Mathematics Part I Solutions Solutions for Class 9 Math Chapter 3 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among Class 9 students for Math Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 46:

Question 1:

AB and AC are two chords of a circle and the bisector of BAC is a diameter of the circle. Prove that AB = AC.


 

Answer:

Given: AB and AC are two chords of a circle and diametre of the circle is the bisector of BAC

To prove: AB = AC

Construction: Draw OF AB and OE AC.

Proof:

In ΔAFO and ΔAEO:

FAO = EAO (As AD is the bisector of BAC)

AFO = AEO = 90° (By construction)

AO = AO (Common)

∴ ΔAFO ΔAEO (By Right-angle, hypotenuse and side criterion)

AF = AE (By c.p.c.t.)

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

AF = FB and AE = EC

Now, AB = AF + FB

= 2AF

= 2AE

= AC (AE = )

Thus, chords AB and AC are equal.


 

Page No 46:

Question 2:

Prove that the lengths of all chords at equal distances from the centre of a circle are equal.


 

Answer:

Given: Chords AB and CD which are at an equal distance from the centre, O of the circle

To prove: AB = DC

Construction: Join AO and OD.

Proof:

OF and OE are the respective perpendicular distances of the chords AB and DC from the centre of the circle.

OF = OE (Given)

In ΔOAF and ΔODE, OF AB and OE DC

By Pythagoras theorem, we have:

OA2 = OF2 + FA2

FA2 = OA2 OF2 ... (1)

Also, OD2 = DE2+ OE2

DE2 = OD2 OE2 ... (2)

However, OA = OD = Radii of the same circle 

Also, OF = OE

So, from equations (1) and (2):

FA2= DE2

FA = DE

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

FA = FB and DE = EC

Now, AB = AF + FB

= 2AF

= 2DE

= DC (DE = )

Thus, the lengths of all chords at equal distances from the centre of the circle are equal.


 

Page No 46:

Question 3:

In the figure below, AB is a chord and CD is the diameter perpendicular to it. Prove that ABC is an isosceles triangle.

Answer:

Given: AB is a chord and CD is the diametre of the circle which is perpendicular to AB.

To prove: ΔABC is an isosceles triangle.

Proof:

As CE is perpendicular to AB, AEC = BEC = 90°.

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

AE = EB

Now, in ΔACE and ΔBCE:

AE = EB (Proved above)

AEC = BEC = 90° (Proved above)

CE = CE (Common)

∴ ΔACE ΔBCE (By side angle side criterion)

AC = BC (By c.p.c.t.)

Thus, ΔABC is an isosceles triangle.


 

Page No 46:

Question 4:

In the figure below, O is the centre of the circle and AB, CD are chords, with OAB = OCD,


 


 

Prove that AB = CD.


 

Answer:

Given: Chords AB and CD form angles with centre O of the circle such that OAB = OCD.

To prove: AB = CD

Construction: Draw OF AB and OE CD.

Proof:

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

AF = FB and CE = ED

In ΔOAF and ΔOCE:

OAF = OCE (Given) 

AFO = CEO = 90° (By construction)

OA = OC (Radii of the same circle)

∴ ΔOAF ΔOCE (By angle angle side criterion)

AF = CE (By c.p.c.t.) 

Now, AB = AF + FB

= 2AF

= 2CE

= CD (CE = )

Thus, AB = C.


 

Page No 46:

Question 5:

In the question above, instead of assuming OAB = OCD, assume that AB = CD and then prove that OAB = OCD.


 

Answer:

Given: Chords AB and CD are two equal chords of a circle.

To prove: OAB = OCD

Construction: Draw OF AB and OE CD.

Proof:

We know that equal chords are equidistant from the centre.

As AB = CD, OF = OE. 

In ΔOAF and ΔOCE:

OF = OE (Proved above)

OA = OC (Radii of the same circle)

AFO = CEO = 90° (By construction) 

∴ ΔOAF ΔOCE (By right-angle, hypotenuse and side criterion) 

⇒∠OAB = OCD (By c.p.c.t.)


 



Page No 49:

Question 1:

In a circle of radius 5 centimetres, two chords of lengths 6 centimetres and 8 centimetres are drawn parallel to each other, on either side of the centre. What is the distance between these chords?


 

Answer:

Given: A circle of radius 5 cm and two chords AB and CD of lengths 8 cm and 6 cm respectively. Also, AB || CD

Construction: Draw OF AB and OE CD.

We know that AB = 2AF = 2, where r is the radius of the circle and d1 is the perpendicular distance of AB from the centre of the circle.

8 = 2

= 4

52 d12 = 42

d12 = 25 − 16 = 9

d1= 3

Similarly, CD = 2CE = 2, where r is the radius of the circle and d2 is the perpendicular distance of CD from the centre of the circle.

6 = 2

= 3

52 d22 = 32

d22 = 25 − 9 = 16

d2 = 4

d1 + d2 = 3 + 4 = 7

Thus, the distance between the chords AB and CD is 7 cm.


 



Page No 50:

Question 1:

In a circle, a chord 3 centimetres away from the centre is 8 centimetres long. How long is a chord, 1.4 centimetres away from the centre in this circle?


 

Answer:

Let AB and CD are the two chords in which AB = 8 cm, OF = 3 cm and OE = 1.4 cm.

Construction: Join OA and OC.

We know that AB = 2AF = 2, where r is the radius of the circle and d1 is the perpendicular distance of AB from the centre of the circle.

8 = 2

= 4

r2 32 = 42

r2 = 16 + 9 = 25

r = 5

Therefore, the radius of the circle is 5 cm.

Similarly, CD = 2CE = 2, where r is the radius of the circle and d2 is the perpendicular distance of CD from the centre of the circle.

CD = 2

2 = CD

CD = 2 ×

CD = 2 × 4.8 = 9.6

Thus, a chord of length 9.6 cm is at a distance of 1.4 cm from the centre of the circle.


 

Page No 50:

Question 2:

In the figure below, AB is a diameter of the circle and PQ is a chord parallel to it.


 


 

Compute the area of the quadrilateral ABQP.


 

Answer:

Given: Two parallel chords PQ and AB of lengths 8 cm and 16 cm

Also, AB is the diameter of the given circle.

Construction: Draw OD PQ

Radius of the circle = = 8 cm

We know that PQ = 2PD = 2, where r is the radius of the circle and d is the perpendicular distance of PQ from the centre of the circle.

8 = 2

= 4

82 d2 = 42

d2 = 64 16 = 48

d =

OD = cm

We know that area of a trapezium = × Sum of parallel sides × Distance between them

Area of trapezium, ABQP = × (8 + 16) cm × cm

= cm2


 

Page No 50:

Question 3:

Find the radius of the circle shown below:


 


 

Answer:

Construction: Extend ED to the centre O of the circle. 

Join OB.

Let the length of OD be x cm.

OE = OD + DE

= (x + 8) cm

OB = (x + 8) cm

We know that AB = 2DB = 2, where r is the radius of the circle and d is the perpendicular distance of AB from the centre of the circle.

24 = 2

12 =

(x + 8)2 x2 = 122

x2 + 64 + 16x x2 = 144

64 + 16x = 144

16x = 144 − 64 

16x = 80

x = 5

OB = (5 + 8) cm = 13 cm

Thus, the radius of the circle is 13 cm.


 



Page No 52:

Question 1:

Draw a triangle with sides 4 centimetres, 5 centimetres and 6 centimetres and draw its circum-circle.

 

Answer:

The dimensions of the required triangle are AB = 4 cm, BC = 5 cm and CA = 6 cm. 

The steps of construction are as follows:

1) Draw a line segment AB of length 4 cm.

2) With A as the centre and 5 cm as the radius, draw an arc on the upper side of AB.

3) With B as the centre and 6 cm as the radius, draw an arc cutting the previously drawn arc at point C.

4) Join AC and BC.

ΔABC is the required triangle.

Circumcircle of the triangle can be constructed as follows:

1) Draw perpendicular bisectors of AB and BC such that they intersect at point O.

2) With O as the centre and radius, OA (or OB or OC), draw a circle passing through the points A, B and C.

The given figure shows the circumcircle of ΔABC.

 

Page No 52:

Question 2:

Draw triangle ΔABC with measurements as given below and draw the circum-circle of each:

(a) AB = 4 cm, AC = 5 cm, A = 60°

(b) AB = 4 cm, AC = 5 cm, A = 120°

(c) AB = 4 cm, AC = 5 cm, A = 90°

Note the position of the circum-centre of each triangle. Draw more triangles and see whether any general conclusion can be formed.

 

Answer:

(a)

The dimensions of the required triangle are AB = 4 cm, A = 60° and CA = 5 cm. 

The steps of construction are as follows:

1) Draw a line segment AB of length 4 cm.

2) Draw TAB of measure 60° at point A.

3) Mark an arc on AT by taking A as the centre and 5 cm as the radius. Name the point of intersection as C.

4) Join CB.

 

ΔABC is the required triangle.

Circumcircle of the triangle can be constructed as follows:

1) Draw perpendicular bisectors of AB and BC such that they intersect at point O.

2) With O as the centre and radius OA (or OB or OC), draw a circle passing through the points A, B and C.

The given figure shows the circumcircle of ΔABC.

 

(b)

The dimensions of the required triangle are AB = 4 cm, A = 120° and CA = 5 cm. 

The steps of construction are as follows:

1) Draw a line segment AB of length 4 cm.

2) Draw TAB of measure 120° at point A.

3) Mark an arc on AT by taking A as the centre and 5 cm as the radius. Name the point of intersection as C.

4) Join CB.

ΔABC is the required triangle.

Circumcircle of the triangle can be constructed as follows:

1) Draw perpendicular bisectors of AB and BC such that they intersect at point O.

2) With O as the centre and radius OA (or OB or OC), draw a circle passing through the points A, B and C.

The given figure shows the circumcircle of ΔABC.

(c)

The dimensions of the required triangle are AB = 4 cm, A = 90° and CA = 5 cm.

The steps of construction are as follows:

1) Draw a line segment AB of length 4 cm.

2) Draw TAB of measure 90° at point A.

3) Mark an arc on AT by taking A as the centre and 5 cm as the radius. Name the point of intersection as C.

4) Join BC.

ΔABC is the required triangle.

Circumcircle of the triangle can be constructed as follows:

1) Draw perpendicular bisectors of AB and BC such that they intersect at point O.

2) With O as the centre and radius OA (or OB or OC), draw a circle passing through the points A, B and C.

The given figure shows the circumcircle of ΔABC.



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