Well-designed AP 10th Class Maths Guide Chapter 11 Areas Related to Circles Exercise 11.1 offers step-by-step explanations to help students understand problem-solving strategies.

## Areas Related to Circles Class 10 Exercise 11.1 Solutions – 10th Class Maths 11.1 Exercise Solutions

Unless stated otherwise, use π = \(\frac{22}{7}\)

Question 1.

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Solution:

Given sector is OAPB.

Radius r = 6 cm

Angle of the sector θ = 60°

Therefore, area of sector OAPB

= 18.84 sq. cm (or)

= \(\frac{132}{7}\) sq. cm

Question 2.

Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:

Given quadrant is OAPB

Circumference C = 2πr = 22 cm

∴ Radius of the circle (r) = \(\frac{7}{2}\) cm.

Angle of the sector 0 = 90° (quadrant) Area of the sector

Therefore, area of the sector OAPB

= \(\frac{77}{8}\) cm²

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.

Solution:

Given length of minute hand of clock (radius) r = 14 cm

Angle made by the minute hand in five minutes θ = 30°

Area of the sector

Therefore area swept by the minute hand in five minutes is 154/3 sq.cm.

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)

Solution:

Given radius of the circle r = 10 cm

Angle at centre θ = 90°

Let AB be the chord which subtends an angle 90° at the centre O

i) Area of minor segment = Area of sec tor – Area of ∆AOB

= \(\frac{\pi \theta}{360^{\circ}}\) × OA × OB – \(\frac{1}{2}\) × OA × OB

= \(\frac{3.14 \times 90^{\circ}}{360}\) × 10 × 10 – 0.5 × 10 × 10

= 78.5 – 50 = 28.5 cm²

∴ Area of minor segment = 28.5 cm²

ii) Radius of circle = OA = OB = 10 cm.

Area of major segment = Area of circle – Area of minor sector.

= πr – \(\frac{\pi \theta}{360^{\circ}}\) × OA × OB

= 3.14 × 10 × 10 – 3.14 × \(\frac{90^{\circ}}{360^{\circ}}\) × 10 × 10

= 314 – 78.5 = 235.5 cm²

∴ Area of major segment = 235.5 cm²

Question 5.

In a circle of radius 21 cm, an arc sub-tends an angle of 60° at the centre. Find:

i) the length of the arc

ii) area of the sector formed by the arc

iii) area of the segment formed by the corresponding chord

Solution:

Given radius of the circle r = 21 cm

angle subtended by an arc θ = 60°

Therefore, the length of arc = 22 cm

ii) Area of the sector OAXB

= \(\frac{x}{360}\) × πr²

Therefore, area of sector OAXB = 231 cm²

iii) Area of the segment AXB = Area of sector – Area of triangle

= \(\frac{x}{360}\) × πr² – \(\frac{1}{2}\) × AB × OM

In AAOB, OM ⊥ AB

∠AOB = 60°, OM bisects ∠AOB.

that is ∠AOM = ∠BOM

= \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 60 = 30°

So, in AAOM, \(\frac{OM}{OA}\) = cos 30°

OM = OA ⋅ cos 30°

= 21 × \(\frac{\sqrt{3}}{2}\) = \(\frac{21 \sqrt{3}}{2}\) cm²

\(\frac{AM}{OA}\) = sin 30° = \(\frac{1}{2}\)

AM = OA⋅sin 30°

= 21 × \(\frac{1}{2}\) = \(\frac{21}{2}\) cm

Therefore AB = 2 × AM

= 2 × \(\frac{21}{2}\) = 21 cm

Area of segment AXB

Question 6.

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor and major segments of the circle. (Use π = 3.14 and \(\sqrt{3}\) = 1.73)

Solution:

Radius (r) of circle = 15 cm.

Area of sector OPRQ = \(\frac{60^{\circ}}{360^{\circ}}\) × πr²

= \(\frac{1}{6}\) × 3.14 × (15)²

= 117.75 cm²

In ∆OPQ, ∠OPQ = ∠OQP (As OP = OQ).

∠OPQ + ∠OQP + ∠POQ = 180°

2∠OPQ = 180°- 60° = 120°.

∠OPQ = 60°

∆OPQ is an equilateral triangle.

Area of ∆OPQ = \(\frac{\sqrt{3}}{4}\) × (side)²

= \(\frac{\sqrt{3}}{4}\) × (15)²

= \(\frac{225 \sqrt{3}}{4}\) cm²

= 56.257\(\sqrt{3}\) = 97.3125 cm²

Area of segment PRQ = Area of sector OPRQ – Area of ∆OPQ.

= 117.75 – 97.3125 = 20.4375 cm²

Area of major segment PSQ = Area of circle – Area of segment PRQ.

= π(15)² – 20.4375

= 3.14 × 225 – 20.4375

= 706.5 – 20.4375 = 686.0625 cm².

Question 7.

A chord of a circle of radius 12 cm sub-tends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and \(\sqrt{3}\) = 1.73)

Solution:

Given radius of the circle r = 12 cm

angle subtended by the chord at the centre ∠AOB (θ) = 120°

In ∆AOB, OM ⊥ AB

∠AOB = 120°, OM bisects ∠AOB

that is ∠AOM = ∠BOM

= \(\frac{1}{2}\) ∠AOM = \(\frac{1}{2}\) × 120 = 60°

So, in ∆AOM,

\(\frac{OM}{OA}=\cos 60^{\circ}=\frac{1}{2}\)

OM = OA⋅cos60°= 12 × \(\frac{1}{2}\) = 6 cm

\(\frac{AM}{OA}=\sin 60^{\circ}\)

AM = OA⋅sin 60°

= 12 × \(\frac{\sqrt{3}}{2}\) = 6\(\sqrt{3}\) cm

Therefore AB = 2 × AM

= 2 × 6\(\sqrt{3}\) = 12\(\sqrt{3}\) cm

Area of minor sgment AXB = area of OAXB – ∆AOB

= 3.14 × 48 – 36 × 1.73

= 150.72 – 62.28 = 88.44 cm²

Therefore, area of minor segment = 88.44 cm²

Question 8.

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope as shown in figure. Find.

i) the area of that part of the field in which the horse can graze.

ii) the increaseng in the grazing area if the rope were 10 m long instead of 5 m. (Use n = 3.14)

Solution:

Given side of the grass field = 15 m

length of rope OX (or) OY = 5 m

r = 5 cm

i) ∠XOY = θ = 90°

Area of the sector XOY

\(\frac{x}{360}\) × πr²

∴ Area of the part grazed by the horse = 19.625 m²

ii) radius r = 10 cm, ∠XOY (θ) = 90°

Then area of the sector

Increase in grazing area = 78.5 – 19.625 = 58.875 m²

Question 9.

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find :

i) the total length of the silver wire required.

ii) the area of each sector of the brooch.

Solution:

Given diameter of the circle = 35 mm 35

i) Length of the silver wire

= Circumference of circle + length of 5 diameters

= 2πr + 5 × d

= 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\) + 5 × 35

= 110+ 175 = 285 mm

ii) Given the circle is divided into 10 equal sectors.

So, area of each sector of brooch

Question 10.

An umbrella has 8 ribs which are equally spaced as shown in figure. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

Given radius of the circle r = 45 cm

Area between the two consecutive ribs = area of the sector

Angle made by two consecutive ribs at the centre = \(\frac{360^{\circ}}{8}\) = 45°

θ = 45°

Area between two consecutive ribs = \(\frac{22275}{28}\)

Question 11.

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Solution:

Given length of the blade of wiper = 25 cm

Wiper sweeps a sector of a circle,

So, radius of circle r = 25 cm

Angle of the sector 9 = 115°

Area of sector = \(\frac{\theta}{360}\) × πr²

= \(\frac{115}{360}\) × \(\frac{22}{7}\) × 25 × 25

Area swept by two wipers = 2 × Area of sectors

= 2 × \(\frac{115}{360}\) × \(\frac{22}{7}\) × 25 × 25

= \(\frac{158125}{126}\)

= 1254.96 cm²

Question 12.

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Solution:

Given radius of sector r = 16.5 km

Angle of the sector θ = 80°

Area of the sea over which the ships are warned = Area of the sector

Area of the sector = \(\frac{\theta}{360}\) × πr²

= 2 × 3.14 × 5.5 × 5.5 = 189.97 km²

Therefore, area of the sea over which the ships are warned = 189.97 km2

Question 13.

A round table cover has six equal de-signs as shown in Figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use \(\sqrt{3}\) = 1.7)

Solution:

Given in the design form 6 segments of circle of radius 28 cm.

Angle of the sector = \(\frac{360}{6}\) = 60°

θ = 60°

Area of six designs

= 6 (area of minor segment)

= 6 (area of sector – area of triangle)

In ∆AOB, OM ⊥ AB,

∠AOB = 60°, OM bisects ∠AOB

that is ∠AOM = ∠BOM

= \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 60 = 30°

So, in ∆AOM,

\(\frac{OM}{OA}\) = cos 30° = \(\frac{\sqrt{3}}{2}\)

OM = OA cos 30°

= 28 × \(\frac{\sqrt{3}}{2}\) =\(14 \sqrt{3}\)cm

\(\frac{AM}{OA}\) = sin 30°

AM = OA⋅sin30°

= 410.66 – 333.2 = 77.46 cm²

Area of minor segment = 77.46 cm²

Area of 6 designs = 6 × area of minor segment

= 6 × 77.46 = 464.76 cm²

Cost of designing per cm² = ₹ 0.35

Cost of designing for 464.76 cm²

= 464.76 × 0.35

= ₹ 162.66

Question 14.

Tick the correct answer in the following :

Area of a sector of angle p (in degrees) of a circle with radius R is

A) \(\frac{p}{180}\) × 2πR

B) \(\frac{p}{180}\) × πR²

C) \(\frac{p}{360}\) × 2πR

D) \(\frac{p}{720}\) × 2πR²

Solution:

Given radius of the circle R

Angle of the sector θ = p

Area of the sector

= \(\frac{\theta}{360}\) × πr²

= \(\frac{p}{360}\) πr² (or) \(\frac{p}{720}\) × 2πR²

So, Answer D.