180° – *∠AOB*

= (180° – 80°) = 100°

i.e., *x* = 100°

The opposite angles of a cyclic quadrilateral are supplementary.

*ABCD* is a cyclic quadrilateral.

Thus,* ∠DAB *+* ∠BCD *= 180°

*∠BCD* = (180° – 50°) = 130°

∴ *y* = 130°

Hence, *x *= 100° and *y* = 130°

#### Page No 483:

#### Answer:

*ABCD *is a cyclic quadrilateral.

We know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.

∴ *∠CBF *=* ∠CDA*

⇒ *∠CBF* = (180° – *x*)

⇒ 130° = 180° – *x* [∵ *∠CBF *= 130°]

⇒* x* = (180° – 130°) = 50°

Hence, *x* = 50°

#### Page No 484:

#### Answer:

We have,

*AB* is a diameter of the circle where *O* is the centre, *DO || BC and ∠BCD *= 120°.

(i)

Since *ABCD* is a cyclic quadrilateral, we have:

*∠BCD + ∠BAD* = 180°

⇒ 120° + *∠BAD *= 180°

⇒ *∠BAD* = (180° – 120°) = 60°

∴ *∠BAD* = 60°

(ii)

*∠BDA *= 90° (Angle in a semicircle)

In Δ *ABD, *we have:

*∠BDA + ∠BAD + ∠ABD* = 180°

⇒ 90° + 60° + *∠ABD *= 180°

⇒ *∠ABD* = (180° – 150°) = 30°

∴ *∠ABD* = 30°

(iii)

*OD = OA* (Radii of a circle)

*∠ODA = ∠OAD
= ∠BAD *= 60°

*∠ODB*= 90° - ∠

*ODA*= (90° - 60°) = 30°

Here, DO || BC (Given; alternate angles)

*∠CBD = ∠ODB*= 30°

∴

*∠CBD*= 30°

(iv)

*∠ADC = ∠ADB*+

*∠CDB*

= 90° + 30° = 120°

In Δ

*AOD*, we have:

*∠ODA*+

*∠OAD*+

*∠AOD*= 180°

⇒ 60° + 60° + ∠

*AOD*= 180°

⇒ ∠

*AOD*

*=*180° – 120° = 60°

Since all the angles of Δ

*AOD*are of 60° each, Δ

*AOD*is an equilateral triangle.

#### Page No 484:

#### Answer:

*AB* and *CD* are two chords of a circle which intersect each other at* P* outside the circle.

*AB *= 6 cm, *BP *= 2 cm and *PD* = 2.5 cm

∴ *AP × BP = CP × DP*

⇒ 8 × 2 = (CD + 2.5) × 2.5 [∵* CP = CD + DP*]

Let *CD *= *x* cm

Thus, 8 × 2 = (*CD* + 2.5) × 2.5

⇒ 16 = 2.5*x* + 6.25

⇒ 2.5*x* = (16 - 6.25) = 9.75

⇒ $x=\frac{9.75}{2.5}=3.9$

Hence, *CD* = 3.9 cm

#### Page No 484:

#### Answer:

O is the centre of the circle where *∠AOD* = 140° and ∠*CAB* = 50°.

(i) *∠BOD *= 180° – *∠AOD*

= (180° – 140°) = 40°

We have the following:

*OB = OD* (Radii of a circle)

*∠OBD = ∠ODB*

In Δ*OBD*, we have:

*∠BOD + ∠OBD + ∠ODB *= 180°

⇒ *∠BOD + ∠OBD + ∠OBD =* 180° [∵ *∠OBD = ∠ODB*]

⇒ 40° +2∠*OBD* = 180°

⇒ 2∠*OBD* = (180° – 40°) = 140°

⇒ ∠*OBD* = 70°

Since *ABCD* is a cyclic quadrilateral, we have:

∠*CAB* + ∠*BDC* = 180°

⇒ ∠*CAB* + ∠*ODB** *+ ∠*ODC* = 180°

⇒ 50° + 70° + ∠*ODC* = 180°

⇒ ∠*ODC* = (180° – 120°) = 60°

∴ ∠*ODC* = 60°

∠*EDB** *= (180° – (∠*ODC** + ∠ODB*)

= 180° – (60° + 70°)

= 180° – 130° = 50°

∴ ∠*EDB* = 50°

(ii) *∠EBD* = 180° - *∠OBD*

= 180° - 70°

= 110°

#### Page No 484:

#### Answer:

ABC is an isosceles triangle.

Here, AB = AC

∴ ∠ACB = ∠ABC ...(i)

So, exterior ∠ADE = ∠ACB

= ∠ABC [from(i)]

∴ ∠ADE = ∠ABC (Corresponding angles)

Hence, DE || BC

#### Page No 484:

#### Answer:

AB and CD are two parallel chords of a circle. BDE and ACE are two straight lines that intersect at E.

If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.

∴ Exterior ∠EDC = ∠A ...(i)

Exterior ∠DCE = ∠B ...(ii)

Also, AB parallel to CD.

Then, ∠EDC = ∠B (Corresponding angles)

and ∠DCE = ∠A (Corresponding angles)

∴ ∠A = ∠B [From(i) amd (ii)]

Hence, ΔAEB is isosceles.

#### Page No 484:

#### Answer:

We know that if one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.

i.e., *∠BAD* = *∠DCF = 75*°

⇒ *∠DCF = x* = 75°

Again, the sum of opposite angles in a cyclic quadrilateral is 180°.

Thus, *∠DCF + ∠DEF = 180*°

⇒ 75° + *y **= 180*°

⇒ *y** = (180*° - 75°) = 105°

Hence, *x* = 75° and *y* = 105°

#### Page No 485:

#### Answer:

ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD.

Draw DE ⊥ AB and CF ⊥ AB.

In ΔADE and ΔBCF, we have:

∠ADE = ∠ADC - 90° = ∠BCD - 90° = ∠BCF (Given: ∠ADC = ∠BCD)

AD = BC (Given)

and ∠AED = ∠BCF = 90°

∴ ΔADE ≅ ΔBCF (By AAS congruency)

⇒ ∠A = ∠B

Now, ∠A + ∠B + ∠C + ∠D = 360°

⇒ 2∠B + 2∠D = 360°

⇒ ∠B + ∠D = 180°

Hence, ABCD is a cyclic quadrilateral.

#### Page No 485:

#### Answer:

Let ABCD be the cyclic quadrilateral and PO, QO, RO and SO be the perpendicular bisectors of sides AB, BC, CD and AD.

We know that the perpendicular bisector of a chord passes through the centre of the circle.

Since, AB, BC, CD and AD are the chords of a circle, PO, QO, RO and SO pass through the centre.

i.e., PO, QO, RO and SO are concurrent.

Hence, the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

#### Page No 485:

#### Answer:

Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O.

The diagonals of a rhombus bisect each other at right angles.

i.e., ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Now, circles with AB, BC, CD and DA as diameter passes through O (angle in a semi-circle is a right angle).

Hence, the circle with four sides of a rhombus as diameter, pass through O, i.e., the point of intersection of its diagonals.

#### Page No 485:

#### Answer:

**Given:** ABCD is a cyclic rectangle whose diagonals intersect at O.

**To prove:** O is the centre of the circle.

**Proof:**

Here, ∠BCD = 90° [Since it is a rectangle]

So, BD is the diameter of the circle** **(if the angle made by the chord at the circle is right angle, then the chord is the diameter).

Also, diagonals of a rectangle bisect each other and are equal.

∴ OA = OB = OC = OD

BD is the diameter.

∴ BO and OD are the radius.

Thus, O is the centre of the circle.

Also, the centre of the circle is circumscribing the cyclic rectangle.

Hence, O is the point of intersection of the diagonals of ABCD.

#### Page No 485:

#### Answer:

Let A, B and C be the given points.

With B as the centre and a radius equal to AC, draw an arc.

With C as the centre and AB as radius, draw another arc intersecting the previous arc at D.

Then D is the desired point.

**Proof: **Join BD and CD.

In ΔABC and ΔDCB, we have:

AB = DC

AC = DB

BC = CB

i.e., ΔABC ≅ ΔDCB

⇒ ∠BAC = ∠CDB

Thus, BC subtends equal angles ∠BAC and ∠CDB on the same side of it.

∴ Points A, B, C and D are cyclic.

#### Page No 485:

#### Answer:

In cyclic quadrilateral ABCD, we have:

∠B + ∠D = 180° ...(i) (Opposite angles of a cyclic quadrilateral )

∠B - ∠D = 60° ...(ii) (Given)

From (i) and (ii), we get:

2∠B = 240°

⇒ ∠B = 120°

∴ ∠D = 60°

Hence, the smaller of the two angles is 60°.

#### Page No 485:

#### Answer:

Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angles.

Let OL ⊥ AB such that LO produced meets CD at M.

Then we have to prove that CM = MD

Clearly, ∠1 = ∠2 [Angles in the same segment]

∠2 + ∠3 = 90° [∵ ∠OLB = 90°]

∠3 + ∠4= 90° [∵ LOM is a straight line and ∠BOC = 90°]

∴ ∠2 + ∠3 = ∠3 + ∠4 ⇒∠2 = ∠4

Thus, ∠1 = ∠2 and ∠2 = ∠4 ⇒ ∠1 = ∠4

∴ OM = CM and, similarly, OM = MD

Hence, CM = MD

#### Page No 485:

#### Answer:

Draw two right triangles *ACB *and *ADB *in a circle with centre *O*, where *AB* is the diameter of the circle.

Join *CO*.

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc *CB *subtends ∠*COB* at the centre and ∠*CAB* at *A* on the circle.

∴ ∠*COB* = 2∠*CAB * ...(1)

Also, arc *CB *subtends ∠*COB* at the centre and ∠*CDB* at *D* on the circle.

∴ ∠*COB* = 2∠*CDB * ...(2)

Equating (1) and (2),

2∠*CAB *= 2∠*CDB
⇒ *∠

*CAB*= ∠

*CDB*

Hence, ∠

*BAC*= ∠

*BDC*.

#### Page No 485:

#### Answer:

In the given figure, *ABCD *is a quadrilateral such that *A* is the centre of the circle passing through *B, C *and* D.*

Join *AC *and *BD*.

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc *CD *subtends ∠*CAD* at the centre and ∠*CBD* at *B* on the circle.

∴ ∠*CAD* = 2∠*CBD * ...(1)

Also, arc *CB *subtends ∠*CAB* at the centre and ∠*CDB* at *D* on the circle.

∴ ∠*CAB* = 2∠*CDB * ...(2)

Adding (1) and (2), we get

∠*CAD + *∠*CAB *= 2(∠*CBD + *∠*CDB*)

*⇒ *∠*BAD *= 2(∠*CBD + *∠*CDB*)

*⇒ *∠*CBD *+ ∠*CDB *= $\frac{1}{2}$∠*BAD*

Hence, ∠*CBD *+ ∠*CDB *= $\frac{1}{2}$∠*BAD*.

#### Page No 489:

#### Answer:

(b) 12 cm

Let AB be the chord of the given circle with centre O and a radius of 13 cm.

Then, AB = 10 cm and OB = 13 cm

From O, draw OM perpendicular to AB.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ BM = $\left(\frac{10}{2}\right)\mathrm{cm}=5\mathrm{cm}$

From the right ΔOMB, we have:

OB^{2}^{ }= OM^{2} + MB^{2}

⇒ 13^{2} = OM^{2} + 5^{2}

⇒ 169 = OM^{2} + 25

⇒ OM^{2} = (169 - 25) = 144

⇒ $\mathrm{OM}=\sqrt{144}\mathrm{cm}=12\mathrm{cm}$

Hence, the distance of the chord from the centre is 12 cm.

#### Page No 489:

#### Answer:

(c) 30 cm

Let AB be the chord of the given circle with centre O and a radius of 17 cm.

From O, draw OM perpendicular to AB.

Then OM = 8 cm and OB = 17 cm

From the right ΔOMB, we have:

OB^{2}^{ }= OM^{2} + MB^{2}

⇒ 17^{2} = 8^{2} + MB^{2}

⇒ 289 = 64 + MB^{2}

⇒ MB^{2} = (289 - 64) = 225

⇒ $\mathrm{MB}=\sqrt{225}\mathrm{cm}=15\mathrm{cm}$

The perpendicular from the centre of a circle to a chord bisects the chord.

∴ AB = 2 × MB = (2 x 15) cm = 30 cm

Hence, the required length of the chord is 30 cm.

#### Page No 489:

#### Answer:

(b) 45°

Since an angle in a semicircle is a right angle, ∠BAC = 90°

∴ ∠ABC + ∠ACB = 90°

Now, AB = AC (Given)

⇒ ∠ABC = ∠ACB = 45°

#### Page No 489:

#### Answer:

(c) 60°

We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference.

Thus, ∠AOB = (2 × ∠ACB) = (2 × 30°) = 60°

#### Page No 489:

#### Answer:

(b) 50°

OA = OB

⇒ ∠OBA = ∠OAB = 40°

Now, ∠AOB = 180° - (40° + 40°) = 100°

∴ $\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 100\right)\xb0=50\xb0$

#### Page No 489:

#### Answer:

(a) 8 cm

Join OC. Then OC = radius = 17 cm

$\mathrm{CL}=\frac{1}{2}\mathrm{CD}=\left(\frac{1}{2}\times 30\right)\mathrm{cm}=15\mathrm{cm}$

In right ΔOLC, we have:

OL^{2} = OC^{2} - CL^{2} = (17)^{2} - (15)^{2} = (289 - 225) = 64

$\Rightarrow \mathrm{OL}=\sqrt{64}=8\mathrm{cm}$

∴ Distance of CD from AB = 8 cm

#### Page No 489:

#### Answer:

(b) 80°

Given: AB = CD

We know that equal chords of a circle subtend equal angles at the centre.

∴ ∠COD = ∠AOB = 80°

#### Page No 490:

#### Answer:

(c) 7.5 cm

Let OA = OC = *r* cm.

Then OE = (*r* - 3) cm and $\mathrm{AE}=\frac{1}{2}\mathrm{AB}=6\mathrm{cm}$

Now, in right ΔOAE, we have:

OA^{2} = OE^{2} +AE^{2}

⇒ (*r*)^{2} = (*r* - 3)^{2} + 6^{2}

⇒ *r*^{2} = *r*^{2} + 9 - 6*r* + 36

⇒ 6*r* = 45

⇒ $r=\frac{45}{6}=7.5$ cm

Hence, the required radius of the circle is 7.5 cm.

#### Page No 490:

#### Answer:

(a) 10 cm

Let the radius of the circle be *r* cm.

Let OD = OB = *r* cm.

Then OE = (*r* - 4) cm and ED = 8 cm

Now, in right ΔOED, we have:

OD^{2} = OE^{2} +ED^{2}

⇒ (*r*)^{2} = (*r* - 4)^{2} + 8^{2}

⇒ *r*^{2} = *r*^{2} + 16 - 8*r* + 64

⇒ 8*r* = 80

⇒ *r* = 10 cm

Hence, the required radius of the circle is 10 cm.

#### Page No 490:

#### Answer:

(d) 10 cm

Draw OE ⊥ AB and OF ⊥ CD.

In Δ OEB and ΔOFC, we have:

OB = OC (Radius of a circle)

∠BOE = ∠COF (Vertically opposite angles)

∠OEB = ∠OFC (90° each)

∴ ΔOEB ≅ ΔOFC (By AAS congruency rule)

∴ OE = OF

Chords equidistant from the centre are equal.

∴ CD = AB = 10 cm

#### Page No 490:

#### Answer:

(b) 75°

OB = BC (Given)

⇒ ∠BOC = ∠BCO = 25°

Exterior ∠OBA = ∠BOC + ∠BCO = (25° + 25°) = 50°

OA = OB (Radius of a circle)

⇒ ∠OAB = ∠OBA = 50°

In Δ AOC, side CO has been produced to D.

∴ Exterior ∠AOD = ∠OAC + ∠ACO

= ∠OAB + ∠BCO

= (50° + 25°) = 75°

#### Page No 490:

#### Answer:

(b) 12 cm

OD ⊥ AB

i.e., D is the mid point of AB.

Also, O is the mid point of BC.

Now, in Δ BAC, D is the mid point of AB and O is the mid point of BC.

∴ $\mathrm{OD}=\frac{1}{2}\mathrm{AC}$ (By mid point theorem)

⇒ AC = 2OD = (2 × 6) cm = 12 cm

#### Page No 490:

#### Answer:

(c) $3\sqrt{3}\mathrm{cm}$

Let ΔABC be an equilateral triangle of side 9 cm.

Let AD be one of its medians.

Then AD ⊥ BC and BD = 4.5 cm

∴ $\mathrm{AD}=\sqrt{{\mathrm{AB}}^{2}-{\mathrm{BD}}^{2}}=\sqrt{{\left(9\right)}^{2}-{\left(\frac{9}{2}\right)}^{2}}=\sqrt{81-\frac{81}{4}}=\sqrt{\frac{324-81}{4}}=\sqrt{\frac{243}{4}}=\frac{9\sqrt{3}}{2}\mathrm{cm}$

Let G be the centroid of ΔABC.

Then AG : GD = 2 : 1

∴ Radius = AG = $\frac{2}{3}\mathrm{AD}=\left(\frac{2}{3}\times \frac{9\sqrt{3}}{2}\right)\mathrm{cm}=3\sqrt{3}\mathrm{cm}$

#### Page No 490:

#### Answer:

(c) 90°

The angle in a semicircle measures 90°.

#### Page No 490:

#### Answer:

(a) equal

The angles in the same segment of a circle are equal.

#### Page No 491:

#### Answer:

(c) 70°

∠BDC = ∠BAC = 60° (Angles in the same segment of a circle)

In Δ BDC, we have:

∠DBC + ∠BDC + ∠BCD = 180° (Angle sum property of a triangle)

∴ 50° + 60° + ∠BCD = 180°

⇒ ∠BCD = 180° - (50° + 60°) = (180° - 110°) = 70°

#### Page No 491:

#### Answer:

(c) 60°

Angles in a semi circle measure 90°.

∴ ∠BAC = 90°

In Δ ABC, we have:

∠BAC + ∠ABC + ∠BCA = 180° (Angle sum property of a triangle)

∴ 90° + ∠ABC + 30° = 180°

⇒ ∠ABC = (180° - 120°) = 60°

∴ ∠CDA = ∠ABC = 60° (Angles in the same segment of a circle)

#### Page No 491:

#### Answer:

(b) 50°

*∠ODB =∠OAC* = 50° (Angles in the same segment of a circle)

#### Page No 491:

#### Answer:

(c) 100°

In Δ OAB, we have:

OA = OB (Radii of a circle)

⇒ ∠OAB = ∠OBA = 20°

In ΔOAC, we have:

OA = OC (Radii of a circle)

⇒ ∠OAC = ∠OCA = 30°

Now, ∠BAC = (20° + 30°) = 50°

∴ ∠BOC = (2 × ∠BAC) = (2 × 50°) = 100°

#### Page No 491:

#### Answer:

(a) 85°

We have:

∠BOC + ∠BOA + ∠AOC = 360°

⇒ ∠BOC + 100° + 90° = 360°

⇒ ∠BOC = (360° - 190°) = 170°

∴ $\angle \mathrm{BAC}=\left(\frac{1}{2}\times \angle \mathrm{BOC}\right)=\left(\frac{1}{2}\times 170\xb0\right)=85\xb0$

#### Page No 491:

#### Answer:

(d) 65°

We have:

OA = OB (Radii of a circle)

Let *∠** *OAB = ∠ OBA = *x*°

In Δ OAB, we have:

*x*° + *x*° + 50° = 180° (Angle sum property of a triangle)

⇒ 2*x*° = (180° - 50°) = 130°

⇒ $x=\left(\frac{130}{2}\right)\xb0=65\xb0$

Hence,* **∠*OAB = 65°

#### Page No 491:

#### Answer:

(c) 30°

∠COB = 180° - 120° = 60° (Linear pair)

Now, arc BC subtends ∠COB at the centre and ∠BDC at the point D of the remaining part of the circle.

∴ ∠COB = 2∠BDC

⇒ $\angle \mathrm{BDC}=\frac{1}{2}\angle \mathrm{COB}=\left(\frac{1}{2}\times 60\xb0\right)=30\xb0$

#### Page No 492:

#### Answer:

(b) 50°

We have:

OA = OB (Radii of a circle)

⇒ ∠OBA = ∠OAB = 50°

∴ ∠CDA = ∠OBA = 50° (Angles in the same segment of a circle)

#### Page No 492:

#### Answer:

(b) 60°

We have:

*∠CDB = ∠CAB = 40°* (Angles in the same segment of a circle)

In Δ CBD, we have:

*∠CDB + ∠BCD +∠CBD = 180°* (Angle sum property of a triangle)

⇒ *40° + 80° *+* ∠CBD = 180°*

⇒ *∠CBD = (180° - 120°) = 60°*

#### Page No 492:

#### Answer:

(c) 80°

We have:

*∠AEB + ∠CEB* = 180° (Linear pair angles)

⇒ 110° + *∠CEB* = 180°

⇒* ∠CEB = (180*° - 110°) = 70°

In Δ*CEB**, we have:
∠CEB + ∠EBC + ∠ECB = 180*° (Angle sum property of a triangle)

⇒ 70° + 30° +

*∠ECB = 180*°

⇒

*∠ECB = (180*° - 100°) = 80°

The angles in the same segment are equal.

Thus, ∠

*ADB*=

*∠ECB =*80°

#### Page No 492:

#### Answer:

(d) 60°

We have:

OA = OB (Radii of a circle)

⇒ ∠OBA= ∠OAB = 20°

In ΔOAB, we have:

∠OAB + ∠OBA + ∠AOB = 180° (Angle sum property of a triangle)

⇒ 20° + 20° + ∠AOB = 180°

⇒ ∠AOB = (180° - 40°) = 140°

Again, we have:

OB = OC (Radii of a circle)

⇒ ∠OBC = ∠OCB = 50°

In ΔOCB, we have:

∠OCB + ∠OBC + ∠COB = 180° (Angle sum property of a triangle)

⇒ 50° + 50° + ∠COB = 180°

⇒ ∠COB = (180° - 100°) = 80°

Since ∠AOB = 140°, we have:

∠AOC + ∠COB = 140°

⇒∠AOC + 80° = 140°

⇒ ∠AOC = (180° - 80°) = 60°

#### Page No 492:

#### Answer:

(b) 30°

We have:

∠ABC + ∠ADC = 180° (Opposite angles of a cyclic quadrilateral)

⇒ ∠ABC + 120° = 180°

⇒ ∠ABC = (180° - 120°) = 60°

Also, ∠ACB = 90° (Angle in a semicircle)

In ΔABC, we have:

∠BAC + ∠ACB + ∠ABC = 180° (Angle sum property of a triangle)

⇒ ∠BAC + 90° + 60° = 180°

⇒ ∠BAC = (180° - 150°) = 30°

#### Page No 492:

#### Answer:

(b) 100°

Since ABCD is a cyclic quadrilateral, we have:

∠BAD + ∠BCD = 180° (Opposite angles of a cyclic quadrilateral)

⇒ 100° + ∠BCD = 180°

⇒ ∠BCD = (180° - 100°) = 80°

Now, AB || DC and CB is the transversal.

∴ ∠ABC + ∠BCD = 180°

⇒ ∠ABC + 80° = 180°

⇒ ∠ABC = (180° - 80°) = 100°

#### Page No 492:

#### Answer:

(c) 115°

Take a point D on the remaining part of the circumference.

Join AD and CD.

Then $\angle \mathrm{ADC}=\frac{1}{2}\angle \mathrm{AOC}=\left(\frac{1}{2}\times 130\xb0\right)=65\xb0$

In cyclic quadrilateral ABCD, we have:

∠ABC + ∠ADC = 180° (Opposite angles of a cyclic quadrilateral)

⇒ ∠ABC + 65° = 180°

⇒ ∠ABC = (180° - 65°) = 115°

#### Page No 493:

#### Answer:

(a) 30°

∠ADC = ∠BAD = 30° (Alternate angles)

∠ADB = 90° (Angle in semicircle)

∴ ∠CDB = (90° + 30°) = 120°

But ABCD being a cyclic quadrilateral, we have:

∠BAC + ∠CDB = 180°

⇒ ∠BAD + ∠CAD + ∠CDB = 180°

⇒ 30° + ∠CAD + 120° = 180°

⇒ ∠CAD = (180° - 150°) = 30°

#### Page No 493:

#### Answer:

(a) 50°

Take a point E on the remaining part of the circumference.

Join AE and CE.

Then $\angle \mathrm{AEC}=\frac{1}{2}\angle \mathrm{AOC}=\left(\frac{1}{2}\times 100\xb0\right)=50\xb0$

Now, side AB of the cyclic quadrilateral ABCE has been produced to D.

∴ Exterior ∠CBD = ∠AEC = 50°

#### Page No 493:

#### Answer:

(c) 100°

OA = OB (Radii of a circle)

⇒ ∠OBA = ∠OAB = 50°

In Δ OAB, we have:

∠ OAB + ∠OBA + ∠AOB = 180° (Angle sum property of a triangle)

⇒ 50° + 50° + ∠AOB = 180°

⇒ ∠AOB = (180° - 100°) = 80°

Since ∠AOB + ∠BOD = 180° (Linear pair)

∴ ∠BOD = (180° - 80°) = 100°

#### Page No 493:

#### Answer:

(b) 70°

BC = CD (given)

⇒ ∠BDC = ∠CBD = 35°

In Δ BCD, we have:

∠BCD + BDC + ∠CBD = 180° (Angle sum property of a triangle)

⇒ ∠BCD + 35° + 35° = 180°

⇒ ∠BCD = (180° - 70°) = 110°

In cyclic quadrilateral ABCD, we have:

∠BAD + ∠BCD = 180°

⇒ ∠BAD + 110° = 180°

∴ ∠BAD = (180° - 110°) = 70°

#### Page No 493:

#### Answer:

(c) 120°

Since ΔABC is an equilateral triangle, each of its angle is 60°.

∴ ∠BAC = 60°

In a cyclic quadrilateral ABCD, we have:

∠BAC + ∠BDC = 180°

⇒ 60° + ∠BDC = 180°

⇒ ∠BDC = (180° - 60°) = 120°

#### Page No 493:

#### Answer:

(b) 80°

In a cyclic quadrilateral ABCD, we have:

Interior opposite angle, ∠ADC = exterior ∠CBE = 100°

∴ ∠CDF = (180° - ∠ADC) = (180° - 100°) = 80° (Linear pair)

#### Page No 493:

#### Answer:

(c) 110°

Join AB.

Then chord AB subtends ∠AOB at the centre and ∠ADB at a point D of the remaining parts of a circle.

∴∠AOB = 2∠ADB

$\Rightarrow \angle \mathrm{ADB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 140\xb0\right)=70\xb0$

In the cyclic quadrilateral, we have:

∠ADB + ∠ACB = 180°

⇒ 70° + ∠ACB = 180°

∴∠ACB = (180° - 70°) = 110°

#### Page No 494:

#### Answer:

(c) 115°

Join AB.

Then chord AB subtends ∠AOB at the centre and ∠ADB at a point D of the remaining parts of a circle.

∴∠AOB = 2∠ADB

$\Rightarrow \angle \mathrm{ADB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 130\xb0\right)=65\xb0$

In cyclic quadrilateral, we have:

∠ADB + ∠ACB = 180°

⇒ 65° + ∠ACB = 180°

∴∠ACB = (180° - 65°) = 115°

#### Page No 494:

#### Answer:

(d) 110°

Since ABCD is a cyclic quadrilateral, we have:

∠BAD + ∠BCD = 180°

⇒ ∠BAD + 110° = 180°

⇒∠BAD = (180° - 110°) = 70°

Similarly in ABEF, we have:

∠BAD + ∠BEF = 180°

⇒ 70° + ∠BEF = 180°

⇒ ∠BEF = (180° - 70°) = 110°

#### Page No 494:

#### Answer:

(c) 105°

We have:

∠ABC + ∠ADC = 180°

⇒ ∠ABC + 95° = 180°

⇒∠ABC = (180° - 95°) = 85°

Now, CF || AB and CB is the transversal.

∴ ∠BCF = ∠ABC = 85° (Alternate interior angles)

⇒ ∠BCE = (85° + 20°) = 105°

⇒ ∠DCB = (180° - 105°) = 75°

Now, ∠BAD + ∠BCD = 180°

⇒ ∠BAD + 75° = 180°

⇒ ∠BAD = (180° - 75°) = 105°

#### Page No 494:

#### Answer:

(c) 8.5 cm

Join AC.

Then AE : CE = DE : BE (Intersecting secant theorem)

∴ AE × BE = DE × CE

Let CD = *x* cm

Then AE = (AB + BE) = (11 + 3) cm = 14 cm; BE = 3cm; CE = (*x* + 3.5) cm; DE = 3.5 cm

∴ 14 × 3 = (*x* + 3.5) × 3.5

$\Rightarrow x+3.5=\frac{14\times 3}{3.5}=\frac{42}{3.5}=12$

⇒ *x* = (12 - 3.5) cm = 8.5 cm

Hence, CD = 8.5 cm

#### Page No 494:

#### Answer:

(b) 6 cm

We know that the line joining their centres is the perpendicular bisector of the common chord.

Join AP.

Then AP = 5 cm; AB = 4 cm

Also, AP^{2}^{ }= BP^{2}^{ }+ AB^{2}

Or BP^{2}^{ }= AP^{2} - AB^{2}

Or BP^{2}^{ }= 5^{2} - 4^{2}

Or BP = 3 cm

∴ ΔABP is a right angled and PQ = 2 × BP = (2 × 3) cm = 6 cm

#### Page No 494:

#### Answer:

(c) 60°

We have:

∠AOB = 2∠ACB

$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 90\xb0\right)=45\xb0$

∠COA = 2∠CBA = (2 × 30°) = 60°

∴ ∠COD = 180° - ∠COA = (180° - 60°) = 120°

$\Rightarrow \angle \mathrm{CAO}=\frac{1}{2}\angle \mathrm{COD}=\left(\frac{1}{2}\times 120\xb0\right)=60\xb0$

View NCERT Solutions for all chapters of Class 9