**Chapter 1 Angle and its Measurement Ex 1.2**

## Chapter 1 Angle and its Measurement Ex 1.2

**Question 1.Find the length of an arc of a circle which subtends an angle of 108° at the centre, if the radius of the circle is 15 cm.Solution:**

Here, r = 15cm and

= 9π cm.

**Question 2.The radius of a circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length equal to length of radius.Solution:**

Here, r = 9cm

Let the arc AB cut off a chord equal to the radius of the circle.

Since OA = OB = AB,

ΔOAB is an equilateral triangle.

m∠AOB = 60°

θ = 60°

**Question 3.Find the angle in degree subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of the circle is 25 cm.Solution:**

Here, r = 25 cm and S = 15 cm

Since S = r.θ,

15 = 25 x θ

**Question 4.A pendulum of length 14 cm oscillates through an angle of 18°. Find the length of its path.Solution:**

**Question 5.Two arcs of the same length subtend angles of 60° and 75° at the centres of the two circles. What is the ratio of radii of two circles?Solution:**

Let , and be the radii of the two circles and let their arcs of same length S subtend angles of 60° and 75° at their centres.

Angle subtended at the centre of the first circle,

Angle subtended at the centre of the second circle,

**Question 6.The area of the circle is 2571 sq.cm. Find the length of its arc subtending an angle of 144° at the centre. Also find the area of the corresponding sector.Solution:**

Area of circle = πr

^{2}

But area is given to be 25 π sq.cm

∴ 25π = πr

^{2}

∴ r

^{2}= 25

∴ r = 5 cm

**Question 7.OAB is a sector of the circle having centre at O and radius 12 cm. If m∠AOB = 45°, find the difference between the area of sector OAB and ΔAOB.Solution:**

Here, r = 12 cm

Draw AM ⊥ OB

In ΔOAM,

[Note: The question has been modified.]

**Question 8.OPQ is the sector of a circle having centre at O and radius 15 cm. If m∠POQ = 30°, find the area enclosed by arc PQ and chord PQ.Solution:**

Here, r = 15 cm

m∠POQ = 30°

\(\left(30 \times \frac{\pi}{180}\right)^{c}[/larex]

∴ θ = [latex]\left(\frac{\pi}{6}\right)^{c}\)

Draw QM ⊥ OP

In ΔOQM,

Shaded portion indicates the area enclosed by arc PQ and chord PQ.

∴ A(shaded portion)

= A(sector OPQ) – A(ΔOPQ)

**Question 9.The perimeter of a sector of the circle of area 25π sq.cm is 20 cm. Find the area of sector.Solution:**

Area of circle = π

But area is given to be 25π sq.cm.

∴ 25π = π

∴ = 25

∴ r = 5 cm

Perimeter of sector = 2r + S

But perimeter is given to be 20 cm.

∴ 20 = 2(5) + S

∴ 20 = 10 + S

∴ S = 10 cm

Area of sector =1/2 x r x S

=1/2 x 5 x 10

= 25sq.cm.

**Question 10.The perimeter of a sector of the circle of area 64 7i sq.cm is 56 cm. Find the area of the sector.Solution:**

Area of circle = π

But area is given to be 25π sq.cm.

∴ 64π = π

∴ = 64

∴ r = 8 cm

Perimeter of sector = 2r + S

But perimeter is given to be 20 cm.

∴ 56 = 2(5) + S

∴ 56 = 16 + S

∴ S = 40 cm

Area of sector =1/2 x r x S

=1/2 x 8 x 40

= 160sq.cm.