NCERT Solutions for Class 9 Maths Chapter 10 Circles

NCERT Solutions for Class 9 Maths Chapter 9 Circles - Exercise: 9.2
Page number: 122, Total questions: 6

Q1. Two circles of radii $5cm$ and $3cm$ intersect at two points, and the distance between their centres is $4cm$. Find the length of the common chord.
Answer:

Given: Two circles of radii $5cm$ and $3cm$ intersect at two points and the distance between their centres is $4cm$ .

To find the length of the common chord.

Construction: Join OP and draw $OM\perp ABandON\perp CD.$

Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.

$\therefore PM\perp AB$

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

Let, PM = x , so QM=4-x

In $\mathrm{△}$ APM, using Pythagoras' theorem

$A{M}^2=A{P}^2-P{M}^2$ ...........................(1)

Also,

In $\mathrm{△}$ AQM, using Pythagoras' theorem

$A{M}^2=A{Q}^2-M{Q}^2$ ...........................(2)

From 1 and 2, we get

$A{P}^2-P{M}^2=A{Q}^2-M{Q}^2$

$\Rightarrow 3^2-x^2=5^2-(4-x{)}^2$

$\Rightarrow 9-x^2=25-16-x^2+8x$

$\Rightarrow 9=9+8x$

$\Rightarrow 8x=0$

$\Rightarrow x=0$

Put,x=0 in equation 1

$A{M}^2=3^2-0^2=9$

$\Rightarrow AM=3$

$\Rightarrow AB=2AM=6$

Q2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of the other chord.
Answer:

Given: two equal chords of a circle intersect within the circle

To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP = DP.

Construction : Join OP and draw $OM\perp ABandON\perp CD.$

Proof:

In $\mathrm{△}$ OMP and $\mathrm{△}$ ONP,

AP = AP (Common)

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

$\mathrm{\angle }$ OMP = $\mathrm{\angle }$ ONP (Both are right angled)

Thus, $\mathrm{△}$ OMP $\cong$ $\mathrm{△}$ ONP (By SAS rule)

PM = PN..........................1 (CPCT)

AB = CD ............................2(Given )

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AM=CN$ ......................3

Adding 1 and 3, we have

AM + PM = CN + PN

$\Rightarrow AP=CP$

Subtract 4 from 2, we get

AB-AP = CD - CP

$\Rightarrow PB=PD$

Q3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Answer:

Given: two equal chords of a circle intersect within the circle.

To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. $\mathrm{\angle }$ OPM= $\mathrm{\angle }$ OPN

Proof :

Construction: Join OP and draw $OM\perp ABandON\perp CD.$

In $\mathrm{△}$ OMP and $\mathrm{△}$ ONP,

AP = AP (Common)

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

$\mathrm{\angle }$ OMP = $\mathrm{\angle }$ ONP (Both are right-angled)

Thus, $\mathrm{△}$ OMP $\cong$ $\mathrm{△}$ ONP (By RHS rule)

$\mathrm{\angle }$ OPM= $\mathrm{\angle }$ OPN (CPCT)

Q4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that $AB=CD$ (see Fig. $10.25$ ).

Answer:

Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.

To prove: AB = CD

Construction: Draw $OM\perp AD$

Proof :

BC is a chord of the inner circle and $OM\perp BC$

So, BM = CM .................1

(Perpendicular OM bisect BC)

Similarly,

AD is a chord of the outer circle and $OM\perp AD$

So, AM = DM .................2

(Perpendicular OM bisect AD )

Subtracting 1 from 2, we get

AM-BM = DM - CM

$\Rightarrow AB=CD$

Q5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius $5m$ drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is $6m$ each, what is the distance between Reshma and Mandip?

Answer:

Given: From the figure, R, S, M are the position of Reshma, Salma, Mandip respectively.

So, RS = SM = 6 cm

Construction: Join OR,OS,RS,RM and OM.Draw $OL\perp RS$ .

Proof:

In $\mathrm{△}$ ORS,

OS = OR and $OL\perp RS$ (by construction )

So, RL = LS = 3cm (RS = 6 cm )

In $\mathrm{△}$ OLS, by pytagoras theorem,

$O{L}^2=O{S}^2-S{L}^2$

$\Rightarrow O{L}^2=5^2-3^2=25-9=16$

$\Rightarrow OL=4$

In $\mathrm{△}$ ORK and $\mathrm{△}$ OMK,

OR = OM (Radii)

$\mathrm{\angle }$ ROK = $\mathrm{\angle }$ MOK (Equal chords subtend equal angle at centre)

OK = OK (Common)

$\mathrm{△}$ ORK $\cong$ $\mathrm{△}$ OMK (By SAS)

RK = MK (CPCT)

Thus, $OK\perp RM$

area of $\mathrm{△}$ ORS = $\frac{1}{2}\times RS\times OL$ ...............................1

area of $\mathrm{△}$ ORS = $\frac{1}{2}\times OS\times KR$ .............................2

From 1 and 2, we get

$\frac{1}{2}\times RS\times OL$ $=\frac{1}{2}\times OS\times KR$

$\Rightarrow RS\times OL=OS\times KR$

$\Rightarrow 6\times 4=5\times KR$

$\Rightarrow KR=4.8cm$

Thus, $RM=2KR=2\times 4.8cm=9.6cm$

Q6. A circular park of radius $20m$ is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer:

Given: In the figure, A, S, D are positioned Ankur, Syed and David respectively.

So, AS = SD = AD

Radius of circular park = 20 m

so, AO=SO=DO=20 m

Construction: AP $\perp$ SD

Proof :

Let AS = SD = AD = 2x cm

In $\mathrm{△}$ ASD,

AS = AD and AP $\perp$ SD

So, SP = PD = x cm

In $\mathrm{△}$ OPD, by Pythagoras,

$O{P}^2=O{D}^2-P{D}^2$

$\Rightarrow O{P}^2={20}^2-x^2=400-x^2$

$\Rightarrow OP=\sqrt{400-x^2}$

In $\mathrm{△}$ APD, by Pythagoras,

$A{P}^2=A{D}^2-P{D}^2$

$\Rightarrow (AO+OP{)}^2+x^2=(2x{)}^2$

$\Rightarrow (20+\sqrt{400-x^2}{)}^2+x^2=4x^2$

$\Rightarrow 400+400-x^2+40\sqrt{400-x^2}+x^2=4x^2$

$\Rightarrow 800+40\sqrt{400-x^2}=4x^2$

$\Rightarrow 200+10\sqrt{400-x^2}=x^2$

$\Rightarrow 10\sqrt{400-x^2}=x^2-200$

Squaring both sides,

$\Rightarrow 100(400-x^2)=(x^2-200{)}^2$

$\Rightarrow 40000-100x^2=x^4-40000-400x^2$

$\Rightarrow x^4-300x^2=0$

$\Rightarrow x^2(x^2-300)=0$

$\Rightarrow x^2=300$

$\Rightarrow x=10\sqrt{3}$

Hence, the length of the string of each phone $=2x=20\sqrt{3}$ m.

NCERT Solutions for Class 9 Maths Chapter 9 Circles - Exercise: 9.3
Page number: 127-129, Total questions: 12

Q1. In Fig. $10.36$ , A,B and C are three points on a circle with centre O such that $\mathrm{\angle }BOC={30}^{\circ }$ and $\mathrm{\angle }AOB={60}^{\circ }$ . If D is a point on the circle other than the arc ABC, find $\mathrm{\angle }ADC$.

Answer:

$\mathrm{\angle }$ AOC = $\mathrm{\angle }$ AOB + $\mathrm{\angle }$ BOC= ${60}^{\circ }+{30}^{\circ }={90}^{\circ }$

$\mathrm{\angle }$ AOC = 2 $\mathrm{\angle }$ ADC (angle subtended by an arc at the centre is double the angle subtended by it at any)

$\mathrm{\angle }ADC=\frac{1}{2}\mathrm{\angle }AOC$

$\Rightarrow \mathrm{\angle }ADC=\frac{1}{2}{90}^{\circ }={45}^{\circ }$

Q2. 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:

Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.

To find: ADB and $\mathrm{\angle }$ ACB.

Solution:

In $\mathrm{△}$ OAB,

OA = AB (Given)

OA = OB (Radii of circle)

So, OA=OB=AB

$\Rightarrow$ ABC is an equilateral triangle.

So, $\mathrm{\angle }$ AOB = ${60}^{\circ }$

$\mathrm{\angle }$ AOB = 2 $\mathrm{\angle }$ ADB

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

$\Rightarrow \mathrm{\angle }ADB=\frac{1}{2}{60}^{\circ }=30$

ACBD is a cyclic quadrilateral.

So, $\mathrm{\angle }$ ACB+ $\mathrm{\angle }$ ADB = ${180}^{\circ }$

$\Rightarrow \mathrm{\angle }ACB+{30}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ACB={180}^{\circ }-{30}^{\circ }={150}^{\circ }$

Q3. In Fig. $10.37$ , $\mathrm{\angle }PQR={100}^{\circ }$ , where P, Q and R are points on a circle with centre O. Find $\mathrm{\angle }OPR$.

Answer:

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, $\mathrm{\angle }$ PSR + $\mathrm{\angle }$ PQR = ${180}^{\circ }$

$\Rightarrow \mathrm{\angle }PSR+{100}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }PSR={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

Here, $\mathrm{\angle }$ POR = 2 $\mathrm{\angle }$ PSR

$\Rightarrow \mathrm{\angle }POR=2\times {80}^{\circ }={160}^{\circ }$

In $\mathrm{△}$ OPR ,

OP=OR (Radii )

$\mathrm{\angle }$ ORP = $\mathrm{\angle }$ OPR (the angles opposite to equal sides)

In $\mathrm{△}$ OPR ,

$\mathrm{\angle }$ OPR+ $\mathrm{\angle }$ ORP+ $\mathrm{\angle }$ POR= ${180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }OPR+{160}^{\circ }={180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }OPR={180}^{\circ }-{160}^{\circ }$

$\Rightarrow 2\mathrm{\angle }OPR={20}^{\circ }$

$\Rightarrow \mathrm{\angle }OPR={10}^{\circ }$

Q4. In Fig. $10.38$ , $\mathrm{\angle }ABC={69}^{\circ },\mathrm{\angle }ACB={31}^{\circ },$ find $\mathrm{\angle }BDC$

Answer:

In $\mathrm{△}$ ABC,

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

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

$\Rightarrow \mathrm{\angle }A+{100}^{\circ }={180}^{\circ }$

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

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

$\mathrm{\angle }$ A = $\mathrm{\angle }$ BDC = ${80}^{\circ }$ (Angles in same segment)

Q5. In Fig. $10.39$ , A, B, C and D are four points on a circle. AC and BD intersect at a point E such that $\mathrm{\angle }BEC={130}^{\circ }$ and $\mathrm{\angle }ECD={20}^{\circ }$ . Find $\mathrm{\angle }BAC$

.

Answer:

$\mathrm{\angle }$ DEC+ $\mathrm{\angle }$ BEC = ${180}^{\circ }$ (linear pairs)

$\Rightarrow$ $\mathrm{\angle }$ DEC+ ${130}^{\circ }$ = ${180}^{\circ }$ ( $\mathrm{\angle }$ BEC = ${130}^{\circ }$ )

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

$\Rightarrow$ $\mathrm{\angle }$ DEC = ${50}^{\circ }$

In $\mathrm{△}$ DEC,

$\mathrm{\angle }$ D+ $\mathrm{\angle }$ DEC+ $\mathrm{\angle }$ DCE = ${180}^{\circ }$

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

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

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

$\mathrm{\angle }$ D = $\mathrm{\angle }$ BAC (angles in same segment are equal )

$\mathrm{\angle }$ BAC = ${110}^{\circ }$

Q6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $\mathrm{\angle }DBC={70}^{\circ }$ , $\mathrm{\angle }BAC$ is ${30}^{\circ }$ , find $\mathrm{\angle }BCD$ . Further, if $AB=BC$ , find $\mathrm{\angle }ECD$ .

Answer:

$\mathrm{\angle }BDC=\mathrm{\angle }BAC$ (angles in the same segment are equal )

$\mathrm{\angle }BDC={30}^{\circ }$

In $\mathrm{△}BDC,$

$\mathrm{\angle }BCD+\mathrm{\angle }BDC+\mathrm{\angle }DBC={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BCD+{30}^{\circ }+{70}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BCD+{100}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BCD={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

If AB = BC,then

$\mathrm{\angle }BCA=\mathrm{\angle }BAC$

$\Rightarrow \mathrm{\angle }BCA={30}^{\circ }$

Here, $\mathrm{\angle }ECD+\mathrm{\angle }BCE=\mathrm{\angle }BCD$

$\Rightarrow \mathrm{\angle }ECD+{30}^{\circ }={80}^{\circ }$

$\Rightarrow \mathrm{\angle }ECD={80}^{\circ }-{30}^{\circ }={50}^{\circ }$

Q7. If the diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer:

AC is the diameter of the circle.

Thus, $\mathrm{\angle }ADC={90}^{\circ }$ and $\mathrm{\angle }ABC={90}^{\circ }$ ............................(1)(Angle in a semi-circle is a right angle)

Similarly, BD is the diameter of the circle.

Thus, $\mathrm{\angle }BAD={90}^{\circ }$ and $\mathrm{\angle }BCD={90}^{\circ }$ ............................(2)(Angle in a semi-circle is a right angle)

From (1) and (2), we get

$\mathrm{\angle }BCD=\mathrm{\angle }ADC=\mathrm{\angle }ABC=\mathrm{\angle }BAD={90}^{\circ }$

Hence, ABCD is a rectangle.

Q8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer:

Given: ABCD is a trapezium.

Construction: Draw AD || BE.

Proof: In quadrilateral ABED,

AB || DE (Given )

AD || BE ( By construction )

Thus, ABED is a parallelogram.

AD = BE (Opposite sides of a parallelogram )

AD = BC (Given )

so, BE = BC

In $\mathrm{△}$ EBC,

BE = BC (Proved above )

Thus, $\mathrm{\angle }C=\mathrm{\angle }2$ ...........(1)(angles opposite to equal sides )

$\mathrm{\angle }A=\mathrm{\angle }1$ ...............(2)(Opposite angles of the parallelogram )

From (1) and (2), we get

$\mathrm{\angle }1+\mathrm{\angle }2={180}^{\circ }$ (linear pair)

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

Thus, ABED is a cyclic quadrilateral.

Q9. Two circles intersect at two points, B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. $10.40$ ). Prove that $\mathrm{\angle }ACP=\mathrm{\angle }QCD$.

Answer:

$\mathrm{\angle }ABP=\mathrm{\angle }QBD$ ................(1)(vertically opposite angles)

$\mathrm{\angle }ACP=\mathrm{\angle }ABP$ ..................(2)(Angles in the same segment are equal)

$\mathrm{\angle }QBD=\mathrm{\angle }QCD$ .................(3)(angles in the same segment are equal)

From (1), (2), (3), we get

$\mathrm{\angle }ACP=\mathrm{\angle }QCD$

Q10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.

Answer:

Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

Proof: AB is the diameter of the circle and $\mathrm{\angle }$ ADB is formed in a semi-circle.

$\mathrm{\angle }$ ADB = ${90}^{\circ }$ ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and $\mathrm{\angle }$ ADC is formed in a semi-circle.

$\mathrm{\angle }$ ADC = ${90}^{\circ }$ ........................2(angle in a semi-circle)

From 1 and 2, we have

$\mathrm{\angle }$ ADB+ $\mathrm{\angle }$ ADC= ${90}^{\circ }$ + ${90}^{\circ }$ = ${180}^{\circ }$

$\mathrm{\angle }$ ADB and $\mathrm{\angle }$ ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

Q11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that
$\mathrm{\angle }CAD=\mathrm{\angle }CBD$.

Answer:

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove : $\mathrm{\angle }CAD=\mathrm{\angle }CBD$

Proof :

Triangle ABC and ADC are on common base BC and $\mathrm{\angle }$ BAC = $\mathrm{\angle }$ BDC.

Thus, point A,B,C,D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, four points lie on the circle.)

$\mathrm{\angle }$ CAD = $\mathrm{\angle }$ CBD (Angles in same segment are equal)

Q12. Prove that a cyclic parallelogram is a rectangle.
Answer:

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

In cyclic quadrilateral ABCD.

$\mathrm{\angle }A+\mathrm{\angle }C={180}^{\circ }$ .......................(1)(sum of either pair of opposite angles of a cyclic quadrilateral)

$\mathrm{\angle }A=\mathrm{\angle }C$ ........................................(2)(opposite angles of a parallelogram are equal )

From (1) and (2),

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

$\Rightarrow 2\mathrm{\angle }A={180}^{\circ }$

$\Rightarrow \mathrm{\angle }A={90}^{\circ }$

We know that a parallelogram with one angle right angle is a rectangle.

Hence, ABCD is a rectangle.