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 Meritnationâ€™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 Δ*AFO *and Δ*AEO*:

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

∠*AFO* = ∠*AEO* = 90° (By construction)

*AO* = *AO* (Common)

∴ Δ*AFO *≅ Δ*AEO* (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*

= 2*AF*

= 2*AE*

= *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 Δ*OAF* and Δ*ODE*, *OF* ⊥* AB *and *OE* ⊥ *DC*

By Pythagoras theorem, we have:

*OA*^{2} = *OF*^{2 }+ *FA*^{2}

⇒ *FA*^{2 }= *OA*^{2} − *OF*^{2 } ... (1)

Also, *OD*^{2} = *DE*^{2}+ *OE*^{2}

⇒ *DE*^{2 }= *OD*^{2} − *OE*^{2 }... (2)

However, *OA* = *OD* = Radii of the same circle

Also, *OF* = *OE*

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

*FA*^{2}= *DE*^{2}

⇒ *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*

= 2*AF*

= 2*DE*

= *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: Δ*ABC *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 Δ*ACE* and Δ*BCE*:

*AE* = *EB *(Proved above)

∠*AEC* = ∠*BEC* = 90° * *(Proved above)

*CE *= *CE * (Common)

∴ Δ*ACE* ≅ Δ*BCE* (By side angle side criterion)

⇒ *AC* = *BC * (By c.p.c.t.)

Thus, Δ*ABC* 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 Δ*OAF* and Δ*OCE*:

∠*OAF* = ∠*OCE** * (Given)

∠*AFO** *= ∠*CEO* = 90°* * (By construction)

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

∴ Δ*OAF* ≅ Δ*OCE* (By angle angle side criterion)

⇒ *AF* = *CE* (By c.p.c.t.)

Now, *AB* = *AF* + *FB*

= 2*AF*

= 2*CE*

= *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 Δ*OAF* and Δ*OCE*:

*OF* = *OE * (Proved above)

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

∠*AFO** *= ∠*CEO* = 90° (By construction)

∴ Δ*OAF* ≅ Δ*OCE* (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 *= 2*AF* = 2, where *r *is the radius of the circle and *d*_{1} is the perpendicular distance of *AB* from the centre of the circle.

⇒ 8 = 2

⇒ = 4

⇒ 5^{2} − *d*_{1}^{2} = 4^{2}

⇒ *d*_{1}^{2} = 25 − 16 = 9

⇒ *d*_{1}= 3

Similarly, *CD *= 2*CE* = 2, where *r *is the radius of the circle and *d*_{2} is the perpendicular distance of *CD *from the centre of the circle.

⇒ 6 = 2

⇒ = 3

⇒ 5^{2} − *d*_{2}^{2} = 3^{2}

⇒ *d*_{2}^{2} = 25 − 9 = 16

⇒ *d*_{2}_{ }= 4

*d*_{1}_{ }+ *d*_{2}_{ }= 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 *= 2*AF* = 2, where *r *is the radius of the circle and *d*_{1} is the perpendicular distance of *AB* from the centre of the circle.

⇒ 8 = 2

⇒ = 4

⇒ *r*^{2} − 3^{2} = 4^{2}

⇒ *r*^{2} = 16 + 9 = 25

⇒ *r *= 5

Therefore, the radius of the circle is 5 cm.

Similarly, *CD *= 2*CE* = 2, where *r *is the radius of the circle and *d*_{2} 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 *= 2*PD* = 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

⇒ 8^{2} − *d*^{2} = 4^{2}

⇒ *d*^{2} = 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

= cm^{2}

#### 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 *= 2*DB* = 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} − *x*^{2} = 12^{2}

⇒ *x*^{2} + 64 + 16*x** *− *x*^{2} = 144

⇒ 64 + 16*x** *= 144

⇒ 16*x** *= 144 − 64

⇒ 16*x** *= 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*.

Δ*ABC** *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 Δ*ABC*.

#### 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*.

Δ*ABC** *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 Δ*ABC*.

(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*.

Δ*ABC** *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 Δ*ABC*.

(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*.

Δ*ABC** *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 Δ*ABC*.

View NCERT Solutions for all chapters of Class 9