NCERT Solutions Class 10 Mathematics. 648 Pages. FREE FLIP-BOOK by Study Innovations

Class X Chapter 10 – Circles Maths
(C) 80 (D) 90

Answer:
It is given that TP and TQ are tangents.
Therefore, radius drawn to these tangents will be perpendicular to the tangents.
Thus, OP ⊥ TP and OQ ⊥ TQ
∠OPT = 90º
∠OQT = 90º
In quadrilateral POQT,
Sum of all interior angles = 360
∠OPT + ∠POQ +∠OQT + ∠PTQ = 360
⇒ 90 + 110º + 90 + PTQ = 360
⇒ PTQ = 70
Hence, alternative (B) is correct
Question 3:
If tangents PA and PB from a point P to a circle with centre O are inclined to each
other an angle of 80 , then ∠POA is equal to

(A) 50 (B) 60
(C) 70 (D) 80
Answer:
It is given that PA and PB are tangents.

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Class X Chapter 10 – Circles Maths

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ⊥ PA and OB ⊥ PB
∠OBP = 90º
∠OAP = 90º
In AOBP,

Sum of all interior angles = 360
∠OAP + ∠APB +∠PBO + ∠BOA = 360

90 + 80 +90º + BOA = 360
∠BOA = 100
In ∆OPB and ∆OPA,
AP = BP (Tangents from a point)
OA = OB (Radii of the circle)
OP = OP (Common side)
Therefore, ∆OPB ≅ ∆OPA (SSS congruence criterion)
A ↔ B, P ↔ P, O ↔ O
And thus, ∠POB = ∠POA

Hence, alternative (A) is correct.

Question 4:
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

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Class X Chapter 10 – Circles Maths
Answer:

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A
and B respectively.
Radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ⊥ RS and OB ⊥ PQ
∠OAR = 90º
∠OAS = 90º
∠OBP = 90º
∠OBQ = 90º
It can be observed that
∠OAR = ∠OBQ (Alternate interior angles)
∠OAS = ∠OBP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.

Question 5:
Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the centre.
Answer:
Let us consider a circle with centre O. Let AB be a tangent which touches the circle at
P.

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Class X Chapter 10 – Circles Maths

We have to prove that the line perpendicular to AB at P passes through centre O. We
shall prove this by contradiction method.
Let us assume that the perpendicular to AB at P does not pass through centre O. Let
it pass through another point O’. Join OP and O’P.

As perpendicular to AB at P passes through O’, therefore,
∠O’PB = 90° … (1)
O is the centre of the circle and P is the point of contact. We know the line joining
the centre and the point of contact to the tangent of the circle are perpendicular to
each other.
∴ ∠OPB = 90° … (2)
Comparing equations (1) and (2), we obtain
∠O’PB = ∠OPB … (3)
From the figure, it can be observed that,
∠O’PB < ∠OPB … (4)
Therefore, ∠O’PB = ∠OPB is not possible. It is only possible, when the line O’P
coincides with OP.
Therefore, the perpendicular to AB through P passes through centre O.

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Class X Chapter 10 – Circles Maths

Question 6:
The length of a tangent from a point A at distance 5 cm from the centre of the circle
is 4 cm. Find the radius of the circle.
Answer:

Let us consider a circle centered at point O.
AB is a tangent drawn on this circle from point A.
Given that,
OA = 5cm and AB = 4 cm
In ∆ABO,
OB ⊥ AB (Radius ⊥ tangent at the point of contact)
Applying Pythagoras theorem in ∆ABO, we obtain
AB2 + BO2 = OA2
42 + BO2 = 52
16 + BO2 = 25
BO2 = 9
BO = 3
Hence, the radius of the circle is 3 cm.

Question 7:
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.
Answer:

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Class X Chapter 10 – Circles Maths

Let the two concentric circles be centered at point O. And let PQ be the chord of the
larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to
the smaller circle.
OA ⊥ PQ (As OA is the radius of the circle)
Applying Pythagoras theorem in ∆OAP, we obtain
OA2 + AP2 = OP2
32 + AP2 = 52
9 + AP2 = 25
AP2 = 16
AP = 4
In ∆OPQ,
Since OA ⊥ PQ,
AP = AQ (Perpendicular from the center of the circle bisects the chord)

PQ = 2AP = 2 × 4 = 8
Therefore, the length of the chord of the larger circle is 8 cm.

Question 8:
A quadrilateral ABCD is drawn to circumscribe a circle (see given figure) Prove that
AB + CD = AD + BC

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Class X Chapter 10 – Circles Maths

Answer:
It can be observed that
DR = DS (Tangents on the circle from point D) … (1)
CR = CQ (Tangents on the circle from point C) … (2)
BP = BQ (Tangents on the circle from point B) … (3)
AP = AS (Tangents on the circle from point A) … (4)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC

Question 9:
In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O
and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.
Prove that ∠AOB=90 .

Answer:
Let us join point O to C.

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Class X Chapter 10 – Circles Maths

In ∆OPA and ∆OCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∆OPA ∆OCA (SSS congruence criterion)
Therefore, P ↔ C, A ↔ A, O ↔ O
∠POA = ∠COA … (i)
Similarly, ∆OQB ∆OCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º
From equations (i) and (ii), it can be observed that
2∠COA + 2 ∠COB = 180º
∠COA + ∠COB = 90º
∠AOB = 90°

Question 10:
Prove that the angle between the two tangents drawn from an external point to a
circle is supplementary to the angle subtended by the line-segment joining the points
of contact at the centre.

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Class X Chapter 10 – Circles Maths
Answer:

Let us consider a circle centered at point O. Let P be an external point from which
two tangents PA and PB are drawn to the circle which are touching the circle at point
A and B respectively and AB is the line segment, joining point of contacts A and B
together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA (radius) ⊥ PA (tangent)
Therefore, ∠OAP = 90°
Similarly, OB (radius) ⊥ PB (tangent)
∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB+∠PBO +∠BOA = 360º
90º + ∠APB + 90º + ∠BOA = 360º
∠APB + ∠BOA = 180º
Hence, it can be observed that the angle between the two tangents drawn from an
external point to a circle is supplementary to the angle subtended by the line-
segment joining the points of contact at the centre.

Question 11:
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:
Since ABCD is a parallelogram,
AB = CD …(1)

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Class X Chapter 10 – Circles Maths
BC = AD …(2)

It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.

Question 12:
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the
segments BD and DC into which BC is divided by the point of contact D are of
lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.

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Class X Chapter 10 – Circles Maths

Answer:

Let the given circle touch the sides AB and AC of the triangle at point E and F
respectively and the length of the line segment AF be x.

In ABC,
CF = CD = 6cm (Tangents on the circle from point C)
BE = BD = 8cm (Tangents on the circle from point B)
AE = AF = x (Tangents on the circle from point A)
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
s = 14 + x

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Class X Chapter 10 – Circles Maths

Area of ∆OBC =

Area of ∆OCA =

Area of ∆OAB =
Area of ∆ABC = Area of ∆OBC + Area of ∆OCA + Area of ∆OAB

Either x+14 = 0 or x − 7 =0
Therefore, x = −14and 7
However, x = −14 is not possible as the length of the sides will be negative.
Therefore, x = 7

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Class X Chapter 10 – Circles Maths

Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm

Question 13:
Prove that opposite sides of a quadrilateral circumscribing a circle subtend
supplementary angles at the centre of the circle.
Answer:

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches
the circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the
center of the circle.
Consider ∆OAP and ∆OAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the same circle)
OA = OA (Common side)
∆OAP ≅ ∆OAS (SSS congruence criterion)
Therefore, A ↔ A, P ↔ S, O ↔ O
And thus, ∠POA = ∠AOS
∠1 = ∠8
Similarly,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7

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Class X Chapter 10 – Circles Maths

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 360º
( 1 + 8) + ( 2 + 3) + ( 4 + 5) + ( 6 + 7) = 360º
2 1 + 2 2 + 2 5 + 2 6 = 360º
2( 1 + 2) + 2( 5 + 6) = 360º
( 1 + 2) + ( 5 + 6) = 180º

AOB + COD = 180º
Similarly, we can prove that BOC + DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend
supplementary angles at the centre of the circle

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Class X Chapter 11 – Constructions Maths

Exercise 11.1
Question 1:
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two
parts. Give the justification of the construction.
Answer:
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle
with line segment AB.
Step 2 Locate 13 (= 5 + 8) points, A1, A2, A3, A4 …….. A13, on AX such that AA1 =
A1A2 = A2A3 and so on.
Step 3 Join BA13.
Step 4 Through the point A5, draw a line parallel to BA13 (by making an angle equal
to ∠AA13B) at A5 intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm
respectively.

Justification
The construction can be justified by proving that

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Class X Chapter 11 – Constructions Maths

By construction, we have A5C || A13B. By applying Basic proportionality theorem for
the triangle AA13B, we obtain

… (1)
From the figure, it can be observed that AA5 and A5A13 contain 5 and 8 equal
divisions of line segments respectively.

… (2)
On comparing equations (1) and (2), we obtain

This justifies the construction.

Question 2:
Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it

whose sides are of the corresponding sides of the first triangle.
Give the justification of the construction.
Answer:
Step 1
Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm
radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs
will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ∆ABC is the
required triangle.
Step 2
Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
Step 3

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Class X Chapter 11 – Constructions Maths

Locate 3 points A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that AA1
= A1A2 = A2A3.
Step 4
Join BA3 and draw a line through A2 parallel to BA3 to intersect AB at point B'.
Step 5
Draw a line through B' parallel to the line BC to intersect AC at C'.
∆AB'C' is the required triangle.

Justification
The construction can be justified by proving that

By construction, we have B’C’ || BC

∴ ∠A = ∠ABC (Corresponding angles)

In ∆AB'C' and ∆ABC,

∠ = ∠ABC (Proved above)

∠ = ∠BAC (Common)

∴∆ ∼ ∆ABC (AA similarity criterion)

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Class X Chapter 11 – Constructions Maths

… (1)
In ∆AA2B' and ∆AA3B,
∠A2AB' = ∠A3AB (Common)
∠AA2B' = ∠AA3B (Corresponding angles)
∴ ∆AA2B' ∼ ∆AA3B (AA similarity criterion)

From equations (1) and (2), we obtain

This justifies the construction.

Question 3:
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose

sides are of the corresponding sides of the first triangle.
Give the justification of the construction.
Answer:
Step 1
Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5
cm radius respectively. Let these arcs intersect each other at point C. ∆ABC is the
required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.
Step 2
Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
Step 3

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Class X Chapter 12 – Areas Related to Circles Maths

Exercise 12.1

Question 1:

The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle
which has circumference equal to the sum of the circumferences of the two circles.
Answer:
Radius (r1) of 1st circle = 19 cm
Radius (r2) or 2nd circle = 9 cm
Let the radius of 3rd circle be r.
Circumference of 1st circle = 2πr1 = 2π (19) = 38π
Circumference of 2nd circle = 2πr2 = 2π (9) = 18π
Circumference of 3rd circle = 2πr
Given that,
Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle
2πr = 38π + 18π = 56π

Therefore, the radius of the circle which has circumference equal to the sum of the
circumference of the given two circles is 28 cm.
Question 2:
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle
having area equal to the sum of the areas of the two circles.
Answer:
Radius (r1) of 1st circle = 8 cm
Radius (r2) of 2nd circle = 6 cm
Let the radius of 3rd circle be r.

Area of 1st circle

Area of 2nd circle
Given that,

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Class X Chapter 12 – Areas Related to Circles Maths

Area of 3rd circle = Area of 1st circle + Area of 2nd circle

However, the radius cannot be negative. Therefore, the radius of the circle having
area equal to the sum of the areas of the two circles is 10 cm.

Question 3:
Given figure depicts an archery target marked with its five scoring areas from the
centre outwards as Gold, Red, Blue, Black and White. The diameter of the region
representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find

the area of each of the five scoring regions.

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Class X Chapter 12 – Areas Related to Circles Maths
Answer:

Radius (r1) of gold region (i.e., 1st circle)
Given that each circle is 10.5 cm wider than the previous circle.
Therefore, radius (r2) of 2nd circle = 10.5 + 10.5
21 cm
Radius (r3) of 3rd circle = 21 + 10.5
= 31.5 cm
Radius (r4) of 4th circle = 31.5 + 10.5
= 42 cm
Radius (r5) of 5th circle = 42 + 10.5
= 52.5 cm
Area of gold region = Area of 1st circle
Area of red region = Area of 2nd circle − Area of 1st circle

Area of blue region = Area of 3rd circle − Area of 2nd circle

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Class X Chapter 12 – Areas Related to Circles Maths

Area of black region = Area of 4th circle − Area of 3rd circle

Area of white region = Area of 5th circle − Area of 4th circle

Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm2, 1039.5
cm2, 1732.5 cm2, 2425.5 cm2, and 3118.5 cm2 respectively.
Question 4:
The wheels of a car are of diameter 80 cm each. How many complete revolutions
does each wheel make in 10 minutes when the car is traveling at a speed of 66 km

per hour?
Answer:
Diameter of the wheel of the car = 80 cm
Radius (r) of the wheel of the car = 40 cm
Circumference of wheel = 2πr
= 2π (40) = 80π cm
Speed of car = 66 km/hour

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Class X Chapter 12 – Areas Related to Circles Maths

Distance travelled by the car in 10 minutes
= 110000 × 10 = 1100000 cm
Let the number of revolutions of the wheel of the car be n.
n × Distance travelled in 1 revolution (i.e., circumference)
= Distance travelled in 10 minutes

Therefore, each wheel of the car will make 4375 revolutions.

Question 5:
Tick the correct answer in the following and justify your choice: If the perimeter and
the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D)7 units
Answer:
Let the radius of the circle be r.
Circumference of circle = 2πr
Area of circle = πr2
Given that, the circumference of the circle and the area of the circle are equal.
This implies 2πr = πr2
2=r
Therefore, the radius of the circle is 2 units.
Hence, the correct answer is A.

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Class X Chapter 12 – Areas Related to Circles Maths

Exercise 12.2
Question 1:
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Answer:

Let OACB be a sector of the circle making 60° angle at centre O of the circle.
Area of sector of angle θ =
Area of sector OACB =

Therefore, the area of the sector of the circle making 60° at the centre of the circle
is
Question 2:
Find the area of a quadrant of a circle whose circumference is 22 cm.
Answer:

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Class X Chapter 12 – Areas Related to Circles Maths

Let the radius of the circle be r.
Circumference = 22 cm
2πr = 22
Quadrant of circle will subtend 90° angle at the centre of the circle.
Area of such quadrant of the circle

Question 3:
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute
hand in 5 minutes.

Page 7 of 48

Class X Chapter 12 – Areas Related to Circles Maths
Answer:

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.

In 5 minutes, minute hand will rotate =
Therefore, the area swept by the minute hand in 5 minutes will be the area of a
sector of 30° in a circle of 14 cm radius.

Area of sector of angle θ =

Area of sector of 30°

Therefore, the area swept by the minute hand in 5 minutes is

Question 4:
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area
of the corresponding:
(i) Minor segment
(ii) Major sector
[Use π = 3.14]

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Class X Chapter 12 – Areas Related to Circles Maths
Answer:

Let AB be the chord of the circle subtending 90° angle at centre O of the circle.
Area of major sector OADB =

Area of minor sector OACB =

Area of ∆OAB =
= 50 cm2
Area of minor segment ACB = Area of minor sector OACB −
Area of ∆OAB = 78.5 − 50 = 28.5 cm2
Question 5:
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) The length of the arc
(ii) Area of the sector formed by the arc

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