NCERT Solution for Areas Related to Circles

r_{2} = 9 cm

∴ Circumference of circle-I = 2π r_{1} = 2π (19) cm

Circumference of circle-II = 2π r_{1} = 2π (19) cm

Sum of the circumferences of circle-I and circle-II

= 2π (19) + 2π (9) = 2π (19 + 9) cm = 2π (28) cm

Let R be the radius of the circle-III.

∴ Circumference of circle-III = 2π R

According to the condition,

2π R = 2π (28)

Thus, the radius of the new circle = 28 cm.

Radius of circle-I, r_{1} = 8 cm

∴ Area of circle-I = πr_{1} 2 = π(8)^{2} cm^{2}

Area of circle-II = r_{2}2 = π(6)2 cm^{2}

Let the area of the circle-III be R

∴ Area of circle-III = πr_{2}

Now, according to the condition,

πr_{1}^{2} + πr_{2}^{2} = πr_{2}

i.e. π(8)^{2} + π(6)^{2} = πr_{2}

⇒ π(8^{2} + 6^{2}) = πr_{2}

⇒ 8^{2} + 6^{2} = r_{2}

⇒ 64 + 36 = r_{2}

⇒ 100 = r_{2}

⇒ 10^{2} = r_{2} ⇒ R = 10

Thus, the radius of the new circle = 10 cm.

Radius of the innermost (Gold Scoring) region

∴ Area of Gold region = π (10.5)^{2} cm^{2}

Distance travelled by the car in 1hr

= 66 km = 66 × 1000 × 100 cm

∴ Distance travelled in 10 minutes

Thus, the required number of revolutions = 4375.

(A) 2 units

(B) π units

(C) 4 unites

(D) 7 units

[Numerical area of the circle] = [Numerical circumference of the circle]

⇒ πr_{2} = 2π r

⇒ πr_{2} – 2π r = 0

⇒ r_{2} – 2 r = 0

⇒ r(r – 2) = 0

⇒ r = 0 or r = 2

But r cannot be zero

∴ r = 2 units.

Thus, the option (A) 2 units is correct.

• Area of Sector and Segment of a Circle

The portion (or part) of the circular region enclosed by two radii and the corresponding are is called a sector of the circle.

The portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.

θ = 60°

∴ 2πr = 22

∴ Area of the quadrant of the circle,

⇒ r = 14 cm

θ Angle swept by the minute hand in 60 minutes = 360°

Sector angle θ = 90°

Area of the sector with θ = 90° and r = 10 cm

Now,

(i) Area of the minor segment

= [Area of minor sector] – [Area of rt. ΔAOB]

(ii) Area of major segment

= [Area of the circle] – [Area of the minor segment]

= πr^{2} – 78.5 cm^{2}

(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

(i) Circumference of the circle = 2 πr

(ii) Area of the sector with sector angle 60°

(iii) Area of the segment APQ

= [Area of the sector AOB] – [Area of ΔAOB] ...(1)

In ΔAOB, OA = OB = 21 cm

∴∠A = ∠B = 60° [θ∠O = 60°]

⇒AOB is an equilateral Δ,

∴AB = 21 cm

Draw OM ⊥ AB such that

Sector angle θ = 60°

∴ Area of the sector with θ = 60°

Since ∠O = 60° and OA = OB = 15 cm

∠AOB is an equilateral triangle.

⇒ AB = 15 cm and ∠A = 60°

Now area of the minor segment

= (Area of minor sector) – (ar ΔAOB)

= (117.75 – 97.3125) cm^{2} = 20.4375 cm^{2}

Area of the major segment

= [Area of the circle] – [Area of the minor segment]

= 706.5 – 20.4375 cm^{2} = 686.0625 cm^{2}.

From (1) and (3)

Area of the minor segment

= [Area of minor segment] – [Area of ΔAOB]

= [150.72 cm^{2}] – [62.28 cm^{2}] = 88.44 cm^{2}.

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

(ii) the increase in the grazing area if the rope were 10m long instead of 5m.

∴ Radius of the circular region grazed by the horse = 5 m

(i) Area of the circular portion grazed

(ii) When length of the rope is increased to 10 m,

∴ r = 10 m

⇒ Area of the circular region where θ = 90°.

∴ Increase in the grazing area

= 78.5 – 19.625 m^{2} = 58.875 m^{2}.

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors = 35 mm

∴ Length of 5 pieces = 5 × 35 = 175 mm

∴ Total length of the silver wire = 110 + 175 mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,

Assuning umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Since circle is divided in 8 equal parts,

∴ Sector angle corresponding to each part

Sector angle (θ) = 115°

∴ Area cleaned by each sweep of the blades

Sector angle (θ) = 80°

∴ Area of the sea surface over which the ships are warned

Since, the circle is divided into six equal sectors.

Now, area of 1 design

= Area of segment APB

= Area of sector – Area of ΔAOB ...(2)

In ΔAOB, ∠AOB = 60°, OA = PN = 28 cm

∴ ∠OAB = 60° and ∠OBA = 60°

⇒ ΔAOB is an equilateral triangle.

⇒ AB = AO = BO

⇒ AB = 28 cm

Draw OM ⊥AB

∴ In right ΔAOM, we have

Now, from (1), (2) and (3), we have:

Area of segment APQ = 410.67 cm^{2} – 333.2 cm^{2} = 77.47 cm^{2}

⇒ Area of 1 design = 77.47 cm^{2}

∴ Area of the 6 equal designs = 6 × (77.47) cm^{2}

= 464.82 cm^{2}

Cost of making the design at the rate of Rs. 0.35 per cm^{2},

= Rs. 0.35 × 464.82

= Rs. 162.68.

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

Angle of sector (θ) = p°

∴ QOR is a diameter.

⇒ ∠RPQ = 90° [Angle in a semi-circle]

Now, in right DRPQ,

RQ^{2} = PQ^{2} + PR^{2}

⇒ RQ^{2} = 24^{2} + 7^{2} = 756 + 49 = 625

∴ Area of the shaded portion = 245.54 cm^{2} – 84 cm^{2} = 161.54 cm^{2}.

Here, θ = 40°

∴ Area of the sector BOD

Now, area of the shaded region

= [Area of sector AOC] – [Area of sector BOD]

∴ Area of the square ABCD = 14 × 14 cm^{2} = 196 cm^{2}

Now, diameter of the circle = (Side of the square) = 14 cm

∴ Area the shaded region

= [Area of the square] – [Area of semi-circle APD + Area of semi-circle BPC]

= 196 – [77 + 77] cm^{2} = 196 – 154 cm^{2} = 42 cm^{2}.

Area of equilateral triangle, having side a = 12 cm, is given by

∴ Area of the square ABCD = 4 × 4 cm^{2} = 16 cm^{2}

θ Each corner has a quadrant circle of radius 1 cm.

∴ Area of all the 4 quadrant squares

Diameter of the middle circle = 2 cm

⇒ Radius fo the middle circle = 1 cm

∴ Area of the middle circle = πr^{2}

Now, area of the shaded region

= [Area pf tje sqiare ABCD] – [(Area of the 4 quadrant circles) + (Area of the middle circle)]

‘O’ is the centre of the circle,

∴ AO = OB = OC = 32 cm

⇒ ∠AOB = ∠BOC = ∠AOC = 120°

Now, in ΔAOB, ∠1 = 30°

∵ ∠1 + ∠2 = 60°

Also OA = OB ⇒ ∠1 = ∠2

If OM ⊥AB, then

∴ Area of the square ABCD = 14 × 14 cm^{2} = 196 cm^{2}.

θ Circles touch each other

∴ Area of the shaded region = [Area of the square ABCD] – [Area of the 4 sectors]

= 196 cm^{2} – 154 cm^{2} = 42 cm^{2}.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long.

If the track is 10 m wide, find:

(i) the distance around the track along its inner edge

(ii) the area of the track.

(ii) Now, area of the track

= Area of the shaded region

= (Area of rectangle ABCD) + (Area of rectangle EFGH)

+ 2 [(Area of 2 semi-circles each of radius 40 m) – (Area of 2 sem-circles each of radius 30 cm)]

⇒ Area of the track

OA = 7 cm

⇒ AB = 2 OA = 2 × 7 = 14 cm

OC = OA = 7 cm

θ AB and CD are perpendicular to each other

⇒ OC ⊥ AB

Now, Area of the shaded region

= [Area of the small circle] + [Area of the big semi-circle OABC] – [Area of ΔABC]

∵ΔABC is an equilateral triangle and area of an equilateral

Now, area of the shaded regino

= [Area of the equilateral triangle ABC] – [Area of 3 equal sectors]

= 17320.5 cm^{2} – 15700 cm^{2} = 1620.5 cm^{2}.

∴ The side of the square ABCD

= 3 × diameter of a circle

= 3 × (2 × radius of a circle) = 3 × (2 × 7 cm)

= 42 cm

⇒ Area of the square ABCD = 42 × 42 cm^{2} = 1764 cm^{2}.

∵ There are 9 squares

∴ Total area of 9 circles = 154 × 9 = 1386 cm^{2}

∴ Area of the remaining portion of the handkerchief = 1764 – 1386 cm^{2} = 378 cm^{2}.

(i) quadrant OACB, (ii) shaded region.

∴ OB^{2} = OA^{2} + OB^{2}

= [20^{2} + 20^{2}] = [400 + 400] = [800]

Area of the square OABC = 20 × 20 cm^{2} = 400 cm^{2}

∴ Area of the shaded region = 268 cm^{2} – 400 cm^{2} = 228 cm^{2}.

R = 21 cm

Therefore, area of the quadrant ABPC

⇒ Area of segment BPC = 154 cm^{2} – 98 cm^{2} = 56 cm^{2}

Now, in right ΔABC,

AC^{2} + AB^{2} = BC^{2}

⇒ 142 + 142 = BC^{2}

⇒ 196 + 196 = BC^{2}

Now, area of the shaded region

= [Area of segment BQC] – [Area of segment BPC]

= 154 cm^{2} – 56 cm^{2} = 98 cm^{2}.

∴ Area of the square (ABCD) = 8 × 8 cm^{2}

= 64 cm^{2}

Now, radius of the quadrant ADQB = 8 cm

Now, area of design

= [Sum of the areas of the two quadrants] – [Area of the square ABCD]