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#### Page No 436:

#### Question 1:

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the circle.

#### 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:

#### Question 2:

Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.

#### 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:

#### Question 3:

A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle.

#### 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:

#### Question 4:

In a circle of radius 5 cm, *AB* and *CD* are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the distance between the chords if they are

(i) on the same side of the centre

(ii) on the opposite sides of the centre.

#### 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:

#### Question 5:

Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. Find the distance between the chords.

#### 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:

#### Question 6:

In the given figure, the diameter *CD* of a circle with centre *O* is perpendicular to chord *AB*. If *AB* = 12 cm and *CE* = 3 cm, calculate the radius of the circle.

#### 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:

#### Question 7:

In the given figure, a circle with centre *O* is given in which a diameter *AB* bisects the chord *CD* at a point *E* such that *CE* = *ED* = 8 cm and *EB* = 4 cm. Find the radius of the circle.

#### 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:

#### Question 8:

In the adjoining figure, *OD* is perpendicular to the chord *AB* of a circle with centre *O*. If *BC* is a diameter, show that *AC* || *CD* and *AC* = 2 × *OD*.

#### 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:

#### Question 9:

In the given figure, *O* is the centre of a circle in which chords *AB* and *CD* intersect at *P* such that *PO* bisects ∠*BPD*. Prove that *AB* = *CD*.

#### 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:

#### Question 10:

Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.

#### 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:

#### Question 11:

Prove that two different circles cannot intersect each other at more than two points.

#### 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:

#### Question 12:

Two circles of radii 10 cm and 8 cm intersect each other, and the length of the common chord is 12 cm. Find the distance between their centres.

#### 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:

#### Question 13:

Two equal circles intersect in *P* and *Q*. A straight line through *P* meets the circles in *A* and *B*. Prove that *QA* = *QB*.

#### 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:

#### Question 14:

If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.

#### 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:

#### Question 15:

In the adjoining figure, two circles with centres at *A* and *B*, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of *AB* meets the bigger circle in *P* and *Q*, find the length of *PQ*.

#### 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:

#### Question 16:

In the given figure, *AB* is a chord of a circle with centre *O* and *AB* is produced to *C* such that *BC* = *OB*. Also, *CO* is joined and produced to meet the circle in *D*. If ∠*ACD* = *y*° and ∠*AOD* = *x*°, prove that *x* = 3*y*.

#### 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:

#### Question 17:

*AB *and *AC *are two chords of a circle of radius *r* such that *AB *= 2*AC*. If *p* and *q* are the distances of *AB *and *AC *from the centre then prove that 4*q*^{2} = *p*^{2} + 3*r*^{2}.

#### 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:

#### Question 18:

In the adjoining figure, *O* is the centre of a circle. If *AB* and *AC* are chords of the circle such that *AB* = *AC*, *OP* ⊥ *AB* and *OQ* ⊥ *AC*, prove that *PB* = *QC*.

#### 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:

#### Question 19:

In the adjoining figure, *BC* is a diameter of a circle with centre *O*. If *AB* and *CD* are two chords such that *AB* || *CD*, prove that *AB* = *CD*.

#### 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:

#### Question 20:

An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

#### 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:

#### Question 21:

In the adjoining figure, *AB* and *AC* are two equal chords of a circle with centre *O*. Show that *O* lies on the bisector of ∠*BAC*.

#### 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:

#### Question 22:

In the adjoining figure, *OPQR* is a square. A circle drawn with centre *O* cuts the square at *X* and *Y*. Prove that *QX* = *QY*.

#### 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:

#### Question 23:

Two circles with centres *O* and *O*' intersect at two points *A* and *B*. A line *PQ *is drawn parallel to *OO*' through *A* or *B*, intersecting the circles at *P* and *Q*. Prove that *PQ *= 2*OO*'.

#### 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:

#### Question 1:

(i) In Figure (1), *O* is the centre of the circle. If ∠*OAB* = 40° and ∠*OCB* = 30°, find ∠*AOC*.

(ii) In Figure (2), *A*, *B* and *C* are three points on the circle with centre *O* such that ∠*AOB* = 90° and ∠*AOC* = 110°. Find ∠*BAC*.

#### 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:

#### Question 2:

In the given figure, *O* is the canter of the circle and ∠*AOB* = 70°.

Calculate the values of (i) ∠*OCA*, (ii) ∠*OAC*.

#### 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:

#### Question 3:

In the given figure, *O* is the centre of the circle. if ∠*PBC* = 25° and ∠*APB* = 110°, find the value of ∠*ADB*.

#### 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:

#### Question 4:

In the given figure, *O* is the centre of the circle. If ∠*ABD* = 35° and ∠*BAC* = 70°, find ∠*ACB*.

#### 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:

#### Question 5:

In the given figure, *O* is the centre of the circle. If ∠*ACB* = 50°, find ∠*OAB*.

#### 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:

#### Question 6:

In the given figure, ∠*ABD* = 54° and ∠*BCD* = 43°, calculate (i) ∠*ACD* (ii) ∠*BAD* (iii) ∠*B**DA*.

#### 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:

#### Question 7:

In the adjoining figure, *DE* is a chord parallel to diameter *AC* of the circle with centre *O*. If ∠*CBD* = 60°, calculate ∠*CDE*.

#### 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:

#### Question 8:

In the adjoining figure, *O* is the centre of a circle. Chord *CD* is parallel to diameter *AB*. If ∠*ABC* = 25°, calculate ∠*CED*.

#### 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:

#### Question 9:

In the given figure, *AB* and *CD* are straight lines through the centre *O* of a circle. If ∠*AOC* = 80° and ∠*CDE* = 40°, find (i) ∠*DCE*, (ii) ∠*ABC*.

#### 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:

#### Question 10:

In the given figure, *O* is the centre of a circle, ∠*AOB* 40° and ∠*BDC* = 100°, find ∠*OBC*.

#### 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:

#### Question 11:

In the adjoining figure, chords *AC* and *BD* of a circle with centre *O*, intersect at right angles at *E*. If ∠*OAB* = 25°, calculate ∠*EBC*.

#### 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:

#### Question 12:

In the given figure, *O* is the centre of a circle in which ∠*OAB* = 20° and ∠*OCB* = 55°. Find (i) ∠*B**OC*, (ii) ∠*AOC*

#### 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:

#### Question 13:

In the given figure, *O* is the centre of the circle and ∠*BCO *= 30°. Find *x* and *y*.

#### 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:

#### Question 14:

In the given figure, *O* is the centre of the circle, *BD *= *OD *and *CD *⊥ *AB*. Find ∠*CAB*.

#### 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:

#### Question 15:

In the given figure, *PQ* is a diameter of a circle with centre *O*. If ∠*PQR* = 65°, ∠*SPR* = 40° and ∠*PQM* = 50°, find ∠*QPR*, ∠*QPM* and ∠*PRS*.

#### 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:

#### Question 16:

In the figure given below, *P* and *Q* are centres of two circles, intersecting at *B* and *C,* and *ACD *is a straight line.

If ∠*APB* = 150° and ∠*BQD *= *x*°, find the value of* x*.

#### 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:

#### Question 17:

In the given figure, ∠*BAC* = 30°. Show that *BC* is equal to the radius of the circumcircle of ∆*ABC* whose centre is *O*.

#### 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:

#### Question 18:

In the given figure, *AB *and *CD *are two chords of a circle, intersecting each other at a point *E*.

Prove that ∠*AEC *= $\frac{1}{2}$(angle subtended by arc *CXA* at the centre + angle subtended by arc *DYB *at the centre).

#### 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:

#### Question 1:

In the given figure, *ABCD* is a cyclic quadrilateral whose diagonals intersect at *P* such that ∠*DBC* = 60° and ∠*BAC* = 40°. Find (i) ∠*BCD*, (ii) ∠*CAD*.

#### 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:

#### Question 2:

In the given figure, *POQ* is a diameter and *PQRS* is a cyclic quadrilateral. If ∠*PSR* = 150°, find ∠*RPQ*.

#### 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:

#### Question 3:

In the given figure, *O* is the centre of the circle and arc *ABC *subtends an angle of 130° at the centre. If *AB *is extended to *P*, find ∠*PBC*.

#### 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:

#### Question 4:

In the given figure, *ABCD* is a cyclic quadrilateral in which *AE* is drawn parallel to *CD*, and *BA* is produced. If ∠*ABC* = 92° and ∠*FAE* = 20°, find ∠*BCD*.

#### 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:

#### Question 5:

In the given figure, *BD* = *DC* and ∠*CBD* = 30°, find m(∠*BAC*).

#### 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:

#### Question 6:

In the given figure, *O* is the centre of the given circle and measure of arc *ABC* is 100°. Determine ∠*ADC* and ∠*ABC*.

#### 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:

#### Question 7:

In the given figure, ∆*ABC* is equilateral. Find (i) ∠*B**DC*, (ii) ∠*BEC*.

#### 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:

#### Question 8:

In the adjoining figure, *ABCD* is a cyclic quadrilateral in which ∠*BCD* = 100° and ∠*ABD* = 50°. Find ∠*A**DB*.

#### 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:

#### Question 9:

In the given figure, *O* is the centre of a circle and ∠*BOD* = 150°. Find the values of *x* and *y*.

#### 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:

#### Question 10:

In the given figure, *O* is the centre of the circle and ∠*DAB* = 50°. Calculate the values of *x* and *y*.

#### 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:

#### Question 11:

In the given figure, sides *AD* and *AB* of cyclic quadrilateral *ABCD* are produced to *E* and *F* respectively. If ∠*CBF* = 130° and ∠*CDE* = *x*°, find the value of *x*.

#### 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:

#### Question 12:

In the given figure, *AB* is a diameter of a circle with centre *O* and *DO* || *CB*.

If ∠*BCD* = 120°, calculate

(i) ∠*BAD*

(ii) ∠*ABD*

(iii) ∠*CBD*

(iv) ∠*ADC*.

Also, show that ∆*OAD* is an equilateral triangle.

#### 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:

#### Question 13:

Two chords *AB* and *CD* of a circle intersect each other at *P* outside the circle. If *AB* = 6 cm, *BP* = 2 cm and *PD* = 25 cm, find *CD*.

#### 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:

#### Question 14:

In the given figure, *O* is the centre of a circle. If ∠*AOD* = 140° and ∠*CAB* = 50°, calculate

(i) ∠*EDB*,

(ii) ∠*E**BD*.

#### 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:

#### Question 15:

In the given figure ∆*ABC* is an isosceles triangle in which *AB* = *AC* and a circle passing through *B* and* C* intersects* AB* and *AC* at *D* and *E* respectively. Prove that *DE* || *BC*.

#### 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:

#### Question 16:

In the given figure, *AB* and *CD* are two parallel chords of a circle. If *BDE* and *ACE* are straight lines, intersecting at *E*, prove that ∆*AEB* is isosceles.

#### 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:

#### Question 17:

In the given figure, ∠*BAD* = 75°, ∠*D**CF* = *x*° and ∠*D**EF* = *y*°. Find the values of *x* and *y*.

#### 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:

#### Question 18:

In the given figure, *ABCD* is a quadrilateral in which *AD* = *BC* and ∠*ADC* = ∠*BCD*. Show that the points *A*, *B*, *C*, *D* lie on a circle.

#### 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:

#### Question 19:

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

#### 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:

#### Question 20:

Prove that the circles described with the four sides of a rhombus, as diameters, pass through the point of intersection of its diagonals.

#### 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:

#### Question 21:

*ABCD* is a rectangle. Prove that the centre of the circle thought *A*, *B*, *C*, *D* is the point of intersection of its diagonals.

#### 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:

#### Question 22:

Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction.

#### 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:

#### Question 23:

In a cyclic quadrilateral *ABCD*, if (∠*B* − ∠*D*) = 60°, show that the smaller of the two is 60°.

#### 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:

#### Question 24:

The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

#### 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:

#### Question 25:

On a common hypotenuse *AB*, two right triangles *ACB *and *ADB *are situated on opposite sides. Prove that ∠*BAC *= ∠*BDC*.

#### 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:

#### Question 26:

*ABCD *is a quadrilateral such that *A* is the centre of the circle passing through *B, C *and* D.* Prove that ∠*CBD *+ ∠*CDB *= $\frac{1}{2}$∠*BAD*.

#### 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:

#### Question 1:

The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is

(a) 11.5 cm

(b) 12 cm

(c) $\sqrt{69}\mathrm{cm}$

(d) 23 cm

#### 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:

#### Question 2:

A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is

(a) 25 cm

(b) 12.5 cm

(c) 30 cm

(d) 9 cm

#### 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:

#### Question 3:

In the given figure,* BOC* is a diameter of a circle and* AB* = *AC*. Then, ∠*ABC* = ?

(a) 30°

(b) 45°

(c) 60°

(d) 90°

#### 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:

#### Question 4:

In the given figure, *O* is the centre of a circle and ∠*ACB* = 30°. Then, ∠*AOB* = ?

(a) 30°

(b) 15°

(c) 60°

(d) 90°

Figure

#### 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:

#### Question 5:

In the given figure, *O* is the centre of a circle. If ∠*OAB* = 40° and *C* is a point on the circle, then ∠*ACB* = ?

(a) 40°

(b) 50°

(c) 80°

(d) 100°

#### 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:

#### Question 6:

In the given figure, *AOB* is a diameter of a circle with centre *O* such that *AB* = 34 cm and *CD* is a chord of length 30 cm. Then the distance of *CD* from *AB* is

(a) 8 cm

(b) 15 cm

(c) 18 cm

(d) 6 cm

#### 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:

#### Question 7:

*AB* and *CD* are two equal chords of a circle with centre *O* such that ∠*AOB* = 80°, then ∠*COD* = ?

(a) 100°

(b) 80°

(c) 120°

(d) 40°

#### 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:

#### Question 8:

In the given figure, *CD* is the diameter of a circle with centre *O* and *CD* is perpendicular to chord *AB*. If *AB* = 12 cm and *CE* = 3 cm, then radius of the circles is

(a) 6 cm

(b) 9 cm

(c) 7.5 cm

(d) 8 cm

#### 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:

#### Question 9:

In the given figure, *O* is the centre of a circle and diameter *AB* bisects the chord *CD* at a point *E* such that *CE* = *ED* = 8 cm and *EB* = 4 cm. The radius of the circle is

(a) 10 cm

(b) 12 cm

(c) 6 cm

(d) 8 cm

#### 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:

#### Question 10:

In the given figure, *BOC* is a diameter of a circle with centre *O*. If *AB* and *CD* are two chords such that *AB* || *CD*. If *AB* = 10 cm, then *CD* = ?

(a) 5 cm

(b) 12.5 cm

(c) 15 cm

(d) 10 cm

#### 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:

#### Question 11:

In the given figure, *AB* is a chord of a circle with centre *O* and *AB* is produced to *C* such that *BC* = *OB*. Also, *CO* is joined and produced to meet the circle in *D*. If ∠*ACD* = 25°, then ∠*AOD* = ?

(a) 50°

(b) 75°

(c) 90°

(d) 100°

#### 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:

#### Question 12:

In the given figure, *AB* is a chord of a circle with centre *O* and *BOC* is a diameter. If *OD* ⊥ *AB* such that *OD* = 6 cm, then *AC* = ?

(a) 9 cm

(b) 12 cm

(c) 15 cm

(d) 7.5 cm

#### 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:

#### Question 13:

An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is

(a) 3 cm

(b) $3\sqrt{2}\mathrm{cm}$

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

(d) 6 cm

Figure

#### 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:

#### Question 14:

The angle in a semicircle measures

(a) 45°

(b) 60°

(c) 90°

(d) 36°

Figure

#### Answer:

(c) 90°

The angle in a semicircle measures 90°.

#### Page No 490:

#### Question 15:

Angles in the same segment of a circle area are

(a) equal

(b) complementary

(c) supplementary

(d) none of these

Figure

#### Answer:

(a) equal

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

#### Page No 491:

#### Question 16:

In the given figure, ∆*ABC* and ∆*DBC* are inscribed in a circle such that ∠*BAC* = 60° and ∠*DBC* = 50°.

(a) 50°

(b) 60°

(c) 70°

(d) 80°

#### 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:

#### Question 17:

In the given figure, *BOC* is a diameter of a circle with centre *O*. If ∠*BCA* = 30°, then ∠*CDA* = ?

(a) 30°

(b) 45°

(c) 60°

(d) 50°

#### 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:

#### Question 18:

In the given figure, *O* is the centre of a circle. If ∠*OAC* = 50°, then ∠*ODB* = ?

(a) 40°

(b) 50°

(c) 60°

(d) 75°

#### Answer:

(b) 50°

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

#### Page No 491:

#### Question 19:

In the given figure, *O* is the centre of a circle in which ∠*OBA* = 20° and ∠*OCA* = 30°. Then, ∠*BOC* = ?

(a) 50°

(b) 90°

(c) 100°

(d) 130°

#### 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:

#### Question 20:

In the given figure, *O* is the centre of a circle. If ∠*AOB* = 100° and ∠*AOC* = 90°, then ∠*BAC* = ?

(a) 85°

(b) 80°

(c) 95°

(d) 75°

#### 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:

#### Question 21:

In the given figure, *O* is the centre of a circle. Then, ∠*OAB* = ?

(a) 50°

(b) 60°

(c) 55°

(d) 65°

#### 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:

#### Question 22:

In the given figure, *O* is the centre of a circle and ∠*AOC* = 120°. Then, ∠*BDC* = ?

(a) 60°

(b) 45°

(c) 30°

(d) 15°

#### 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:

#### Question 23:

In the given figure, *O* is the centre of a circle and ∠*OAB* = 50°. Then , ∠*CDA* = ?

(a) 40°

(b) 50°

(c) 75°

(d) 25°

#### 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:

#### Question 24:

In the give figure, *AB* and *CD* are two intersecting chords of a circle. If ∠*CAB* = 40° and ∠*BCD* = 80°, then ∠*CBD* = ?

(a) 80°

(b) 60°

(c) 50°

(d) 70°

Figure

#### 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:

#### Question 25:

In the given figure, *O* is the centre of a circle and chords *AC* and *BD* intersect at *E*. If ∠*AEB* = 110° and ∠*CBE* = 30°, then ∠*ADB* = ?

(a) 70°

(b) 60°

(c) 80°

(d) 90°

#### 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:

#### Question 26:

In the given figure, *O* is the centre of a circle in which ∠*OAB* = 20° and ∠*OCB* = 50°. Then, ∠*AOC* = ?

(a) 50°

(b) 70°

(c) 20°

(d) 60°

#### 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:

#### Question 27:

In the given figure, *AOB* is a diameter and *ABCD* is a cyclic quadrilateral. If ∠*ADC* = 120°, then ∠*BAC* = ?

(a) 60°

(b) 30°

(c) 20°

(d) 45°

#### 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:

#### Question 28:

In the given figure *ABCD* is a cyclic quadrilateral in which *AB* || *DC* and ∠*BAD* = 100°. Then, ∠*ABC* = ?

(a) 80°

(b) 100°

(c) 50°

(d) 40°

#### 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:

#### Question 29:

In the given figure, *O* is the centre of a circle and ∠*AOC* = 130°. Then, ∠*ABC* = ?

(a) 50°

(b) 65°

(c) 115°

(d) 130°

#### 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:

#### Question 30:

In the given figure, *AOB* is a diameter of a circle and *CD* || *AB*. If ∠*BAD* = 30°, then ∠*CAD* = ?

(a) 30°

(b) 60°

(c) 45°

(d) 50°

#### 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:

#### Question 31:

In the given figure, *O* is the centre of a circle in which ∠*AOC* = 100°. Side *AB* of quad. *OABC* has been produced to *D*. Then, ∠*CBD* = ?

(a) 50°

(b) 40°

(c) 25°

(d) 80°

#### 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:

#### Question 32:

In the given figure, *O* is the centre of a circle and ∠*OAB* = 50°. Then, ∠*BOD* = ?

(a) 130°

(b) 50°

(c) 100°

(d) 80°

#### 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:

#### Question 33:

In the give figure, *ABCD* is a cyclic quadrilateral in which *BC* = *CD* and ∠*CBD* = 35°. Then, ∠*BAD* = ?

(a) 65°

(b) 70°

(c) 110°

(d) 90°

#### 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:

#### Question 34:

In the given figure, equilateral ∆*ABC* is inscribed in a circle and *ABCD* is a quadrilateral, as shown. Then, ∠*BDC* = ?

(a) 90°

(b) 60°

(c) 120°

(d) 150°

#### 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:

#### Question 35:

In the given figure, sides* AB* and *AD* of quad. *ABCD* are produced to *E* and *F* respectively. If ∠*CBE* = 100°, then ∠*CDE* = ?

(a) 100°

(b) 80°

(c) 130°

(d) 90°

#### 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:

#### Question 36:

In the given figure, *O* is the centre of a circle and ∠*AOB* = 140°. Then, ∠*ACB* = ?

(a) 70°

(b) 80°

(c) 110°

(d) 40°

#### 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:

#### Question 37:

In the given figure, *O* is the centre of a circle and ∠*AOB* = 130°. Then, ∠*A**CB* = ?

(a) 50°

(b) 65°

(c) 115°

(d) 155°

#### 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:

#### Question 38:

In the given figure, *ABCD* and *ABEF* are two cyclic quadrilaterals. If ∠*BCD* = 110°, then ∠*BEF* = ?

(a) 55°

(b) 70°

(c) 90°

(d) 110°

#### 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:

#### Question 39:

In the given figure, *ABCD* is a cyclic quadrilateral in which *DC* is produced to *E* and *CF* is drawn parallel to *AB* such that ∠*ADC* = 95° and ∠*ECF* = 20°. Then, ∠*BAD* = ?

(a) 95°

(b) 85°

(c) 105°

(d) 75°

#### 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:

#### Question 40:

Two chords *AB* and *CD* of a circle intersect each other at a point *E* outside the circle. If *AB* = 11 cm, *BE* = 3 cm and *DE* = 3.5 cm, then *CD* = ?

(a) 10.5 cm

(b) 9.5 cm

(c) 8.5 cm

(d) 7.5 cm

#### 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:

#### Question 41:

In the given figure, *A* and *B* are the centres of two circles having radii 5 cm and 3 cm respectively and intersecting at points *P* and *Q* respectively. If *AB* = 4 cm, then the length of common chord *PQ* is

(a) 3 cm

(b) 6 cm

(c) 7.5 cm

(d) 9 cm

#### 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:

#### Question 42:

In the given figure, ∠*AOB* = 90° and ∠*ABC* = 30°. Then, ∠*CAO* = ?

(a) 30°

(b) 45°

(c) 60°

(d) 90°

#### 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°

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

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