Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 1 Angle and its Measurement Ex 1.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions 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

θ = 108° = \(\left(108 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{3 \pi}{5}\right)^{\mathrm{c}}\)

Since S = r.θ

S = 15 x \(\frac{3 \pi}{5}\)

= 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°

= \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)

Since S = r.θ,

S = 9 x \(\frac{\pi}{3}\) = 3π cm.

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

∴ The required angle in degree is \(\left(\frac{108}{\pi}\right)^{0}\) or (34.40)°(approx.).

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 r_{1}, and r_{2} 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,

θ_{1} = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)

∴ S = r_{1}θ_{1} = r_{1}(\(\left(\frac{\pi}{3}\right)\))

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

θ = 144° = \(=\left(144 \times \frac{\pi}{180}\right)^{c}=\left(\frac{4 \pi}{5}\right)^{\mathrm{c}}\)

Since s = rθ

S = 5(\(\frac{4 \pi}{5}\)) = 4π

Also, A(sector) = \(\frac{1}{2}\) x r x S = \(\frac{1}{2}\) x 5 x 4π

= 10π sq. 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

\(\theta=45^{\circ}=\left(45 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{4}\right)^{c}\)

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,

sin 30° = \(\frac{\text { QM }}{15}\)

QM= 15 x \(\frac{1}{2}=\frac{15}{2}\)

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 = πr^{2}

But area is given to be 25π sq.cm.

∴ 25π = πr^{2}

∴ r^{2} = 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 = \(\frac{1}{2}\) x r x S

= \(\frac{1}{2}\) 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 = πr^{2}

But area is given to be 25π sq.cm.

∴ 64π = πr^{2}

∴ r^{2} = 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 = \(\frac{1}{2}\) x r x S

= \(\frac{1}{2}\) x 8 x 40

= 160sq.cm.