Rd Sharma 2020 2021 Solutions for Class 9 Maths Chapter 15 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 Rd Sharma 2020 2021 Book of Class 9 Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 15.100:

#### Question 1:

In the given figure, Δ*ABC* is an equilateral triangle. Find *m*∠*BEC*.

#### Answer:

It is given that, is an equilateral triangle

We have to find

Since is an equilateral triangle.

So

And

…… (1)

Since, quadrilateral*BACE*is a cyclic qualdrilateral

So , (Sum of opposite angles of cyclic quadrilateral is.)

Hence

#### Page No 15.100:

#### Question 2:

In the given figure, Δ*PQR* is an isosceles triangle with *PQ* = *PR* and *m* ∠*PQR* = 35°. Find *m* ∠*QSR* and *m *∠*QTR*.

#### Answer:

Disclaimer: Figure given in the book was showing *m*∠*PQR* as *m*∠*SQR. *

It is given that Δ*PQR* is an isosceles triangle with *PQ* = *PR* and *m*∠*PQR* = 35°

We have to find the m∠*QSR* and m∠*QTR*

Since Δ*PQR* is an isosceles triangle

So ∠*PQR* = ∠*PRQ* = 35°

Then

Since *PQTR* is a cyclic quadrilateral

So

In cyclic quadrilateral *QSRT* we have

Hence,

and

#### Page No 15.101:

#### Question 3:

In the given figure, *O* is the centre of the circle. If ∠*BOD* = 160°, find the values of* x* and *y*.

#### Answer:

It is given that *O* is centre of the circle and ∠*BOD* = 160°

We have to find the values of *x *and *y*.

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

Therefore,

Since, quadrilateral *ABCD* is a cyclic quadrilateral.

So,

*x* + *y* = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°.)

Hence and

#### Page No 15.101:

#### Question 4:

In the given figure, *ABCD* is a cyclic quadrilateral. If ∠*BCD* = 100° and ∠*ABD* = 70°, find ∠*ADB*.

#### Answer:

It is given that ∠*BCD* = 100° and ∠*ABD* = 70°

We have to find the ∠*ADB*

We have

∠*A* + ∠*C* = 180° (Opposite pair of angle of cyclic quadrilateral)

So,

Now in is and

Therefore,

Hence,

#### Page No 15.101:

#### Question 5:

If *ABCD* is a cyclic quadrilateral in which *AD* || *BC* (In the given figure). Prove that ∠*B* = ∠*C*.

#### Answer:

It is given that, *ABCD* is cyclic quadrilateral in which *AD* || *BC*

We have to prove

Since, *ABCD* is a cyclic quadrilateral

So,

and ..… (1)

and (Sum of pair of consecutive interior angles is 180°) …… (2)

From equation (1) and (2) we have

…… (3)

…… (4)

Hence Proved

#### Page No 15.101:

#### Question 6:

In the given figure, *O *is the centre of the circle. Find ∠*CBD.*

#### Answer:

It is given that,

We have to find

Since, (Given)

So,

(The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.)

Now,

(Opposite pair of angle of cyclic quadrilateral)

So,

…… (1)

(Linear pair)

()

Hence

#### Page No 15.101:

#### Question 7:

In the given figure, *AB* and *CD* are diameters of a circle with centre *O*. If ∠*OBD* = 50°, find ∠*AOC*.

#### Answer:

It is given that, *AB* and *CD* are diameter with center *O* and

We have to find

Construction: Join the point *A* and *D* to form line *AD*

Clearly arc *AD* subtends at *B* and at the centre.

Therefore, $\angle AOD=2\angle ABD=100\xb0$ …… (1)

Since *CD* is a straight line then

$\angle DOA+\angle AOC=180\xb0\left(\mathrm{Linear}\mathrm{pair}\right)$

Hence

#### Page No 15.101:

#### Question 8:

On a semi-circle with *AB* as diameter, a point *C* is taken, so that m (∠*CAB*) = 30°. Find *m* (∠*ACB*) and m (∠*ABC*).

#### Answer:

It is given that, as diameter, is centre and

We have to find and

Since angle in a semi-circle is a right angle therefore

In we have

(Given)

(Angle in semi-circle is right angle)

Now in we have

Hence and

#### Page No 15.101:

#### Question 9:

In a cyclic quadrilateral *ABCD* if *AB* || *CD* and ∠*B** *= 70°, find the remaining angles.

#### Answer:

It is given that, *ABCD* is a cyclic quadrilateral such that *AB* || *CD *and

Sum of pair of opposite angles of cyclic quadrilateral is 180°.

( given)

So,

Also *AB *|| *CD* and *BC* transversal

So,

Now

#### Page No 15.101:

#### Question 10:

In a cyclic quadrilateral *ABCD*, if *m *∠*A* = 3 (*m* ∠*C*). Find *m* ∠*A*.

#### Answer:

It is given that

*ABCD* is cyclic quadrilateral and

We have to find

Since *ABCD* is cyclic quadrilateral and sum of opposite pair of cyclic quadrilateral is 180°.

So

And

Therefore

Hence

#### Page No 15.102:

#### Question 11:

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

#### Answer:

It is given that, *O* is the centre of the circle and $\angle DAB=50\xb0$.

We have to find the values of *x* and *y*.

*ABCD* is a cyclic quadrilateral and

So,

50° + *y* = 180°

*y* = 180° − 50°

*y *= 130°

Clearly is an isosceles triangle with *OA* = *OB* and

Then,

(Since)

So,

*x* + $\angle AOB$ = 180° (Linear pair)

Therefore,* x* = 180° − 80° = 100°

Hence,

and

#### Page No 15.102:

#### Question 12:

In the given figure, if ∠*BAC** *= 60° and ∠*BCA* = 20°, find ∠*ADC*.

#### Answer:

It is given that, and

We have to find the

In given we have

$\angle ABC+\angle BCA+\angle BAC=180\xb0\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ABC=180\xb0-\left(60\xb0+20\xb0\right)=100\xb0\phantom{\rule{0ex}{0ex}}$

In cyclic quadrilateral we have

(Sum of pair of opposite angles of a cyclic quadilateral is 180º)

Then,

Hence

#### Page No 15.102:

#### Question 13:

In the given figure, if *ABC* is an equilateral triangle. Find ∠*BDC* and ∠*BEC*.

#### Answer:

It is given that, *ABC* is an equilateral triangle

We have to find and

Since is an equilateral triangle

So,

And is cyclic quadrilateral

So (Sum of opposite pair of angles of a cyclic quadrilateral is 180°.)

Then,

Similarly *BECD* is also cyclic quadrilateral

So,

Hence, and .

#### Page No 15.102:

#### Question 14:

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

#### Answer:

It is given that, *O* is the centre of the circle and

We have to find the value of *x*, *y* and *z*.

Since, angle in the same segment are equal

So

And *z* = 30°

As angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Since

Then,

*y *= 2*z*

= 2 × 30°

= 60°

*z*+

*x*= 180°

*x*= 180° − 30°

= 150°

Hence,

*x*= 150°,

*y*= 60° and

*z*= 30°

#### Page No 15.102:

#### Question 15:

In the given figure, ∠*BAD* = 78°, ∠*DCF* = *x*° and ∠*DEF* =* y*°. Find the values of *x* and *y*.

#### Answer:

It is given that, and, are cyclic quadrilateral

We have to find the value of* x *and *y*.

Since , is a cyclic quadrilateral

So (Opposite angle of a cyclic quadrilateral are supplementary)

()

..… (1)

*x* = 180° − 102°

= 78°

Now in cyclic quadrilateral *DCFE
x + y =* 180°

*(Opposite angles of a cyclic quadrilateral are supplementary)*

*y =*180° − 78°

*=*102°

Hence,

*x*= 78° and

*y*= 102°

#### Page No 15.102:

#### Question 16:

In a cyclic quadrilateral ABCD, if ∠*A* − ∠*C* = 60°, prove that the smaller of two is 60°

#### Answer:

It is given that ∠*A* – ∠*C* = 60° and *ABCD* is a cyclic quadrilateral.

We have to prove that smaller of two is 60°

Since *ABCD* is a cyclic quadrilateral

So ∠*A* + ∠*C* = 180° (Sum of opposite pair of angles of cyclic quadrilateral is 180°) ..… (1)

And,

∠*A *– ∠*C* = 60° (Given) ..… (2)

Adding equation (1) and (2) we have

So, ∠*C* = 60°

Hence, smaller of two is 60°.

#### Page No 15.103:

#### Question 17:

In the given figure, *ABCD* is a cyclic quadrilateral. Find the value of *x*.

#### Answer:

Here, *ABCD* is a cyclic quadrilateral, we need to find *x*.

In cyclic quadrilateral the sum of opposite angles is equal to 180°.

Therefore,

$\angle ADC+\angle ABC=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 180\xb0-80\xb0+180\xb0-x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=100\xb0$

Hence, the value of *x* is 100°.

#### Page No 15.103:

#### Question 18:

*ABCD *is a cyclic quadrilateral in which:

(i) *BC* || *AD*, ∠*ADC* = 110° and ∠*BAC* = 50°. Find ∠*DAC*.

(ii) ∠*DBC* = 80° and ∠*BAC* = 40°. Find ∠*BCD*.

(iii) ∠*BCD* = 100° and ∠*ABD* = 70° find ∠*ADB*.

#### Answer:

(i) It is given that , and

We have to find

In cyclic quadrilateral *ABCD*

..… (1)

..… (2)

Since,

So,

Therefore in ,

So , ..… (3)

Now, ( and is transversal)

(ii) It is given that , and

We have to find

_{ (Angle in the same segment are equal)}

_{Hence, }

(iii) It is given that, ∠*BCD *= 100° and ∠*ABD* = 70°

As we know that sum of the opposite pair of angles of cyclic quadrilateral is 180°.

$\angle DAB+\angle BCD=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle DAB=180\xb0-100\xb0\phantom{\rule{0ex}{0ex}}=80\xb0$

In Δ*ABD* we have,

$\angle DAB+\angle ABD+\angle BDA=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle BDA=180\xb0-150\xb0=30\xb0$

_{Hence, }

#### Page No 15.103:

#### Question 19:

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

#### Answer:

Here, *ABCD* is a rhombus; we have to prove the four circles described on the four sides of any rhombus *ABCD *pass through the point of intersection of its diagonals *AC* and *BD*.

Let the diagonals* AC* and *BD* intersect at *O*.

We know that the diagonals of a rhombus intersect at right angle.

Therefore,

Now, means that circle described on *AB* as diameter passes through *O.*

Similarly the remaining three circles with* BC*, *CD* and *AD *as their diameter will also pass through *O.*

Hence, all the circles with described on the four sides of any rhombus *ABCD *pass through the point of intersection of its diagonals *AC* and *BD*.

#### Page No 15.103:

#### Question 20:

If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.

#### Answer:

To prove: *AC* = *BD*

Proof: We know that equal chords subtend equal at the centre of circle and the angle subtended by a chord at the centre is twice the angle subtended by it at remaining part of the circle.

$\angle AOD=\angle BOC\left(O\mathrm{is}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{circle}\right)\phantom{\rule{0ex}{0ex}}\angle AOD=2\angle ACD\phantom{\rule{0ex}{0ex}}\mathrm{and}\angle BOC=2\angle BDC\phantom{\rule{0ex}{0ex}}\mathrm{Since},\angle AOD=\angle BOC\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ACD=\angle BDC.....\left(1\right)\phantom{\rule{0ex}{0ex}}\angle ACB=\angle ADB.....\left(2\right)\left(\mathrm{Angle}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{segment}\mathrm{are}\mathrm{equal}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right)\phantom{\rule{0ex}{0ex}}\angle BCD=\angle ADC.....\left(3\right)\phantom{\rule{0ex}{0ex}}\mathrm{In}\u25b3ACD\mathrm{and}\u25b3BDC\phantom{\rule{0ex}{0ex}}CD=CD\left(\mathrm{common}\right)\phantom{\rule{0ex}{0ex}}\angle BCD=\angle ADC\left[\mathrm{Using}\left(3\right)\right]\phantom{\rule{0ex}{0ex}}AD=BC\left(\mathrm{given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\u25b3ACD\cong BDC\left(\mathrm{SAS}\mathrm{congruency}\mathrm{criterion}\right)\phantom{\rule{0ex}{0ex}}\therefore AC=BD\left(\mathrm{cpct}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{Proved}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 15.103:

#### Question 21:

Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).

#### Answer:

$\angle ADB=90\xb0\left(\mathrm{Angle}\mathrm{in}\mathrm{a}\mathrm{semicircle}\right)\phantom{\rule{0ex}{0ex}}\angle ADC=90\xb0\left(\mathrm{Angle}\mathrm{in}\mathrm{a}\mathrm{semicircle}\right)\phantom{\rule{0ex}{0ex}}\mathrm{So},\angle ADB+\angle ADC=90\xb0+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},BDC\mathrm{is}\mathrm{a}\mathrm{line}.\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{two}\mathrm{circles}\mathrm{lie}\mathrm{on}\mathrm{the}\mathrm{third}\mathrm{side}.$

#### Page No 15.103:

#### Question 22:

*ABCD* is a cyclic trapezium with *AD* || *BC*. If ∠*B* = 70°, determine other three angles of the trapezium.

#### Answer:

If in cyclic quadrilateral , then we have to find the other three angles.

Since, *AD *is parallel to *BC,* So,

(Alternate interior angles)

Now, since *ABCD* is cyclic quadrilateral, so

And,

$\mathrm{Hence},\angle A=110\xb0,\angle C=70\xb0\mathrm{and}\angle D=110\xb0$.

#### Page No 15.103:

#### Question 23:

In the given figure, *ABCD* is a cyclic quadrilateral in which *AC* and *BD* are its diagonals. If ∠*DBC* = 55° and ∠*BAC* = 45°, find ∠*BCD**.*

#### Answer:

It is given that is a cyclic quadrilateral with and as its diagonals.

We have to find

Since angles in the same segment of a circle are equal

So

Since (Opposite angle of cyclic quadrilateral)

Hence

#### Page No 15.103:

#### Question 24:

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

#### Answer:

To prove: Perpendicular bisector of side *AB*, *BC*, *CD* and *DA* are concurrent i.e, passes through the same point.

Proof:

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

Therefore, Perpendicular bisectors of chord *AB*, *BC*, *CD* and *DA* pass through the centre which means they all passes through the same point.

Hence, the perpendicular bisector of *AB*, *BC*, *CD* and* DA* are concurrent.

#### Page No 15.103:

#### Question 25:

Prove that the centre of the circle circumscribing the cyclic rectangle *ABCD* is the point of intersection of its diagonals.

#### Answer:

Here, *ABCD* is a cyclic rectangle; we have to prove that the centre of the corresponding circle is the intersection of its diagonals.

Let *O* be the centre of the circle.

We know that the angle formed in the semicircle is 90°.

Since, *ABCD* is a rectangle, So

Therefore, *AC* and *BD* are diameter of the circle.

We also know that the intersection of any two diameter is the centre of the circle.

Hence, the centre of the circle circumscribing the cyclic rectangle *ABCD* is the point of intersection of its diagonals.

#### Page No 15.103:

#### Question 26:

*ABCD* is a cyclic quadrilateral in which *BA* and *CD* when produced meet in *E* and *EA* = *ED*. Prove that:

(i) *AD* || *BC*

(ii) *EB* = *EC*.

#### Answer:

(i) If *ABCD *is a cyclic quadrilateral in which *AB* and* CD* when produced meet in* E *such that *EA* = *ED*, then we have to prove the following,* AD* || *BC*

(ii) *EB *=* EC*

(i) It is given that *EA *=* ED*, so

$\angle EAD=\angle EDA=x$

Since, *ABCD* is cyclic quadrilateral

Now,

Therefore, the adjacent angles andare supplementary

Hence*, AD* ||* BC*

(ii) Since, *AD* and *BC* are parallel to each other, so,

$\angle ECB=\angle EDA\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle EBC=\angle EAD\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\mathrm{But},\angle EDA=\angle EAD\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\angle ECB=\angle EBC\phantom{\rule{0ex}{0ex}}\Rightarrow EC=EB\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\u25b3ECB\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.$

#### Page No 15.104:

#### Question 27:

Prove that the angle in a segment shorter than a semicircle is greater than a right angle.

#### Answer:

$\stackrel{\u23dc}{QP}\mathrm{is}\mathrm{a}\mathrm{major}\mathrm{arc}\mathrm{and}\angle PSQ\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{formed}\mathrm{by}\mathrm{it}\mathrm{in}\mathrm{the}\mathrm{alternate}\mathrm{segment}.\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is}\mathrm{twice}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{it}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{alternate}\mathrm{segment}\mathrm{of}\mathrm{the}\mathrm{circle}.\phantom{\rule{0ex}{0ex}}\therefore 2\angle PSQ=m\left(\stackrel{\u23dc}{QP}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle PSQ=360\xb0-m\left(\stackrel{\u23dc}{PQ}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle PSQ=360\xb0-\angle POQ\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle PSQ=360\xb0-180\xb0\left(\because \angle POQ180\xb0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle PSQ180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle PSQ90\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.

#### Page No 15.104:

#### Question 28:

Prove that the angle in a segment greater than a semi-circle is less than a right angle.

#### Answer:

$\mathrm{To}\mathrm{prove}:\angle ABC\mathrm{is}\mathrm{an}\mathrm{acute}\mathrm{angle}\phantom{\rule{0ex}{0ex}}\mathrm{Proof}:\phantom{\rule{0ex}{0ex}}AD\mathrm{being}\mathrm{the}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{circle}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ACD=90\xb0\left[\mathrm{Angle}\mathrm{in}\mathrm{a}\mathrm{semicircle}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}\right]\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{in}\u25b3ACD,\angle ACD=90\xb0\mathrm{which}\mathrm{means}\mathrm{that}\angle ADC\mathrm{is}\mathrm{an}\mathrm{acute}\mathrm{angle}......\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Again},\angle ABC=\angle ADC\left[\mathrm{Angle}\mathrm{in}\mathrm{a}\mathrm{same}\mathrm{segment}\mathrm{are}\mathrm{always}\mathrm{equal}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ABC\mathrm{is}\mathrm{also}\mathrm{an}\mathrm{acute}\mathrm{angle}.\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}$

#### Page No 15.104:

#### Question 29:

Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half the hypotenuse.

#### Answer:

We have to prove that

Let be a right angle at B and *P* be midpoint of *AC*

Draw a circle with center at *P* and *AC* diameter

Since therefore circle passing through *B*

So

Hence

Proved.

#### Page No 15.107:

#### Question 1:

If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is

(a) 15 cm

(b) 16 cm

(c) 17 cm

(d) 34 cm

#### Answer:

(c) 17 cm

We will represent the given data in the figure

In the diagram *AB* is the given chord of 16 cm length and *OM* is the perpendicular distance from the centre to *AB*.

We know that perpendicular from the centre to any chord divides it into two equal parts.

So, *AM* = *MB* = = 8 cm.

Now consider right triangle *OMA* and by using Pythagoras theorem

Hence, correct answer is option (c).

#### Page No 15.107:

#### Question 2:

The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is

(a) $\sqrt{5}\mathrm{cm}$

(b) $2\sqrt{5}\mathrm{cm}$

(c) $2\sqrt{7}\mathrm{cm}$

(d) $\sqrt{7}\mathrm{cm}$

#### Answer:

(b) $2\sqrt{5}\mathrm{cm}$

We will represent the given data in the figure

We know that perpendicular drawn from the centre to the chord divides the chord into two equal parts.

So , *AM* = *MB* = $\frac{AB}{2}=\frac{8}{2}$ = 4 cm.

Using Pythagoras theorem in the Δ*AMO, *

Hence, the correct answer is option (b).

#### Page No 15.107:

#### Question 3:

If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ∠BAO =

(a) 60°

(b) 45°

(c) 30°

(d) 15°

#### Answer:

We will associate the given information in the following figure.

Since *AO* = *r* (radius of circle)

*AM* = (given)

Extended *OM* to *D* where *MD* =

Consider the triangles *AOM* and triangle *AMD*

So by SSS property

So *AD* = *AO* = *r* and *OD=OM+MD=r*

Hence Δ*AOD* is equilateral triangle

So

We know that in equilateral triangle altitudes divide the vertex angles

Hence option (*c*) is correct.

#### Page No 15.107:

#### Question 4:

*ABCD* is a cyclic quadrilateral such that ∠*ADB* = 30° and ∠*DCA* = 80°, then ∠*DAB* =

(a) 70°

(b) 100°

(c) 125°

(d) 150°

#### Answer:

(a) 70°

It is given that *ABCD* is cyclic quadrilateral ∠*ADB* = 90° and ∠*DCA* = 80°. We have to find ∠*DAB*

We have the following figure regarding the given information

∠*BDA*= ∠

*BCA*= 30° (Angle in the same segment are equal)

Now, since *ABCD* is a cyclic quadrilateral

So, ∠*DAB* + ∠*BCD* = 180°

Hence the correct answer is option (a).

#### Page No 15.107:

#### Question 5:

A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is

(a) 12 cm

(b) 14 cm

(c) 16 cm

(d) 18 cm

#### Answer:

(d) 18 cm

We are given the chord of length 14 cm and perpendicular distance from the centre to the chord is 6 cm. We are asked to find the length of another chord at a distance of 2 cm from the centre.

We have the following figure

We are given *AB* = 14 cm, *OD* = 6 cm, *MO* = 2 cm, *PQ *= ?

Since, perpendicular from centre to the chord divide the chord into two equal parts

Therefore

Now consider the Δ*OPQ* in which *OM* = 2 cm

So using Pythagoras Theorem in Δ*OPM*

Hence, the correct answer is option (d).

#### Page No 15.108:

#### Question 6:

One chord of a circle is known to be 10 cm. The radius of this circle must be

(a) 5 cm

(b) greater than 5 cm

(c) greater than or equal to 5 cm

(d) less than 5 cm

#### Answer:

(b) greater than 5 cm

We are given length of a chord to be 10 cm and we have to give information about the radius of the circle.

Since in any circle, diameter of the circle is greater then any chord.

So diameter > 10

⇒ 2*r* > 10

⇒ *r* > 5 cm

Hence, the correct answer is option (*b*)

#### Page No 15.108:

#### Question 7:

*ABC* is a triangle with *B* as right angle, *AC *= 5 cm and *AB* = 4 cm. A circle is drawn with* A* as centre and *AC *as radius. The length of the chord of this circle passing through *C* and *B* is

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

#### Answer:

(d) 6 cm

We are given a right triangle *ABC* such that, *AC* = 5 cm, *AB* = 4 cm. A circle is drawn with *A* as centre and *AC* as radius. We have to find the length of the chord of this circle passing through *C* and *B*. We have the following figure regarding the given information.

In the circle produce *CB* to *P*. Here *PC* is the required chord.

We know that perpendicular drawn from the centre to the chord divide the chord into two equal parts.

So, *PC* = 2*BC*

Now in Δ*ABC* apply Pythagoras theorem

So, *PC *= 2 ×* BC*

= 2 × 3

= 6 cm

Hence, the correct answer is option (d).

#### Page No 15.108:

#### Question 8:

If *AB*, *BC* and *CD* are equal chords of a circle with *O* as centre and* AD* diameter, than ∠*AOB* =

(a) 60°

(b) 90°

(c) 120°

(d) none of these

#### Answer:

(a) 60°

As we know that equal chords make equal angle at the centre.

Therefore,

$\angle AOB=\angle BOC=\angle COD\phantom{\rule{0ex}{0ex}}\angle AOB+\angle BOC+\angle COD=180\xb0\left[\mathrm{Linear}\mathrm{pair}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 3\angle AOB=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle AOB=60\xb0$

Hence, the correct answer is option (a).

#### Page No 15.108:

#### Question 9:

Let *C* be the mid-point of an arc *AB* of a circle such that m $\stackrel{\u23dc}{AB}$ = 183°. If the region bounded by the arc *ACB* and the line segment *AB* is denoted by *S*, then the centre *O* of the circle lies

(a) in the interior of *S*

(b) in the exertior of *S*

(c) on the segment *AB*

(d) on AB and bisects *AB*

#### Answer:

(a) in the interior of *S*

Given: *m* $\stackrel{\u23dc}{AB}$ = 183° and *C* is mid-point of arc *ABO* is the centre.

With the given information the corresponding figure will look like the following

So the center of the circle lies inside the shaded region *S*.

Hence, the correct answer is option (a).

#### Page No 15.108:

#### Question 10:

In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are

(a) 90° and 270°

(b) 90° and 90°

(c) 270° and 90°

(d) 60° and 210°

#### Answer:

(c) 270° and 90°

We are given the major arc is 3 times the minor arc. We are asked to find the corresponding central angle.

See the corresponding figure.

We know that angle formed by the circumference at the centre is 360°.

Since the circumference of the circle is divided into two parts such that the angle formed by major and minor arcs at the centre are 3*x* and *x *respectively.

So = 90° and = 3*x* = 270°

Hence, the correct answer is option (c).

#### Page No 15.108:

#### Question 11:

If *A* and *B* are two points on a circle such that *m* $\stackrel{\u23dc}{\left(AB\right)}$ = 260°. A possible value for the angle subtended by arc *BA* at a point on the circle is

(a) 100°

(b) 75°

(c) 50°

(d) 25°

#### Answer:

(c) 50°

We are given

Suppose point *P* is on the circle.

Since

So, * * = 360° − 260° = 100°

We know that angle subtended by chord *AB* at the centre is twice that of subtended at the point *P*

So, = = 50°

Hence, the correct answer is option (*c*).

#### Page No 15.108:

#### Question 12:

An equilateral triangle *ABC *is inscribed in a circle with centre *O*. The measures of ∠*BOC* is

(a) 30°

(b) 60°

(c) 90°

(d) 120°

#### Answer:

(d) 120°

We are given that an equilateral Δ*ABC* is inscribed in a circle with centre *O*. We need to find ∠*BOC*

We have the following corresponding figure:

We are given *AB = BC = AC*

Since the sides *AB*, *BC*, and *AC* are these equal chords of the circle.

So, the angle subtended by these chords at the centre will be equal.

Hence

Hence, the correct answer is option (d).

#### Page No 15.108:

#### Question 13:

If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a

(a) rhombus

(b) rectangle

(c) parallelogram

(d) square

#### Answer:

(d) square

The given information in the form of the following figure is as follows:

Since, four sides of the quadrilateral *ACBD* are four chords which subtend equal angles at the centre. Therefore,

(Since *AB* and *CD* are perpendicular diameters)

So sides* AC,* *BC*, *BD* and *AD *are equal, as equal chords subtend equal angle at the centre.

So , *AC = CB = BD = DA* …… (1)

We know that diameters subtend an angle of measure 90° on the circle.

So, …… (2)

From (1) and (2) we can say that is a square.

Hence, the correct answer is option (d).

#### Page No 15.108:

#### Question 14:

If *ABC* is an arc of a circle and ∠*ABC** *= 135°, then the ratio of arc $\stackrel{\u23dc}{ABC}$ to the circumference is

(a) 1 : 4

(b) 3 : 4

(c) 3 : 8

(d) 1 : 2

#### Answer:

(c) 3 : 8

The length of an arc subtending an angle ‘’ in a circle of radius ‘*r*’ is given by the formula,

Length of the arc =

Here, it is given that the arc subtends an angle of with its centre. So the length of the given arc in a circle with radius ‘*r*’ is given as

Length of the arc =

The circumference of the same circle with radius ‘*r*’ is given as .

The ratio between the lengths of the arc and the circumference of the circle will be,

Hence, the correct answer is option (c).

#### Page No 15.108:

#### Question 15:

The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is

(a) 60°

(b) 75°

(c) 120°

(d) 150°

#### Answer:

(d) 150°

We are given that the chord is equal to its radius.

We have to find the angle subtended by this chord at the minor arc.

We have the corresponding figure as follows:

We are given that

*AO = OB = AB*

So *, $\u25b3$AOB* is an equilateral triangle.

Therefore, we have

∠*AOB* = 60°

Since, the angle subtended by any chord at the centre is twice of the angle subtended at any point on the circle.

Take a point *P* on the minor arc.

Since is a cyclic quadrilateral

So, opposite angles are supplementary. That is

Hence, the correct answer is option (d).

#### Page No 15.108:

#### Question 16:

*PQRS *is a cyclic quadrilateral such that *PR* is a diameter of the circle. If ∠*QPR* = 67° and ∠*SPR* = 72°, then ∠*QRS* =

(a) 41°

(b) 23°

(c) 67°

(d) 18°

#### Answer:

Here we have a cyclic quadrilateral *PQRS* with *PR* being a diameter of the circle. Let the centre of this circle be ‘*O*’.

We are given that and. This is shown in fig (2).

So we see that,

$\angle QPS=\angle QPR+\angle RPS\phantom{\rule{0ex}{0ex}}=67\xb0+72\xb0\phantom{\rule{0ex}{0ex}}=139\xb0$

(a) 41°

In a cyclic quadrilateral it is known that the opposite angles are supplementary.

Hence the correct answer is option (a).

#### Page No 15.108:

#### Question 17:

If *A* , *B*, *C* are three points on a circle with centre *O *such that ∠*AOB* = 90° and ∠*BOC** *= 120°, then ∠*ABC* =

(a) 60°

(b) 75°

(c) 90°

(d) 135°

#### Answer:

(b) 75°

To solve this problem we need to know that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

Here we are given that ‘*A*’, ‘*B*’, ‘*C*’ are three points on a circle with centre ‘*O*’ such that and .

From the figure we see that,

Now, as seen earlier, the angle made by the arc ‘*AC*’ with the centre of the circle will be twice the angle it makes in any point in the remaining part of the circle.

Since the point ‘*C*’ lies on the remaining part of the circle, the angle the arc ‘*AC*’ makes with this point has to be half of the angle ‘*AC*’ makes with the centre.Therefore we have,

Hence the correct answer is option (b).

#### Page No 15.108:

#### Question 18:

The greatest chord of a circle is called its

(a) radius

(b) secant

(c) diameter

(d) none of these

#### Answer:

(c) diameter

The greatest chord in a circle is the diameter of the circle.

Hence the correct answer is option (c).

#### Page No 15.109:

#### Question 19:

Angle formed in minor segment of a circle is

(a) acute

(b) obtuse

(c) right angle

(d) none of these

#### Answer:

(b) obtuse

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment.

The angle formed by the chord in the minor segment will always be obtuse.

Hence the correct answer is option (b).

#### Page No 15.109:

#### Question 20:

Number of circles that can be drawn through three non-collinear points is

(a) 1

(b) 0

(c) 2

(d) 3

#### Answer:

(a) 1

Suppose we are given three non-collinear points as *A, B* and *C*

1. Join *A* and *B.*

2. Join *B* and *C.*

3. Draw perpendicular bisector of *AB* and *BC* which meet at *O* as centre of the circle.

So basically we can only draw one circle passing through three non-collinear points *A, B* and *C.*

Hence, the correct answer is option (a).

#### Page No 15.109:

#### Question 21:

In the given figure, if chords *AB *and *CD* of the circle intersect each other at right angles, then *x* + *y* =

(a) 45°

(b) 60°

(c) 75°

(d) 90°

#### Answer:

(d) 90°

We are given the following figure

*∠**ACD** = **∠**ABD *(Angle in the same segment are equal)

⇒ ∠*ACD* = *y*

Consider the Δ*ACM* in which

Hence, the correct answer is option (d).

#### Page No 15.109:

#### Question 22:

In the given figure, if ∠*ABC* = 45°, then ∠*AOC *=

(a) 45°

(b) 60°

(c) 75°

(d) 90°

#### Answer:

(d) 90°

We have to find ∠*AOC*.

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

Hence, the correct answer is option (d).

#### Page No 15.109:

#### Question 23:

In the given figure, chords *AD* and *BC* intersect each other at right angles at a point P. If ∠*DAB* = 35°, then

∠*ADC* =

(a) 35°

(b) 45°

(c) 55°

(d) 65°

#### Answer:

(c) 55°

(Angle in the same segment are equal.)

Also, since the chords ‘*AD*’ and ‘*BC*’ intersect perpendicularly we have,

Consider the triangle ,

Hence, the correct answer is option (c).

#### Page No 15.109:

#### Question 24:

In the given figure, *O* is the centre of the circle and ∠*BDC* = 42°. The measure of ∠*ACB* is

(a) 42°

(b) 48°

(c) 58°

(d) 52°

#### Answer:

(b) 48°

Construction: Join A and D.

Since *AC* is the diameter. So ∠*ADC* will be 90°.

Therefore,

∠*ACB* = ∠*ADB* = 48° (Angle in the same segment are equal.)

Hence, the correct answer is option (b).

#### Page No 15.109:

#### Question 25:

In a circle with centre O, *AB* and* CD* are two diameters perpendicular to each other. The length of chord* AC* is

(a) 2*AB*

(b) $\sqrt{2}$

(c) $\frac{1}{2}AB$

(d) $\frac{1}{\sqrt{2}}AB$

#### Answer:

(d) $\frac{1}{\sqrt{2}}AB$

We are given a circle with centre at *O* and two perpendicular diameters *AB* and *CD.*

We need to find the length of *AC.*

We have the following corresponding figure:

Since, *AB = CD * (Diameter of the same circle)

Also, ∠*AOC* = 90°

And, *AO* =

Here, *AO = OC* (radius)

In Δ*AOC*

Hence, the correct answer is option (d).

#### Page No 15.109:

#### Question 26:

Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circle is

(a) $\sqrt{r}$

(b) $\sqrt{2}rAB$

(c) $\sqrt{3}r$

(d) $\frac{\sqrt{3}}{2}$

#### Answer:

(c) $\sqrt{3}r$

We are given two circles of equal radius intersect each other such that each passes through the centre of the other.

We need to find the common chord.

We have the corresponding figure as follows:

*AO *= *AO*′ = *r* (radius)

And *OO*′ = *r*

So*, *Δ*OAO*′ is an equilateral triangle.

We know that the attitude of an equilateral triangle with side *r* is given by

That is *AM* =

We know that the line joining centre of the circles divides the common chord into two equal parts.

So we have

Hence, the correct answer is option (c).

#### Page No 15.110:

#### Question 27:

If *AB* is a chord of a circle, *P* and *Q* are the two points on the circle different from *A* and *B*, then

(a) ∠*APB* = ∠*AQB*

(b) ∠*APB* + ∠*AQB* = 180° or ∠*APB* = ∠*AQB*

(c) ∠*APB* + ∠*AQB* = 90°

(d) ∠*APB* + ∠*AQB* = 180°

#### Answer:

(b) ∠*APB* + ∠*AQB* = 180° or ∠*APB* = ∠*AQB*

We are given *AB* is a chord of the circle; *P* and *Q* are two points on the circle different from *A* and *B*.

We have following figure.

Case 1: Consider *P* and *Q* are on the same side of *AB*

We know that angle in the same segment are equal.

Hence, ∠*APB* = ∠*AQB*

Case 2: Now consider *P* and *Q* are on the opposite sides of *AB*

In this case we have the following figure:

Since quadrilateral *APBQ* is a cyclic quadrilateral.

Therefore,

∠*APB* + ∠*AQB* = 180° (Sum of the pair of opposite angles of cyclic quadrilateral is 180°.)

Therefore, ∠*APB* = ∠*AQB* or ∠*APB* + ∠*AQB* = 180°

Hence, the correct answer is option (b).

#### Page No 15.110:

#### Question 28:

*AB* and *CD* are two parallel chords of a circle with centre *O* such that *AB* = 6 cm and *CD* = 12 cm. The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is

(a) 6 cm

(b) $5\sqrt{2}\mathrm{cm}$

(c) 7 cm

(d) $3\sqrt{5}\mathrm{cm}$

#### Answer:

(d) $3\sqrt{5}\mathrm{cm}$

Let distance between the centre and the chord CD be *x* cm and the radius of the circle is* r* cm.

We have to find the radius of the following circle:

In triangle OND,

…… (1)

Now, in triangle AOM,

…… (2)

From (1) and (2), we have,

Hence, the correct answer is option (d).

#### Page No 15.110:

#### Question 29:

In a circle of radius 17 cm, two parallel chords are drawn on opposite side of a diameter. The distance between the chords is 23 cm. If the length of one chord is 16 cm, then the length of the other is

(a) 34 cm

(b) 15 cm

(c) 23 cm

(d) 30 cm

#### Answer:

(d) 30 cm

Given that: Radius of the circle is 17 cm, distance between two parallel chords* AB* and* CD* is 23 cm, where *AB*= 16 cm. We have to find the length of *CD.*

We know that the perpendicular drawn from the centre of the circle to any chord divides it into two equal parts.

So, *AM *=* MB *= 8 cm

Let* OM* = *x* cm

In triangle OMB,

Now, in triangle OND, *ON* = (23 − *x*) cm = (23 − 15) cm = 8 cm

Therefore, the length of the other chord is

Hence, the correct answer is option (d).

#### Page No 15.110:

#### Question 30:

In the given figure, *O* is the centre of the circle such that ∠*AOC* = 130°, then ∠*ABC* =

(a) 130°

(b) 115°

(c) 65°

(d) 165°

#### Answer:

(b) 115°

We have the following information in the following figure. Take a point *P* on the circle in the given figure and join *AP* and *CP.*

Since, the angle subtended by a chord at the centre is twice that of subtended atany part of the circle.

So,

Since is a cyclic quadrilateral and we known that opposite angles are supplementary.

Therefore,

$\angle ABC+\angle APC=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ABC+65\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ABC=180\xb0-65\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ABC=115\xb0$

Hence, the correct answer is optoon (b).

#### Page No 15.110:

#### Question 1:

Fill In The Blanks

*AD* is a diameter, ,of circle and *AB* is chord. If *AD* = 34 cm, *AB* = 30 cm, then *BD* = ____________

#### Answer:

Given:

*AD* is a diameter

*AB* is chord

*AD* = 34 cm

*AB* = 30 cm

Let *O* be the centre of the circle.

*AO* = *OD* = 17 cm ...(1)

Let *OL* is a line perpendicular to *AB*, where *L *is the point on *AB*.

Then, *AL = LB* = 15 cm ...(2) (∵ a perpendicular from the centre of the circle to the chord, bisects the chord)

In ∆*ALO*,

Using pythagoras theorem,

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

⇒ 15^{2} + *LO*^{2} = 17^{2} (From (1) and (2))

⇒ 225 + *LO*^{2} = 289

⇒ *LO*^{2} = 289 − 225

⇒ *LO*^{2} = 64

⇒ *LO* = 8 cm

Now, In ∆*ABD*,

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

Therefore, *LO* = $\frac{1}{2}$*BD*

⇒ *BD* = 2*LO*

⇒ *BD* = 2(8)

⇒ *BD* = 16 cm

Hence, *BD* = __16 cm__.

#### Page No 15.110:

#### Question 2:

Fill In The Blanks

*AD* is a diameter of a circle and *AB* is a chord. If *AD* = 34 cm, *AB* = 30cm, then the distance of *AB* from the centre of the circle is ________.

#### Answer:

Given:

*AD* is a diameter

*AB* is chord

*AD* = 34 cm

*AB* = 30 cm

Let *O* be the centre of the circle.

*AO* = *OD* = 17 cm ...(1)

Let *OL* is a line perpendicular to *AB*, where *L *is the point on *AB*.

Then, *AL = LB* = 15 cm ...(2) (∵ a perpendicular from the centre of the circle to the chord, bisects the chord)

In ∆*ALO*,

Using pythagoras theorem,

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

⇒ 15^{2} + *LO*^{2} = 17^{2} (From (1) and (2))

⇒ 225 + *LO*^{2} = 289

⇒ *LO*^{2} = 289 − 225

⇒ *LO*^{2} = 64

⇒ *LO* = 8 cm

Thus, the distance of the chord from the centre is 8 cm.

Hence, the distance of *AB* from the centre of the circle is __8 cm__.

#### Page No 15.110:

#### Question 3:

Fill In The Blanks

If *AB* = 12 cm, *BC *= 16 cm, and *AB* is perpendicular to *BC*, then the radius of the circle passing through the points *A*,*B* and* C* is _________.

#### Answer:

Given:

*AB* = 12 cm

*BC *= 16 cm

*AB* is perpendicular to *BC*

Since, *AB* is perpendicular to *BC*

Therefore, the circle formed by joining *A*, *B* and* C* is a circle with diameter *AC*.

In ∆*ABC*,

Using pythagoras theorem,

*AB*^{2} +* BC*^{2} = *AC*^{2}

⇒ 12^{2} + 16^{2} = *AC*^{2} (given)

⇒ 144 + 256 = *AC*^{2}

⇒ *AC*^{2}= 400

⇒ *AC* = 20

Thus, the diameter of the circle is 20 cm.

Therefore, the radius of the circle is half of the diameter of the circle.

Radius = $\frac{1}{2}\left(20\right)$ =10 cm

Hence, the radius of the circle passing through the points *A*, *B* and* C* is __10 cm__.

#### Page No 15.110:

#### Question 4:

Fill In The Blanks

*ABCD* is a cyclic quadrilateral such that *AB* is a diameter of the circle circumscibing it and ∠*ADC* = 140^{∘ },then ∠* BAC* = ________.

#### Answer:

Given:

*ABCD* is a cyclic quadrilateral

*AB* is a diameter of the circle circumscribing *ABCD*

∠*ADC* = 140^{∘}

In a cyclic quadrilateral, the sum of opposite angles is 180^{∘}.

Thus, ∠*ADC* + ∠*ABC *= 180°

⇒ 140° + ∠*ABC *= 180°

⇒ ∠*ABC *= 180° − 140°

⇒ ∠*ABC *= 40° ...(1)

Since *AB* is the diameter of the circle

Therefore, ∠*ACB = *90° (angle in the semi circle) ...(2)

In ∆*ABC*,

∠*BAC + *∠*ACB + *∠*ABC *= 180° (angle sum property)

⇒ ∠*BAC +* 90°* + *40°* *= 180° (From (1) and (2))

⇒ ∠*BAC +* 130°* *= 180°

⇒ ∠*BAC =* 180° − 130°

⇒ ∠*BAC =* 50°

Hence, ∠*BAC* = __50°__.

#### Page No 15.110:

#### Question 5:

Fill In The Blanks

Two chords *AB* and *CD* of a circle are each at a distance of 6 cm from the centre. the ratio of their lengths is ________.

#### Answer:

Given:

Two chords *AB* and *CD* of a circle are each at a distance of 6 cm from the centre.

The chords which are equidistant from the centre of the circle are of equal length.

Hence, the ratio of their length is __1 : 1__.

#### Page No 15.111:

#### Question 6:

Fill In The Blanks

If two equals chords *AB* and *AC* of a circle with centre *O* are on the opposite sides of *OA*, then ∠*OAB* = ____________ .

#### Answer:

Given:

*AB =AC*

*AB* and *AC* lies on the opposite sides of *OA*

The chords of equal length, are equidistant from the centre of the circle.

In ∆*OAB* and ∆*OAC*,

*AB = AC* (given)

*OA = OA* (common)

*OB = OC* (radius of the circle)

By SSS property,

∆*OAB* ≅ ∆*OAC*

Therefore, ∠*OAB = *__∠__* OAC* (by C.P.C.T.)

Hence, ∠

*OAB =*

__∠__

__OAC__.#### Page No 15.111:

#### Question 7:

Fill In The Blanks

Two congruent circle with centres O and *O*' intesect at two points *P* and *Q*. then then, ∠*POQ *:∠*PO'Q* = ____________.

#### Answer:

Given:

Two congruent circle with centres *O* and *O*' intersect at two points *P* and *Q.*

In ∆*OPQ* and ∆*O'PQ*,

*OP = O'P* (radius)

*PQ = PQ* (common)

*OQ = O'Q* (radius)

By SSS property,

∆*OPQ* ≅ ∆*O'PQ*

Therefore, ∠*POQ = *__∠__* PO'Q* (by C.P.C.T.)

Hence, ∠

*POQ*: ∠

*PO'Q*=

__1 : 1__.

#### Page No 15.111:

#### Question 8:

Fill In The Blanks

If *AOB* is a diameter of a circle and *C* is a point on the circle, then the *AC*^{2} + *BC*^{2 }= ____________.

#### Answer:

Given:

*AOB* is a diameter of a circle

*C* is a point on the circle

Since, *AOB* is the diameter of the circle

Therefore, ∠*ACB = *90° (angle in the semi circle) ...(2)

In right angled ∆*ABC*,

Using pythagoras theorem.

*AC*^{2} + *BC*^{2} = *AB*^{2}

Hence, *AC*^{2} + *BC*^{2 }= __ AB^{2}__.

#### Page No 15.111:

#### Question 9:

Fill In The Blanks

If *O* is the circumcentre of Δ*ABC* and *D* is the mid-point of the base *BC*, then ∠*BOD* = _______________.

#### Answer:

Given:

*O* is the circumcentre of ∆*ABC*

*D* is the mid-point of the base* BC*

In ∆*BOD* and ∆*COD*,

*OB = OC* (radius)

*BD = CD* (*D* is the mid-point of the base* BC*)

*OD = OD* (common)

By SSS property,

∆*BOD* ≅ ∆*COD*

Therefore, ∠*BOD = *∠*COD* (by C.P.C.T.)

*⇒ *∠*BOC = *∠*BOD + *∠*COD
⇒ *∠

*BOC =*2∠

*BOD*..(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠

*BOC =*2∠

*BAC*

⇒2∠

⇒

*BOD =*2∠

*BAC*

⇒∠

⇒

*BOD =*∠

*BAC*

Hence, ∠

*BOD =*

__∠__.

*BAC*#### Page No 15.111:

#### Question 10:

Fill In The Blanks

If *O* is the circumcentre of Δ*ABC*, then ∠*OBC* + ∠*BAC* = __________.

#### Answer:

Given:

*O* is the circumcentre of ∆*ABC*

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*BOC = *2∠*BAC* ...(1)

In ∆*OBC* ,

*OB = OC* (radius)

Thus*, *∠*OBC = *∠*OCB* ...(2)

Also, ∠*OBC + *∠*OCB + *∠*BOC =* 180° (angle sum property)

*⇒ *2∠*OBC + *2∠*BAC =* 180° (from (1) and (2))

*⇒ *∠*OBC + *∠*BAC =* 90°

Hence, ∠*OBC + *∠*BAC =* __90°__.

#### Page No 15.111:

#### Question 11:

Fill In The Blanks

A chord of a circle is equal to its radius. The angle subtended by this chord at a point in major segment is ___________.

#### Answer:

Given:

A chord of a circle is equal to its radius

Let *AB *is a chord and *O* is the centre of the circle.

*AB = OA = OB* (∵ Chord is equal to the radius)

⇒ ∆*ABO* is equilateral triangle

Thus, ∠*AOB = *60° ...(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*AOB = *2∠*ADB , *where* D *is any point on the major segment of the circle

*⇒ *2∠*ADB =* 60° (from (1) and (2))

*⇒ *∠*ADB =* 30°

Hence, the angle subtended by the chord at a point in major segment is __30°__.

#### Page No 15.111:

#### Question 12:

Fill In The Blanks

If a pair of opposite sides of a quadrilateral are equal, then its diagonals are ___________..

#### Answer:

Given:

A pair of opposite sides of a quadrilateral *ABCD* are equal.

i.e., *AB = CD* and *AD = BD*

Then, the quadrilateral can be a parallelogram, a rectangle, rhombus or a square.

In all the cases the diagonals bisects each other.

Hence, its diagonals are __bisecting__.

#### Page No 15.111:

#### Question 13:

Fill In The Blanks

If arcs *AXB* and *CYD *of a circle are congruent, then *AB* : *CD* = ___________.

#### Answer:

Given:

Arcs *AXB* and *CYD *of a circle are congruent

We know, if any two arcs are congruent, then their corresponding chords are equal.

Thus, Chord *AB* = Chord *CD*

Hence, *AB* : *CD* = __1 : 1__.

#### Page No 15.111:

#### Question 14:

*A*,* B* and *C* are three points on a circle, then the perpendicular bisector of *AB*, *BC* and *CA* are ____________.

#### Answer:

Given:

*A*,* B* and *C* are three points on a circle

Let *ABC* be a triangle.

We know, a circumcentre is the point of intersection of the perpendicular bisectors of the triangle.

Thus, the perpendicular bisector of *AB*, *BC* and *CA* intersect at a point known as circumcentre.

Hence, the perpendicular bisector of *AB*, *BC* and *CA* are __concurrent__.

#### Page No 15.111:

#### Question 15:

Fill In The Blanks

If *AB* and *AC* are equal chords of a circle, then the biesector of ∠*BAC* passes through the ___________.

#### Answer:

Given:

*AB* and *AC* are equal chords of a circle

Let *O* be the centre of the circle.

In ∆*OAB* and ∆*OAC*,

*AB = AC* (given)

*OA = OA* (common)

*OB = OC* (radius of the circle)

By SSS property,

∆*OAB* ≅ ∆*OAC*

Therefore, ∠*OAB = *__∠__* OAC* (by C.P.C.T.)

Thus, ∠

*BAC =*2∠

*OAB.*

Hence, the bisector of ∠

*BAC*passes through the

__centre__.

#### Page No 15.111:

#### Question 16:

Fill In The Blanks

*ABCD* is such a quadrilateral that *A* is the centre of the circle passing through *B*, *C* and *D*. If ∠*CBD* + ∠*CDB* = *k* ∠*BAD*, then *k* = _______.

#### Answer:

Given:

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

∠*CBD + *∠*CDB = k*∠*BAD*

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*CAD = *2∠*CBD* ...(1)

Also, ∠*CAB = *2∠*CDB* ...(2)

Adding (1) and (2), we get

∠*CAD + *∠*CAB **= *2∠*CBD + *2∠*CDB
⇒ *∠

*BAD =*2∠

*CBD +*2∠

*CDB*

⇒∠

⇒

*BAD =*2(∠

*CBD +*∠

*CDB*)

*⇒*∠

*CBD +*∠

*CDB*= $\frac{1}{2}$∠

*BAD*

Hence,

*k*= $\overline{)\frac{1}{2}}$.

#### Page No 15.111:

#### Question 17:

Fill In The Blanks

Two chords *AB* and* AC* of a circle are on the opposite sides of the centre. If *AB* and *AC* subtend angles equal to 90^{∘} and 150^{∘} respectively at the centre, then ∠*BAC* = _________.

#### Answer:

Given:

*AB* and *AC* subtend angles equal to 90° and 150° respectively at the centre

i.e., ∠*AOB *= 90° and ∠*AOC *= 150° ...(1)

In ∆*AOB*,

*OA = OB* (radius)

∴ ∠*OBA = *∠*OAB *(angles opposite to equal sides are equal) ...(2)

Now, ∠*OBA + *∠*OAB + *∠*AOB = *180° (angle sum property)

*⇒* 2∠*OAB + * 90° = 180° (From (1) and (2))

*⇒* 2∠*OAB *= 180° − 90°

*⇒* 2∠*OAB *= 90°

*⇒* ∠*OAB *= 45° ...(3)

In ∆*AOC*,

*OA = OC* (radius)

∴ ∠*OCA = *∠*OAC *(angles opposite to equal sides are equal) ...(4)

Now, ∠*OCA + *∠*OAC + *∠*AOC = *180° (angle sum property)

*⇒* 2∠*OAC + * 150° = 180° (From (1) and (4))

*⇒* 2∠*OAC *= 180° − 150°

*⇒* 2∠*OAC *= 30°

*⇒* ∠*OAC *= 15° ...(5)

Thus,

∠*BAC = *∠*OAC **+ *∠*OAB
⇒ *∠

*BAC =*15° + 45° (From (3) and (5))

*⇒*∠

*BAC =*60°

Hence, ∠

*BAC*=

__60°__.

#### Page No 15.111:

#### Question 18:

Fill In The Blanks

Two congruent circles have centres at O and *O'*. Arc *AXb* of circle centred at *O*, subtends an angle of 75^{∘} at the centre *O *and arc* PYQ* ( or circle centred at *O'*) subtends an angle 25^{∘} at the centre *O'*. The ratio of the arcs *AXB* and *PYQ* is ___________.

#### Answer:

Given:

*O *and* O' *are the centres of two congruent circles

*AXB* of circle centred at *O*, subtends an angle of 75^{∘} at the centre *O *

arc* PYQ* of circle centred at *O'* ,subtends an angle 25^{∘}at the centre *O'*

Since, the circles are congruent

Therefore, they have same radius of measure *r* cm. ...(1)

We know, Length of arc = $\frac{\theta}{360\xb0}\times 2\mathrm{\pi}r$

Thus,

$\mathrm{Length}\mathrm{of}\mathrm{arc}AXB=\frac{75\xb0}{360\xb0}\times 2\mathrm{\pi}r...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Length}\mathrm{of}\mathrm{arc}PYQ=\frac{25\xb0}{360\xb0}\times 2\mathrm{\pi}r...\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{Length}\mathrm{of}\mathrm{arc}AXB}{\mathrm{Length}\mathrm{of}\mathrm{arc}PYQ}=\frac{{\displaystyle \frac{75\xb0}{360\xb0}}\times 2\mathrm{\pi}r}{{\displaystyle \frac{25\xb0}{360\xb0}}\times 2\mathrm{\pi}r}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle 75\xb0}}{{\displaystyle 25\xb0}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle 3}}{{\displaystyle 1}}$

Hence, the ratio of the arcs *AXB* and *PYQ* is __3 : 1__.

#### Page No 15.111:

#### Question 19:

In the given figure, *AB* and *CD* are two equal chords of a circle with centre *O*. *OP* and *OQ* are perpendicular on chords *AB* and *CD*, respectively. If ∠*POQ* = 150^{∘}, then ∠*APQ* = __________.

#### Answer:

Given:

*AB* = *CD*

*OP *⊥* AB* and *OQ* ⊥ *CD*

∠*POQ *= 150° ...(1)

In ∆*POQ*,

*OP = OQ* (equal chords are equidistant from the centre)

∴ ∠*OPQ = *∠*OQP *(angles opposite to equal sides are equal) ...(2)

Now, ∠*OPQ + *∠*OQP + *∠*POQ = *180° (angle sum property)

*⇒* 2∠*OPQ + * 150° = 180° (From (1) and (2))

*⇒* 2∠*OPQ *= 180° − 150°

*⇒* 2∠*OPQ *= 30°

*⇒* ∠*OPQ *= 15° ...(3)

Since, *OP *⊥* AB*

Thus, ∠*OPA *= 90° ....(4)

Now, ∠*OPA = *∠*OPQ + *∠*APQ *

*⇒* 90° = 15° *+ *∠*APQ* (From (3) and (4))

*⇒* ∠*APQ *= 90° − 15°

*⇒* ∠*APQ *= 75°

Hence, ∠*APQ* = __75°__.

#### Page No 15.111:

#### Question 20:

In the given figure, if *OA* = 5cm, *AB *= 8 cm and *OD* is perpendicular to *AB*, then *CD* is equal to _______.

#### Answer:

Given:

*OA* = 5cm ...(1)

*AB *= 8 cm

*OD* is perpendicular to *AB*

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

Therefore, *AC = CB* = $\frac{1}{2}AB$

⇒ *AC* = 4 cm ...(2)

In right angled ∆*OAC*,

Using pythagoras theorem

*OA*^{2}* = AC*^{2}* + OC*^{2}

⇒ 5^{2} = 4^{2 }+ *OC*^{2} (From (1) and (2))

⇒ 25 = 16^{ }+ *OC*^{2}

⇒ *OC*^{2} = 25 − 16

⇒ *OC*^{2} = 9

⇒ *OC* = 3 cm

*OD *= 5 cm (radius)

∴ *CD = OD − OC*

⇒ *CD =* 5 *−* 3

⇒ *CD =* 2 cm

Hence, *CD* is equal to __2 cm__.

#### Page No 15.112:

#### Question 21:

In the given figure, if ∠ABC = 20∘ , then ∠AOC is equal to ____________.

#### Answer:

Given:

∠*ABC* = 20° ...(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*AOC = *2∠*ABC
⇒ *∠

*AOC =*2(20°) (From (1))

*⇒*∠

*AOC =*40°

Hence, ∠

*AOC*is equal to

__40°__.

#### Page No 15.112:

#### Question 22:

In the given figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to ____________.

#### Answer:

Given:

*AOB* is a diameter of the circle

*AC = BC*

We know, the diameter subtends a right angle to any point on the circle.

∴ ∠*ACB = *90° ...(1)

In ∆*ACB*,

*AC = BC* (given)

∴ ∠*CAB = *∠*CBA *(angles opposite to equal sides are equal) ...(2)

Now,

∠*CAB + *∠*CBA + *∠*ACB = *180° (angle sum property)

⇒ 2∠*CAB + *90° *= *180° (From (1) and (2))

⇒ 2∠*CAB **= *180° − * *90°

⇒ 2∠*CAB **= *90°

⇒ ∠*CAB **= *45°

Hence, ∠*CAB* is equal to __45°__.

#### Page No 15.112:

#### Question 23:

In the given figure,∠AOB = 90° and ∠ABC = 30° , then ∠CAO is equal to ___________.

#### Answer:

Given:

∠*AOB* = 90° ...(1)

∠*ABC* = 30° ...(2)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*AOB = *2∠*ACB*

⇒ 90° =* *2∠*ACB*

⇒ ∠*ACB *= 45° ...(3)

In ∆*ACB*,

∠*CAB + *∠*CBA + *∠*ACB = *180° (angle sum property)

⇒ ∠*CAB + *30° + 45° *= *180° (From (2) and (3))

⇒ ∠*CAB + *75°*= *180°

⇒ ∠*CAB **= *180° − * *75°

⇒ ∠*CAB **= *105° ...(4)

Also, in ∆*OAB*,

*OA = OB*

∴ ∠*OAB = *∠*OBA *(angles opposite to equal sides are equal) ...(5)

Now,

∠*OAB + *∠*OBA + *∠*AOB = *180° (angle sum property)

⇒ 2∠*OAB + *90° *= *180° (From (1) and (5))

⇒ 2∠*OAB **= *180° − * *90°

⇒ 2∠*OAB **= *90°

⇒ ∠*OAB **= *45° ...(6)

∠*CAO = *∠*CAB *−* *∠*OAB
= *105° − 45° (From (4) and (6))

*=*60°

Hence, ∠

*CAO*is equal to

__60°__.

#### Page No 15.112:

#### Question 24:

In the given figure, if ∠OAB = 40∘ , then ∠ACB = ____________

#### Answer:

Given:

∠*OAB* = 40° ...(1)

In ∆*OAB*,

*OA = OB*

∴ ∠*OAB = *∠*OBA = *40° (angles opposite to equal sides are equal) ...(2)

Now,

∠*OAB + *∠*OBA + *∠*AOB = *180° (angle sum property)

⇒ 40° + 40°* + *∠*AOB* *= *180° (From (1) and (2))

⇒ ∠*AOB = *180° − 80°

⇒ ∠*AOB = *100° ...(3)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*AOB = *2∠*ACB*

⇒ 100° =* *2∠*ACB* (From (3))

⇒ ∠*ACB *= 50°

Hence, ∠*ACB* = __50°__.

#### Page No 15.112:

#### Question 25:

In the given figure,, if ∠DAB = 60∘ , ∠ABD = 50∘ , then ∠ACB = _________

#### Answer:

Given:

∠*DAB* = 60° ...(1)

∠*ABD* = 50° ...(2)

In ∆*ADB*,

∠*DAB + *∠*DBA + *∠*ADB = *180° (angle sum property)

⇒ 60° + 50°* + *∠*ADB* *= *180° (From (1) and (2))

⇒ ∠*ADB = *180° − 110°

⇒ ∠*ADB = *70° ...(3)

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

Thus, ∠*ADB = *∠*ACB*

⇒ 70° =* *∠*ACB* (From (3))

⇒ ∠*ACB *= 70°

Hence, ∠*ACB* = __70°__.

#### Page No 15.112:

#### Question 26:

In the given figure,, BC is a diameter of circle and ∠BAO = 60∘ . Then, ∠ADC = __________.

#### Answer:

Given:

*BC* is a diameter of circle

∠*BAO* = 60° ...(1)

In ∆*OAB*,

*OA = OB*

∴ ∠*OAB = *∠*OBA = *60° (angles opposite to equal sides are equal) ...(2)

Also,

∠*AOC = *∠*OAB + *∠*OBA* (exterior angle)

⇒ ∠*AOC = *60° + 60° (From (2))

⇒ ∠*AOC = *120° ...(3)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*AOC = *2∠*ADC*

⇒ 120° =* *2∠*ADC* (From (3))

⇒ ∠*ADC *= 60°

Hence, ∠*ADC* = __60°__.

#### Page No 15.113:

#### Question 27:

In the given figure, if AOB is a diameter and ∠ADC = 120° , then ∠CAB = ___________

#### Answer:

Given:

*AOB* is a diameter of circle

∠*ADC *= 120°

Quadrilateral *ADCB* is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is 180^{∘}.

Thus, ∠*ADC* + ∠*CBA *= 180°

⇒ 120° + ∠*CBA *= 180°

⇒ ∠*CBA *= 180° − 120°

⇒ ∠*CBA *= 60° ...(1)

We know, the diameter subtends a right angle to any point on the circle.

∴ ∠*ACB = *90° ...(2)

In ∆*ACB*,

∠*CAB + *∠*CBA + *∠*ACB = *180° (angle sum property)

⇒ ∠*CAB + *60° + 90° *= *180° (From (1) and (2))

⇒ ∠*CAB = *180° − 150°

⇒ ∠*CAB = *30° ...(3)

Hence, ∠*CAB* = __30°__.

#### Page No 15.113:

#### Question 28:

In the given figure, if *AOC* is a diameter of the circle and *AXB* = $\frac{1}{2}$ are *BYC*, then ∠*BOC* = __________.

#### Answer:

Given:

*AOC* is a diameter of circle

arc* AXB* = $\frac{1}{2}$ arc *BYC
⇒ * ∠

*BOA*= $\frac{1}{2}$∠

*BOC*..(1)

Now, ∠

*BOA*+ ∠

*BOC*= 180° (Angles on a straight line)

⇒ $\frac{1}{2}$∠

*BOC*+ ∠

*BOC*= 180° (From (1))

⇒ $\frac{3}{2}$∠

*BOC*= 180°

⇒ ∠

*BOC*= $\frac{2}{3}\times $180°

⇒ ∠

*BOC*= 120°

Hence, ∠

*BOC*=

__120°__.

#### Page No 15.113:

#### Question 29:

In the given figure, ∠*ABC* = 45^{∘} , then ∠*AOC* = _________.

#### Answer:

Given:

∠*ABC* = 45^{∘} ..(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*AOC = *2∠*ABC*

⇒ ∠*AOC = *2(45°) (From (1))

⇒ ∠*AOC *= 90°

Hence, ∠*AOC* = __90°__.

#### Page No 15.113:

#### Question 30:

In the given figure, if ∠*ADC* = 130^{∘} and chord *BC* = chord *BE*, then ∠*CBE* = _______.

#### Answer:

Given:

∠*ADC* = 130^{∘} ...(1)

chord *BC* = chord *BE* ...(2)

Quadrilateral *ADCB* is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is 180^{∘}.

Thus, ∠*ADC* + ∠*CBA *= 180°

⇒ 130° + ∠*CBA *= 180°

⇒ ∠*CBA *= 180° − 130°

⇒ ∠*CBA *= 50° ...(3)

In ∆*CBO* and ∆*EBO*,

*BC = BE* (given)

*OB = OB* (common)

*OC = OE* (radius of the circle)

By SSS property,

∆*OCB* ≅ ∆*OEB*

Therefore, ∠*OBC = *∠*OBE* = 50° (by C.P.C.T.) ...(4)

Thus, ∠*CBE = *∠*OBE + *∠*OBC
= *50° + 50° (From (4))

*=*100°

Hence, ∠

*CBE*=

__100°__.

#### Page No 15.113:

#### Question 31:

In the given figure, if ∠*ACB* = 40^{∘} , then ∠*AOB* = ______ and ∠*OAB* = _____________

#### Answer:

Given:

∠*ACB* = 40^{∘} ...(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

Thus, ∠*AOB = *2∠*ACB*

⇒ ∠*AOB = *2(40°) (From (1))

⇒ ∠*AOB *= 80° ...(2)

In ∆*AOB,*

*OA = OB* (radius)

*∴ *∠*OAB = *∠*OBA *(angles opposite to equal sides are equal) ...(3)

∠*OAB + *∠*OBA + *∠*AOB = *180° (angle sum property)

*⇒ *2∠*OAB + *80° *= *180° (From (2) and (3))

*⇒ *2∠*OAB* *= *180° − 80°

*⇒ *2∠*OAB* *= *100°

*⇒ *∠*OAB* *=* 50°

Hence, ∠*AOB* = __80°__ and ∠*OAB* = __50°__.

#### Page No 15.113:

#### Question 32:

In the given figure, *AOB* is a diameter of the circle and *C*, *D* , *E* are any three points to semi circle then ∠*ACD* + ∠*BED* = _______

#### Answer:

Given:

*AOB* is a diameter of the circle

We know, the diameter subtends a right angle to any point on the circle.

∴ ∠*AEB = *90° ...(1)

Quadrilateral *ACDE* is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is 180^{∘}.

Thus, ∠*ACD* + ∠*DEA *= 180° ...(2)

Adding (1) and (2), we get

∠*ACD* + ∠*DEA + *∠*AEB = *90° + 180°

⇒ ∠*ACD* + ∠*DEB** = *270°

Hence, ∠*ACD* + ∠*BED* = __270°__.

#### Page No 15.113:

#### Question 33:

In the given figure, if ∠*OAB* = 30^{∘} and ∠*OCB* = 57^{∘} , then ∠*BOC* = _______ and ∠*AOC* = _________ .

#### Answer:

Given:

∠*OAB* = 30^{∘} ...(1)

∠*OCB* = 57^{∘} ...(2)

In ∆*COB,*

*OC = OB* (radius)

*∴ *∠*OCB = *∠*OBC = *57^{∘ }(angles opposite to equal sides are equal) ...(3)

∠*OCB + *∠*OBC + *∠*COB = *180° (angle sum property)

*⇒ *57^{∘}* + *57^{∘} *+ *∠*COB = *180° (From (3))

*⇒ *∠*COB* *= *180° − 114°

*⇒ *∠*COB* *= *66° ...(4)

In ∆*AOB,*

*OA = OB* (radius)

*∴ *∠*OAB = *∠*OBA = *30^{∘ }(angles opposite to equal sides are equal) ...(5)

∠*OAB + *∠*OBA + *∠*AOB = *180° (angle sum property)

*⇒ *30^{∘}* + *30^{∘} *+ *∠*AOB = *180° (From (5))

*⇒ *∠*AOB* *= *180° − 60°

*⇒ *∠*AOB* *= *120° ...(6)

Now,

∠*AOB* *= *∠*AOC + *∠*COB*

⇒ 120° = ∠*AOC + *66° (From (4) and (6))

⇒ ∠*AOC = *120° −* *66°

⇒ ∠*AOC = *54°

Hence, ∠*BOC* = __66°__ and ∠*AOC* = __54°__.

#### Page No 15.114:

#### Question 1:

In the given figure, two circles intersect at *A* and *B*. The centre of the smaller circle is* **O* and it lies on the circumference of the larger circle. If ∠*APB* = 70°, find ∠*ACB*.

#### Answer:

Consider the smaller circle whose centre is given as ‘*O*’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

Now consider the larger circle and the points ‘*A*’, ‘*C*’, ‘*B*’ and ‘*O*’ along its circumference. ‘*ACBO*’ form a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

Hence ,the measure of is.

#### Page No 15.114:

#### Question 2:

In the given figure, two congruent circles with centres *O* and *O*' intersect at *A* and *B*. If ∠*AOB* = 50°, then find ∠*APB*.

#### Answer:

Since both the circles are congruent, they will have equal radii. Let their radii be ‘*r*’.

So, from the given figure we have,

Now, since all the sides of the quadrilateral *OBO’A* are equal it has to be a rhombus.

One of the properties of a rhombus is that the opposite angles are equal to each other.

So, since it is given that, we can say that the angle opposite it, that is to say that should also have the same value.

Hence we get

Now, consider the first circle with the centre ‘*O*’ alone. ‘*AB*’ forms a chord and it subtends an angle of 50° with its centre, that is.

A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

This means that,

Hence the measure of is

#### Page No 15.114:

#### Question 3:

In the given figure, *ABCD* is a cyclic quadrilateral in which ∠*BAD* = 75°, ∠*ABD* = 58° and ∠*ADC *= 77°, *AC* and *BD* intersect at *P*. Then, find ∠*DPC*.

#### Answer:

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .

Here we have a cyclic quadrilateral *ABCD*. The centre of this circle is given as ‘*O*’.

Since in a cyclic quadrilateral the opposite angles are supplementary, here

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment. The angle that the chord forms with any point on the circumference of a particular segment is always the same.

Here, ‘*CD*’ is a chord and ‘*A*’ and ‘*B*’ are two points along the circumference on the major segment formed by the chord ‘*CD*’.

So,

Now,

In any triangle the sum of the interior angles need to be equal to .

Consider the triangle *Δ**ABP*,

$\angle PAB+\angle ABP+\angle APB=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle APB=180\xb0-30\xb0-58\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle APB=92\xb0$

From the figure, since ‘*AC*’ and ‘*BD*’ intersect at ‘*P*’ we have,

Hence the measure of is.

#### Page No 15.115:

#### Question 4:

In the given figure, if ∠*AOB* = 80° and ∠*ABC* = 30°, then find ∠*CAO.*

#### Answer:

Consider the given circle with the centre ‘*O*’. Let the radius of this circle be ‘*r*’. ‘*AB*’ forms a chord and it subtends an angle of 80° with its centre, that is.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

In any triangle the sum of the interior angles need to be equal to 180°.

Consider the triangle

Since, , we have. So the above equation now changes to

Considering the triangle *Δ**ABC* now,

Hence, the measure of is.

#### Page No 15.115:

#### Question 5:

In the given figure, *A* is the centre of the circle. *ABCD* is a parallelogram and *CDE *is a straight line. Find ∠*BCD* : ∠*ABE*.

#### Answer:

It is given that ‘*ABCD*’ is a parallelogram. But since ‘*A*’ is the centre of the circle, the lengths of ‘*AB*’ and ‘*AD*’ will both be equal to the radius of the circle.

So, we have.

Whenever a parallelogram has two adjacent sides equal then it is a rhombus.

So ‘*ABCD*’ is a rhombus.

Let.

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

By this property we have

In a rhombus the opposite angles are always equal to each other.

So,

Since the sum of all the internal angles in any triangle sums up to in triangle , we have

In the rhombus ‘*ABCD*’ since one pair of opposite angles are ‘’ the other pair of opposite angles have to be

From the figure we see that,

So now we can write the required ratio as,

Hence the ratio between the given two angles is .

#### Page No 15.115:

#### Question 6:

In the given figure, *AB* is a diameter of the circle such that ∠*A* = 35° and ∠*Q* = 25°, find ∠*PBR*.

#### Answer:

Let us first consider the triangle Δ*ABQ*.

It is known that in a triangle the sum of all the interior angles add up to 180°.

So here in our triangle Δ*ABQ* we have,

By a property of the circle we know that an angle formed in a semi-circle will be 90°..

In the given circle since ‘*AB*’ is the diameter of the circle the angle which is formed in a semi-circle will have to be 90°.

So, we have

Now considering the triangle we have,

From the given figure it can be seen that,

Now, we can also say that,

Hence the measure of the angle is 115°.

#### Page No 15.115:

#### Question 7:

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

#### Answer:

Consider the circle with the centre ‘*P*’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

Since ‘*ACD*’ is a straight line, we have

Now let us consider the circle with centre ‘*Q*’. Here let ‘*E*’ be any point on the circumference along the major arc ‘*BD*’. Now ‘*CBED*’ forms a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

So here,

So, now we have

Hence, the measure of is .

#### Page No 15.115:

#### Question 8:

In the given figure, if *O* is the circumcentre of ∠*ABC*, then find the value of ∠*OBC* + ∠*BAC*.

#### Answer:

Since, O is the circumcentre of $\u25b3ABC$, So, O would be centre of the circle passing through points *A*, *B* and *C*.

$\angle ABC$= 90° (Angle in the semicircle is 90°.)

$\Rightarrow \angle OAB+\angle OBC=90\xb0\phantom{\rule{0ex}{0ex}}$ .....(1)

As *OA* = *OB* (Radii of the same circle)

$\therefore \angle OAB=\angle OBA\left(\mathrm{Angle}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}\right)\phantom{\rule{0ex}{0ex}}\mathrm{or},\angle BAC=\angle OBA\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right)\phantom{\rule{0ex}{0ex}}\mathit{\angle}BAC+\angle OBC=90\xb0$

#### Page No 15.116:

#### Question 9:

If the given figure, AOC is a diameter of the circle and arc AXB = $\frac{1}{2}$ arc BYC. Find ∠BOC.

#### Answer:

We need to find

#### Page No 15.116:

#### Question 10:

In the given figure, *ABCD* is a quadrilateral inscribed in a circle with centre *O*. *CD* is produced to *E* such that ∠*AED* = 95° and ∠*OBA* = 30°. Find ∠*OAC*.

#### Answer:

We are given *ABCD* is a quadrilateral with center *O*, ∠*ADE* = 95° and ∠*OBA* = 30°

We need to find ∠*OAC*

We are given the following figure

Since ∠*ADE* = 95°

⇒ ∠*ADC* = 180 ° − 95° = 85°

Since squo;*ABCD* is cyclic quadrilateral

This means

∠*ABC* + ∠*ADC* = 180°

Since *OB* = *OC* (radius)

⇒ ∠*OBC* = ∠*OCB* = 65°

In Δ*OBC*

Since ÐBAC and ÐBOC are formed on the same base which is chord.

So

Consider Δ*BOA* which is isosceles triangle.

∠OAB = 30°

#### Page No 15.28:

#### Question 1:

The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.

#### Answer:

Let *AB* be a chord of a circle with centre *O* and radius 8 cm such that

*AB* = 12 cm

We draw and join *OA*.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

Now in we have

Hence the distance of chord from the centre .

#### Page No 15.28:

#### Question 2:

Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.

#### Answer:

Given that *OA* = 10 cm and *OL* = 5 cm, we have to find the length of chord *AB*.

Let *AB* be a chord of a circle with centre *O* and radius 10 cm such that *AO* = 10 cm

We draw and join *OA*.

Since, the perpendiculars from the centre of a circle to a chord bisect the chord.

Now in we have

Hence the length of chord

#### Page No 15.28:

#### Question 3:

Find the length of a chord which is at a distance of 4 cm from the centre of the circle of radius 6 cm.

#### Answer:

Given that and , find the length of chord *AB*.

Let *AB* be a chord of a circle with centre *O* and radius 6 cm such that

We draw and join *OA*.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

Now in we have

$\Rightarrow AL=\sqrt{20}=4.47$

$AB=2\times AL\phantom{\rule{0ex}{0ex}}=2\times 4.47\phantom{\rule{0ex}{0ex}}=8.94\mathrm{cm}$

Hence the length of the chord is 8.94 cm.

#### Page No 15.28:

#### Question 4:

Give a method to find the centre of a given circle.

#### Answer:

Let *A*, *B* and *C* are three distinct points on a circle .

Now join *AB* and *BC* and draw their perpendicular bisectors.

The point of intersection of the perpendicular bisectors is the centre of given circle.

Hence *O* is the centre of circle.

#### Page No 15.28:

#### Question 5:

Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

#### Answer:

Let *MN *is the diameter and chord *AB* of circle *C*(*O, r*) then according to the question

*AP* = *BP.*

Then we have to prove that *.*

Join *OA *and* OB.*

In ΔAOP and ΔBOP

(Radii of the same circle)

*AP* = *BP* (P is the mid point of chord *AB*)

*OP* = *OP* (Common)

Therefore,

(by cpct)

Hence, proved.

#### Page No 15.29:

#### Question 6:

A line segment *AB* is of length 5cm. Draw a circle of radius 4 cm passing through *A* and *B*. Can you draw a circle of radius 2 cm passing through *A* and *B*? Give reason in support of your answer.

#### Answer:

Given that a line *AB* = 5 cm, one circle having radius of which is passing through point *A* and *B* and other circle of radius.

As we know that the largest chord of any circle is equal to the diameter of that circle.

So,

There is no possibility to draw a circle whose diameter is smaller than the length of the chord.

#### Page No 15.29:

#### Question 7:

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

#### Answer:

Let *ABC* be an equilateral triangle of side 9 cm and let *AD* be one of its medians. Let *G* be the centroid of. Then

We know that in an equilateral triangle centroid coincides with the circumcentre. Therefore, *G* is the centre of the circumcircle with circumradius *GA*.

As per theorem, *G* is the centre and . Therefore,

In we have

Therefore radius *AG* = $\frac{2}{3}AD=3\sqrt{3}\mathrm{cm}$

#### Page No 15.29:

#### Question 8:

Given an arc of a circle, complete the circle.

#### Answer:

Let *PQ *be an arc of the circle.

In order to complete the circle. First of all we have to find out its centre and radius.

Now take a point *R* on the arc *PQ* and join *PR* and *QR*.

Draw the perpendicular bisectors of *PR* and *QR* respectively.

Let these perpendicular bisectors intersect at point *O*.

Then *OP* = *OQ*, draw a circle with centre *O* and radius *OP* = *OQ* to get the required circle.

#### Page No 15.29:

#### Question 9:

Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

#### Answer:

Given that two different pairs of circles in the figure.

As we see that only two points *A, B *of first pair of circle and* C*, *D* of the second pair of circles are common points.

Thus only two points are common in each pair of circle.

#### Page No 15.29:

#### Question 10:

Suppose you are given a circle. Give a construction to find its centre.

#### Answer:

Given a circle *C*(*O*, *r*).

We take three points *A, B* and *C* on the circle.

Join *AB* and *BC.*

Draw the perpendicular bisector of chord *AB *and *BC.*

Let these bisectors intersect at point *O*.

Hence, *O* is the centre of circle.

#### Page No 15.29:

#### Question 11:

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre?

#### Answer:

Let *AB* and *CD *be two parallel chord of the circle with centre *O* such that *AB* = 6 cm,* CD* = 8 cm and* O**P* = 4 cm. let the radius of the circle be cm.

According to the question, we have to find O*Q*

Draw and as well as point *O*, *Q*, and* P* are collinear.

Let

Join *OA* and *OC*, then

*OA* =* OC* = *r*

Now and

So, *AP* = 3 cm and *CQ* = 4 cm

In we have

And in

#### Page No 15.29:

#### Question 12:

Two chords *AB*,* CD* of lengths 5 cm, 11 cm respectively of a circle are parallel, If the distance between *AB* and *CD* is 3 cm, find the radius of the circle.

#### Answer:

Let *AB* and *CD *be two parallel chord of the circle with centre *O* such that *AB* = 5 cm and *CD* = 11 cm. let the radius of the circle be cm.

Draw and as well as point *O*, *Q* and *P* are collinear.

Clearly,* PQ* = 3 cm

Let then

In we have

…… (1)

And

…… (2)

From (1) and (2) we get

$\Rightarrow 6x+\frac{61}{4}=\frac{121}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 6x=\frac{121-61}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 6x=\frac{60}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5}{2}$

Putting the value of *x* in (2) we get,

#### Page No 15.29:

#### Question 13:

Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.

#### Answer:

Let *P* is the mid point of chord *AB* of circle *C*(*O*, *r*) then according to question, line *OQ *passes through the point *P.*

Then prove that *OQ *bisect the arc *AB.*

Join *OA *and* OB.*

In $\u25b3AOP\mathrm{and}\u25b3BOP$

(Radii of the same circle)

(P is the mid point of chord *AB*)

(Common)

Therefore,

(by cpct)

Thus

Arc *AQ = *arc* BQ*

Therefore,

Hence Proved.

#### Page No 15.29:

#### Question 14:

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

#### Answer:

We have to prove that two different circles cannot intersect each other at more than two points.

Let the two circles intersect in three points *A, B* and *C*.

Then as we know that these three points *A, B* and *C* are non-collinear. So, a unique circle passes through these three points.

This is a contradiction to the fact that two given circles are passing through *A, B, C.*

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

Hence, proved.

#### Page No 15.29:

#### Question 15:

Two chords *AB* and *CD* of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between *AB* and *CD* is 6 cm, find the radius of the circle.

#### Answer:

Let *AB* and *CD *be two parallel chord of the circle with centre* O* such that *AB* = 5 cm,* CD* = 11 cm and *PQ* = 6 cm. Let the radius of the circle be cm.

Draw and as well as point* O,* *Q*, and *P* are collinear.

Clearly, *PQ* = 6 cm

Let *OQ* = *x* cm then

Join *OA* and *OC*, then

*OA* =* OC* = *r*

Nowand

So, and

In we have

…… (1)

And

…… (2)

From (1) and (2) we get

Putting the value of *x* in (1) we get,

#### Page No 15.47:

#### Question 1:

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.

#### Answer:

Using the data given in the question we can draw a diagram that looks like fig (1).

From the figure we see that it is an isosceles triangle that has been circumscribed in a circle of radius *R* = 20 m.

The equal sides of the isosceles triangle measure 24 m in length. The length of the base of the isosceles triangle is what we are required to find out.

Since it is an isosceles triangle the perpendicular dropped from the vertex *A* to the base will pass though the circumcentre of the triangle. Let ‘*h*’ be the height of the triangle.

Since the triangle has been circumscribed by a circle of radius ‘*R*’ the length of the distances from ‘*O*’ to any of the three persons would be ‘*R*’.

Let the positions of the persons Isha, Ishita and Nisha be replaced by ‘*A*’, ‘*B*’ and ‘*C*’ respectively. And let the length of the unknown base be,* BC* = 2*x *m.

This is shown in the fig (2).

Now, consider the triangle Δ*BOD*, we have

At the same time consider, we have

Substitute this value in equation we got for ‘*R*’, we get

Now we have got the value of the height of the triangle as *h* = 14.4 m.

Substituting the value of *h* in the below equation,

Now we have the value of *x* = 19.2 m

We need the value of

Hence, the distance between Ishita and Nisha is .

#### Page No 15.47:

#### Question 2:

A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and hands to talk to each other. Find the length of the string of each phone.

#### Answer:

From the given data, we see that the given situation is equivalent to an equilateral triangle circumscribed by a circle.

Let the positions of the three boys Ankur, Amit and Anand be denoted by the points ‘*A*’,’*B*’ and ‘*C*’. Let ‘*O*’ be the centre of the circle, ‘*a*’ is the sides of the equilateral triangle and ‘*R*’ is its circumradius.

Now, in an equilateral triangle with side ‘*a*’, the height, ‘*h*’ of the equilateral triangle would be,

*AB* =* BC* = *CA*

Therefore, $\u2206ABC$is an equilateral triangle.

*OA *= 40 m

Medians of equilateral triangle pass through the circumcentre (O) of the equilateral triangle* ABC*. We know that medians intersect each other in the ratio 2 : 1. As *AD* is the median of equilateral triangle *ABC*, we can write

$\frac{OA}{OD}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{40}{OD}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow OD=20\mathrm{m}\phantom{\rule{0ex}{0ex}}AD=AO+OD=\left(40+20\right)\mathrm{m}=60\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u2206ADC,\phantom{\rule{0ex}{0ex}}A{C}^{2}=A{D}^{2}+D{C}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A{C}^{2}={\left(60\right)}^{2}+\frac{A{C}^{2}}{4}\left[\because AC=BC,\mathit{}DC=\frac{1}{2}BC\Rightarrow DC=\frac{1}{2}AC\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3A{C}^{2}}{4}=3600\phantom{\rule{0ex}{0ex}}\Rightarrow A{C}^{2}=4800\phantom{\rule{0ex}{0ex}}\Rightarrow AC=40\sqrt{3}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Hence the length of the string of each phone is

#### Page No 15.5:

#### Question 1:

Fill in the blanks:

(i) All points lying inside/outside a circle are called ........ points /.......points.

(ii) Circles having the same centre and different radii are called ...... circles.

(iii) A point whose distance from the centre of a circle is greater than its radius lies in ......of the circle.

(iv) A continuous piece of a circle is ....... of the circle.

(v) The longest chord of a circle is a ....... of the circle.

(vi) An arc is a ..... when its ends are the ends of a diameter.

(vii) Segment of a circle is the region between an arc and ....of the circle.

(viii) A circle divides the plane, on which it lies, in ...... parts.

#### Answer:

(i) interior/exterior

(ii) concentric

(iii) the exterior

(iv) arc

(v) diameter

(vi) semi-circle

(vii) centre

(viii) three

#### Page No 15.6:

#### Question 2:

Write the truth value (T/F) fo the following with suitable reasons:

(i) A circle is a plane figure.

(ii) Line segment joining the centre to any point on the circle is a radius of the circle.

(iii) If a circle is divided into three equal arcs each is a major arc.

(iv) A circle has only finite number of equal chords.

(v) A chord of a circle, which is twice as long is its radius is a diameter of the circle.

(vi) Sector is the region between the chord and its corresponding arc.

(vii) The degree measure of an arc is the complement of the central angle containing the arc.

(viii) The degree measure of a semi-circle is 180°

#### Answer:

(i) Given that a circle is a plane figure.

As we know that a circle is a collection of those points in a plane that are at a given constant distance from a fixed point in the plane.

Thus the given statement is .

(ii) Given that line segment joining the centre to any point on the circle is a radius of the circle.

As we know that line segment joining the centre to any point on the circle is a radius of the circle.

Thus the given statement is .

(iii) Given that if a circle is divided into three equal arcs each is a major arc.

As we know that if points *P*, *Q* and *R* lies on the given circle *C*(*O*, *r*) in such a way that

Then each arc is called major arc.

Thus the given statement is .

(iv) It is given that a circle has only finite number of equal chords.

As we know that a circle having infinite number of unequal chords.

Thus the given statement is.

(v) Given that a chord of the circle, which is twice as long as its radius is diameter of the circle.

As we know that a chord of a circle which is largest to others and passing through the centre of the circle and twice as long as its radius is called diameter of the circle.

Thus the given statement is .

(vi) It is given that sector is the region between the chord and its corresponding arc.

As we know that the region between the chord and its corresponding arc is called sector.

Thus the given statement is.

(vii) Given that the degree measure of an arc is the complement of the central angle containing the arc.

As we know that the degree measure of a minor arc is the measure of the central angle containing the arc and that of a major arc is 360° minus the degree measure of the corresponding minor arc.

Let degree measure of an arc is *θ *of a given circle is denoted by

Thus the given statement is.

(viii) Given that the degree measure of a semi-circle is 180°.

As we know that the diameter of a circle divides into two equal parts and each of these two arcs are known as semi-circle.

and are semi circle

Hence,

Thus the given statement is.

#### Page No 15.72:

#### Question 1:

In the given figure, *O* is the centre of the circle. If $\angle APB$= 50°, find ∠*AOB* and ∠*OAB*.

#### Answer:

This question seems to be incorrect.

#### Page No 15.72:

#### Question 2:

In the given figure, *O* is the centre of the circle. Find ∠*BAC*.

#### Answer:

It is given that

And (given)

We have to find

In given triangle

(Given)

* OB* =* OA* (Radii of the same circle)

Therefore, is an isosceles triangle.

So, $\angle OBA=\angle OAB$ ..… (1)

(Given)

[From (1)]

So

Again from figure, is given triangle and

Now in ,

(Radii of the same circle)

$\angle OAC=\angle OCA$

(Given that)

Then,

Since

Hence

#### Page No 15.72:

#### Question 3:

If *O* is the centre of the circle, find the value of *x* in each of the following figures.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

#### Answer:

We have to find in each figure.

(i) It is given that

$\angle AOC+\angle COB=180\xb0\left[\mathrm{Linear}\mathrm{pair}\right]$

As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Now ,$x=\frac{1}{2}\angle COB=22{\frac{1}{2}}^{\xb0}$

Hence

(ii) As we know that = *x * [Angles in the same segment]

line is diameter passing through centre,

So,

$\angle BCA=90\xb0\left[\mathrm{Angle}\mathrm{inscribed}\mathrm{in}\mathrm{a}\mathrm{semicircle}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}\right]$

$\angle CAB+\angle ABC+\angle BCA=180\xb0\left[\mathrm{Angle}\mathrm{sum}\mathrm{property}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow x+40\xb0+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=50\xb0$

(iii) It is given that

$\angle ABC=\frac{1}{2}\left(\mathrm{Reflex}\angle AOC\right)\phantom{\rule{0ex}{0ex}}$

So

And

Then

Hence

(iv)

(Linear pair)

And

*x* =

Hence,

(v) It is given that

is an isosceles triangle.

Therefore

And,

$\mathrm{In}\u25b3AOB,\phantom{\rule{0ex}{0ex}}\angle AOB+\angle OBA+\angle BAO=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 70\xb0+\angle BAO=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle BAO=110\xb0\phantom{\rule{0ex}{0ex}}$

$\angle AOB=2\left(\mathrm{Reflex}\angle ACB\right)\phantom{\rule{0ex}{0ex}}$

Hence,

(vi) It is given that

And

$\angle COA+\angle AOB=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle COA=180\xb0-60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle COA=120\xb0$

$\u2206OCA\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.$

So

Hence,

(vii) (Angle in the same segment)

In we have

Hence

(viii)

As (Radius of circle)

Therefore, is an isosceles triangle.

So (Vertically opposite angles)

Hence,

(ix) It is given that

…… (1) (Angle in the same segment)

$\angle ADB=\angle ACB=32\xb0$ ......(2) (Angle in the same segment)

Because and are on the same segment of the circle.

Now from equation (1) and (2) we have

Hence,

(x) It is given that

$\angle BAC=\angle BDC=35\xb0$ (Angle in the same segment)

Now in $\u2206BDC$ we have

$\angle BDC+\angle DCB+\angle CBD=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 35\xb0+65\xb0+\angle CBD=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle CBD=180\xb0-100\xb0=80\xb0$

Hence,

(xi)

(Angle in the same segment)

In we have

Hence

(xii)

(Angle in the same segment)

is an isosceles triangle

So, (Radius of the same circle)

Then

Hence

#### Page No 15.73:

#### Question 4:

*O* is the circumcentre of the triangle *ABC* and *OD* is perpendicular on *BC.* Prove that ∠*BOD* = ∠*A*

#### Answer:

We have to prove that

Since, circumcenter is the intersection of perpendicular bisectors of each side of the triangle.

Now according to figure *A*, *B*,* C* are the vertices of Δ*ABC*

In , is perpendicular bisector of *BC*

So, *BD* = *CD *

*OB*=

*OC*(Radius of the same circle)

And,

OD = OD (Common)

Therefore,

$\u25b3BDO\cong \u25b3CDO\left(\mathrm{SSS}\mathrm{congruency}\mathrm{criterion}\right)$

$\angle BOD=\angle COD\left(\mathrm{by}\mathrm{cpct}\right)$

We know that angle formed any chord of the circle at the center is twice of the angle formed at the circumference by same chord

Therefore,

Therefore,

Hence proved

#### Page No 15.73:

#### Question 5:

In the given figure, *O* is the centre of the circle, *BO* is the bisector of ∠*ABC*. Show that *AB *= *AC*.

#### Answer:

It is given that,∠*ABC* is on circumference of circle *BD* is passing through centre.

Construction: Join *A* and *C* to form *AC* and extend* BO* to *D* such that *BD* be the perpendicular bisector of *AC*.

Now in $\u25b3BDA\mathrm{and}\u25b3BDC$ we have

AD = CD (*BD* is the perpendicular bisector)

So $\angle BDA=\angle BDC=90\xb0$

(Common)

$\u25b3BDA\cong \u25b3BDC\left(\mathrm{SAS}\mathrm{congruency}\mathrm{criterion}\right)$

Hence (by cpct)

#### Page No 15.73:

#### Question 6:

In the given figure,* O* and *O*' are centres of two circles intersecting at *B* and *C*. *ACD* is a straight line, find *x*.

#### Answer:

It is given that

Two circles having center *O* and *O'* and ∠*AOB* = 130°

And *AC* is diameter of circle having center *O*

We have

So

Now, reflex

So

$x\xb0=360\xb0-230\xb0=130\xb0$

Hence,

#### Page No 15.74:

#### Question 7:

In the given figure, if ∠*ACB* = 40°, ∠*DPB* = 120°, find ∠*CBD*.

#### Answer:

It is given that ∠*ACB* = 40° and ∠*DPB* = 120°

Construction: Join the point *A* and *B*

(Angle in the same segment)

Now in $\u25b3BDP$ we have

$\angle DPB+\angle PBD+\angle BDP=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 120\xb0+\angle PBD+40\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle PBD=20\xb0$

Hence

#### Page No 15.74:

#### Question 8:

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

#### Answer:

We have to find and

Construction: - *O* is centre and *r* is radius and given that chord is equal to radius of circle

Now in we have

AO = OB = BA ( It is given that chord is equal to radius of circle)

So, is an equilateral triangle

So, $\angle AOB=2\angle ADB$ (The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

Then

So

$\angle AEB=\frac{1}{2}\left(\mathrm{Reflex}\angle AOB\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(360\xb0-60\xb0\right)\phantom{\rule{0ex}{0ex}}=150\xb0$

Therefore,

and

Hence, the angle subtended by the chord at a point on the minor arc is 150° and also at a point on the major arc is 30°.

#### Page No 15.74:

#### Question 9:

In the given figure, it is given that *O* is the centre of the circle and ∠*AOC* = 150°. Find ∠*ABC*.

#### Answer:

It is given that *O* is the centre of circle and *A*, *B* and *C* are points on circumference.

(Given)

We have to find ∠*ABC*

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

$\angle ABC=\frac{1}{2}\left(\mathrm{reflex}\angle AOC\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(360\xb0-150\xb0\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 210\xb0\phantom{\rule{0ex}{0ex}}=105\xb0$

Hence,

#### Page No 15.74:

#### Question 10:

In the given figure, *O *is the centre of the circle, prove that ∠*x* = ∠*y* + ∠*z.*

#### Answer:

It is given that, *O* is the center of circle and *A*, *B* and *C* are points on circumference on triangle

We have to prove that ∠*x* = ∠*y* + ∠*z*

∠4 and ∠3 are on same segment

So, ∠4 = ∠3

(Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

…… (1)

(Exterior angle is equal to the sum of two opposite interior angles) …… (2)

$\angle 4=\angle z+\angle 1$ (Exterior angle is equal to the sum of two opposite interior angles)

…… (3)

Adding (2) and (3)

……(4)

From equation (1) and (4) we have

#### Page No 15.74:

#### Question 11:

In the given figure, *O* is the centre of a circle and *PQ* is a diameter. If ∠*ROS* = 40°, find ∠*RTS*.

#### Answer:

It is given that *O* is the centre and $\angle ROS=40\xb0$

We have

In right angled triangle *RQT* we have

$\angle RQT+\angle QTR+\angle TRQ=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 20\xb0+\angle QTR+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle QTR=70\xb0$

$\mathrm{Hence},\angle RTS=70\xb0$

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