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

## Maharashtra State Board 11th Maths Solutions Chapter 1 Angle and its Measurement Miscellaneous Exercise 1

I. Select the correct option from the given alternatives.

Question 1.

\(\left(\frac{22 \pi}{15}\right)^{c}x\) is equal to

(A) 246°

(B) 264°

(C) 224°

(D) 426°

Answer:

(B) 264°

Question 2.

156° is equal to

Answer:

(B)

Question 3.

A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes 88 metres when it traces the angle of 12° at the centre, then the length of the rope is

(A) 70 m

(B) 55 m

(C) 40 m

(D) 35 m

Answer:

(A) 70 m

Question 4.

A pendulum 14 cm long oscillates through an angle of 12°, then the angle of the path described by its extremities is

Answer:

(D)

Question 5.

Angle between hands of a clock when it shows the time 9 :45 is

(A) (7.5)°

(B) (12.5)°

(C) (17.5)°

(D) (22.5)°

Answer:

(D) (22.5)°

Question 6.

20 metres of wire is available for fencing off a flower-bed in the form of a circular sector of radius 5 metres, then .the maximum area (in sq. m.) of the flower-bed is

(A) 15

(B) 20

(C) 25

(D) 30

Answer:

(C) 25

r + r + rθ = 20m

2r + rθ = 20

Question 7.

If the angles of a triangle are in the ratio 1:2:3, then the smallest angle in radian is

(A) \(\frac{\pi}{3}\)

(B) \(\frac{\pi}{6}\)

(C) \(\frac{\pi}{2}\)

(D) \(\frac{\pi}{9}\)

Answer:

(B) \(\frac{\pi}{6}\)

Question 8.

A semicircle is divided into two sectors whose angles are in the ratio 4:5. Find the ratio of their areas?

(A) 5:1

(B) 4:5

(C) 5:4

(D) 3:4

Answer:

(B) 4:5

Question 9.

Find the measure of the angle between hour- hand and the minute hand of a clock at twenty minutes past two.

(A) 50°

(B) 60°

(C) 54°

(D) 65°

Answer:

(A) 50°

Question 10.

The central angle of a sector of circle of area 9π sq.cm is 60°, the perimeter of the sector is

(A) π

(B) 3 + π

(C) 6 + π

(D) 6

Answer:

(C) 6 + π

II. Answer the following.

Question 1.

Find the number of sides of a regular polygon, if each of its interior angles is \(\frac{3 \pi^{c}}{4}\).

Solution:

Each interior angle of a regular polygon

= \(\frac{3 \pi}{4}=\left(\frac{3 \pi}{4} \times \frac{180}{\pi}\right)^{\circ}\) = 135°

Interior angle + Exterior angle = 180°

∴ Exterior angle = 180° – 135° = 45°

Let the number of sides of the regular polygon be n.

But in a regular polygon, exterior angle = \(\frac{360^{\circ}}{\text { no.of sides }}\)

∴ 45° = \(\frac{360^{\circ}}{\mathrm{n}}\)

∴ n = \(\frac{360^{\circ}}{45^{\circ}}\) = 8

∴ Number of sides of a regular polygon = 8.

Question 2.

Two circles each of radius 7 cm, intersect each other. The distance between their centres is 7√2 cm. Find the area common to both the circles.

Solution:

Let O and O_{1} be the centres of two circles intersecting each other at A and B.

Then OA = OB = O_{1}A = O_{1}B = 7 cm

and OO_{1} = 7√2 cm

OO_{1}^{2} = 98 ………………(i)

Since OA^{2} + O_{1}A^{2} = 7^{2}

= 98

= OO_{1}^{2} …..[ from (i)]

m∠OAO_{1} = 90°

□ OAO_{1}B is a square.

m∠AOB = m∠AO_{1}B = 90°

A(□ OAO_{1}B) = (side)^{2} = (7)^{2} = 49 sq.cm

∴ Required area = area of shaded portion = A(sector OAB) + A(sector O_{1}AB)) – A(□ OAO_{1}B)

Question 3.

∆PQR is an equilateral triangle with side 18 cm. A circle is drawn on segment QR as diameter. Find the length of the arc of this circle within the triangle.

Solution:

Let ‘O’ be the centre of the circle drawn on QR as a diameter.

Let the circle intersect seg PQ and seg PR at points M and N respectively.

Since l(OQ) = l(OM),

m∠OM Q = m∠OQM = 60°

m∠MOQ = 60°

Similarly, m∠NOR = 60°

Given, QR =18 cm.

r = 9 cm

θ = 60° = (60 x \(\frac{\pi}{180}\))^{c}

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

∴ l(arc MN) = S = rθ = 9 x \(\frac{\pi}{3}\) = 3π cm.

Question 4.

Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm.

Solution:

Let S be the length of the arc and r be the radius of the circle.

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

S = 37.4 cm

Since S = rθ,

Question 5.

A wire of length 10 cm is bent so as to form an arc of a circle of radius 4 cm. What is the angle subtended at the centre in degrees?

Solution:

S = 10 cm and r = 4 cm

Since S = rθ,

10 = 4 x θ

Question 6.

If two arcs of the same length in two circles subtend angles 65° and 110° at the centre. Find the ratio of their radii.

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 65° and 110° at their centres.

Angle subtended at the centre of the first circle,

Angle subtended at the centre of the second circle,

Question 7.

The area of a circle is 81TH sq.cm. Find the length of the arc subtending an angle of 300° at the centre and also the area of corresponding sector.

Solution:

Area of circle = πr^{2}

But area is given to be 81 n sq.cm

∴ πr^{2} = 81π

∴ r2 = 81

∴ r = 9 cm

θ = 300° = \(=\left(300 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{5 \pi}{3}\right)^{\mathrm{c}}\)

Since S = rθ

S = 9 x \(\frac{5 \pi}{3}\) = 15π cm

Area of sector = \(\frac{1}{2}\) x r x S

= \(\frac{1}{2}\) x 9 x 15π = \(\frac{135 \pi}{2}\) sq.cm

Question 8.

Show that minute-hand of a clock gains 5° 30′ on the hour-hand in one minute.

Solution:

Angle made by hour-hand in one minute

\(=\frac{360^{\circ}}{12 \times 60}=\left(\frac{1}{2}\right)^{\circ}\)

Angle made by minute-hand in one minute = \(\frac{360^{\circ}}{60}\) = 6°

∴ Gain by minute-hand on the hour-hand in one minute

= \(6^{\circ}-\left(\frac{1}{2}\right)^{\circ}=\left(5 \frac{1}{2}\right)^{\circ}\) = 5°30′

[Note: The question has been modified.]

Question 9.

A train is running on a circular track of radius 1 km at the rate of 36 km per hour. Find the angle to the nearest minute, through which it will turn in 30 seconds.

Solution:

r = 1km = 1000m

l(Arc covered by train in 30 seconds)

= 30 x \(\frac{36000}{60 \times 60}\)m

∴ S = 300 m

Since S = rθ,

300 = 1000 x θ

= (17.18)°

= 17° +(0.18)°

= 17° + (0.18 x 60)’ = 17° + (10.8)’

∴ θ = 17°11′(approx.)

Question 10.

In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Solution:

Let ‘O’ be the centre of the circle and AB be the chord of the circle.

Here, d = 40 cm

∴ r = \(\frac{40}{2}\) = 20 cm

Since OA = OB = AB,

∆OAB is an equilateral triangle.

The angle subtended at the centre by the minor

arc AOB is θ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)

= l(minor arc of chord AB) = rθ = 20 x \(\frac{\pi}{3}\)

= \(\frac{20 \pi}{3}\) cm

Question 11.

The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Find angles of the quadrilateral in radians.

Solution:

Let the measures of the angles of the quadrilateral in degrees be a – 3d, a – d, a + d, a + 3d, where a > d > 0

∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 360°

… [Sum of the angles of a quadrilateral is 360°]

∴ 4a = 360°

∴ a = 90°

According to the given condition, the greatest angle is double the least,

∴ a + 3d = 2.(a – 3d)

∴ 90° + 3d = 2.(90° – 3d)

∴ 90° + 3d = 180° – 6d 9d = 90°

∴ d = 10°

∴ The measures of the angles in degrees are

a – 3d = 90° – 3(10°) = 90° – 30° = 60°,

a – d = 90° – 10° = 80°,

a + d = 90°+ 10°= 100°,

a + 3d = 90° + 3(10°) = 90° + 30° = 120°