SOLUTION:

Soln. : We have, r₁ = 19 cm and r₂ = 9 cm
∴ Circumference of circle-I = 2πr₁ = 2π(19) cm
and circumference of circle-II = 2πr₂ = 2π (9) cm
Sum of the circumference 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 = 2πR
According to the condition, 2πR = 2π(28)

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

SOLUTION:

Soln. : We have,
Radius of circle-I, r₁ = 8 cm
Radius of circle-II, r₂ = 6 cm
∴ Area of circle-I = πr₁² = π(8)² cm²
Area of circle-II = πr₂² = π(6)² cm²
Let the radius of the circle-III be R
∴ Area of circle-III = πR²
Now, according to the condition,
πr₁² + πr₂² = πR²
⇒ π(8)² + π(6)² = πR²
⇒ π(8² + 6²) = πR²
⇒ 8² + 6² = R²
⇒ 64 + 36 = R²
⇒ 100 = R²
⇒ 10² = R² ⇒ R = 10
Thus, the radius of the new circle = 10 cm.

SOLUTION:

Soln. : Diameter of the innermost region = 21 cm
Radius of the innermost (Gold Scoring) region
∴ Area of Gold region = π(10.5)² cm²


Area of the Red region = π(10.5 + 10.5)² - π(10.5)² = π(21)² - π(10.5)² = π[(21)² - (10.5)²]
= [(21 + 10.5) (21 - 10.5)]cm²


Since each band is 10.5 cm wide
∴ Radius of Gold and Red region
= (10.5 + 10.5) = 21 cm.
Area of Blue region
= π[(21 + 10.5)² -(21)²]cm²
[(31.5)² - (21)²] cm²
= [(31.5 + 21) (31.5 - 21)]cm²


Similarly,
Area of Black region
= π[(31.5 + 10.5)² - (31.5)²] cm²
= [(42)² - (31.5)²] cm²
= [(42 - 31.5) (42 + 31.5)] cm²


Area of White region
= π[(42 + 10.5)² - (42)²] cm²
= π[(52.5)² - (42)²] cm²
= π[(52.5 + 42)(52.5 - 42)] cm²

= 3118.5 cm²

SOLUTION:

Soln. : Diameter of a wheel = 80 cm
∴ Radius of the wheel =
So, circumference of the wheel

⇒ Distance covered by a wheel in one revolution

Distance travelled by the car in 1 hour
= 66 km = 66 × 1000 × 100 cm
∴ Distance travelled in 10 minutes

Now, number of revolutions


Thus, the required number of revolutions
= 4375.

SOLUTION:

Soln. : (A) We have
[Numerical area of the circle]
= [Numerical circumference of the circle]
⇒ πr² = 2πr
⇒ πr² - 2πr = 0
⇒ r² - 2r = 0
⇒ r(r - 2) = 0
⇒ r = 0 or r = 2
But r cannot be zero
∴ r = 2 units.
Thus, the radius of circle is 2 units.

SOLUTION:

Soln. : Here r = 6 cm and θ = 60°

∴ The Area of a sector

SOLUTION:

Soln. : Let radius of the circle = r
∴ 2πr = 22


Here θ = 90°
∴ Area of the quadrant of the circle,

SOLUTION:

Soln. : Length of minute hand = radius of the circle ⇒ r = 14 cm
Q Angle swept by the minute hand in
60 minutes = 360°
∴ Angle swept by the minute hand in 5 minutes
Now, area of the sector with r = 14 cm
and θ = 30°


Thus, the required area swept by the minute hand in 5 minutes

SOLUTION:

Soln. : Length of the radius (r) = 10 cm Sector angle θ = 90°
Area of the sector




Now, (i) Area of the minor segment
= [Area of the minor sector] -
[Area of right ∠AOB]

= 78.5 cm² - 50 cm² = 28.5 cm².
(ii) Area of the major segment
= [Area of the circle] -
[Area of the minor segment]
= πr² - 78.5 cm²

= (314 - 78.5) cm² = 235.5 cm².

SOLUTION:

Soln. : Here, radius = 21 cm and θ = 60°
(i) Circumference of the circle = 2πr


∴ Length of arc APB


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

= 11 × 21 cm² = 231 cm²
(iii) Area of the segment APB = [Area of the sector AOB] - [Area of ∠AOB] ...(1)
In ∠AOB, OA = OB = 21 cm
∴ ∠A = ∠B = 60° [Q ∠O = 60°]
⇒ AOB is an equilateral ∠.
∴ AB = 21 cm

...(2)
From (1) and (2), we have
Area of segment

SOLUTION:

Soln. : Here, radius (r) = 15 cm and
Sector angle (θ) = 60°
∴ Area of the sector


Since ∠O = 60° and OA = OB = 15 cm
∴ AOB is an equilateral triangle.
⇒ AB = 15 cm and ∠A = 60°
Draw OM ⊥ AB, in ∠AMO


Now, ar(∠AOB)


Now area of the minor segment
= (Area of minor sector) - (ar ∠AOB)
= (117.75 97.3125) cm² = 20.4375 cm²
Area of the major segment
= [Area of the circle] -
[Area of the minor segment]
= πr² - 20.4375 cm²

= 706.5 - 20.4375 cm² = 686.0625 cm².

SOLUTION:

Soln. : Here θ = 120° and r = 12 cm
∴ Area of the sector


= 150.72 cm² ...(1)

Now, area of ∠AOB
[Q OM ⊥AB] ...(2)
In ∠OAB, ∠O = 120°
⇒ ∠A + ∠B = 180° - 120° = 60° Q OB = OA = 12 cm
⇒ ∠A = ∠B = 30°
So,
⇒



Now, from (2),
Area of ∠AOB

= 36 × 1.73 cm² = 62.28 cm² ...(3)
From (1) and (3)
Area of the minor segment
= [Area of sector] - [Area of ∠AOB]
= [150.72 cm²] - [62.28 cm²] = 88.44 cm².

SOLUTION:

Soln. : Here, Length of the rope = 5 m
∴ 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 new circular portion grazed


∴ Increase in the grazing area
= (78.5 - 19.625) m² = 58.875 m².

SOLUTION:

Soln. : Diameter of the circle = 35 mm
∴ Radius (r) =

(i) Circumference = 2πr

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,
∴ Sector angle θ
⇒ Area of each sector

SOLUTION:

Soln. : Here, radius (r) = 45 cm
Since circle is divided in 8 equal parts,
∴ Sector angle corresponding to each part

⇒ Area of a sector (part)


∴ The required area between the two ribs

SOLUTION:

Soln. : Here, radius (r) = 25 cm.
Sector angle (θ) = 115°
∴ Area cleaned by each sweep of the blades
[Q there are 2 blades]

SOLUTION:

Soln. : Here, Radius (r) = 16.5 km and Sector angle (θ) = 80°
∴ Area of the sea surface over which the ships are warned

SOLUTION:

Soln. : Here, r = 28 cm
Since, the circle is divided into six equal sectors.
∴ Sector angle
∴ Area of each sector


...(1)
Now, area of 1 design
= Area of segment APB
= Area of sector ABO - Area of ∠AOB ...(2)
In ∠AOB, ∠AOB = 60°, OA = OB = 28 cm
∴ ∠OAB = 60° and ∠OBA = 60°
⇒ ∠AOB is an equilateral triangle.
⇒ AB = AO = BO ⇒ AB = 28 cm
Draw OM ⊥ AB
∴ In right ∠AOM, we have


∴ Area of ∠AOB

= 14 × 14 × 1.7 cm² = 333.3 cm² ...(3)
Now, from (1), (2) and (3), we have:
Area of segment APB = 410.67 cm² - 333.2 cm²
= 77.47 cm²
⇒ Area of 1 design = 77.47 cm²
∴ Area of the 6 equal designs
= 6 × (77.47) cm² = 464.82 cm²
Cost of making the design at the rate of
₹0.35 per cm²
= ₹ 0.35 × 464.82 = ₹ 162.68.

SOLUTION:

Soln. : (D) Here, radius (r) = R
Angle of sector (θ) = p°
∴ Area of the sector

SOLUTION:

Soln. : Since O is the centre of the circle,
∴ QOR is a diameter.
⇒ ∠RPQ = 90° [Angle in a semi-circle]
Now, in right ∠RPQ, RQ² = PQ² + PR²
[Pythagoras theorem]

⇒ RQ² = 24² + 7² = 576 + 49 = 625

∴ radius of circle
∴ Area of ∠RPQ

= 12 × 7 cm² = 84 cm²
Now, area of semi-circle


∴ Area of the shaded portion
= 245.54 cm² - 84 cm² = 161.54 cm².

SOLUTION:

Soln. : Radius of the outer circle = 14 cm
and θ = 40°
∴ Area of the sector AOC



Radius of the inner circle = 7 cm and θ = 40°
∴ Area of the sector BOD


Now, area of the shaded region
= Area of sector AOC - Area of sector BOD

SOLUTION:

Soln. : Side of the square = 14 cm ∴ Area of the square ABCD = 14 × 14 cm²
= 196 cm²
Now, diameter of the circle
= (Side of the square) = 14 cm

⇒ Radius of each of the circles
∴ Area of the semi-circle APD

Area of the semi-circle BPC

∴ Area of the shaded region
= Area of the square - [Area of semi-circle APD
+ Area of semi-circle BPC]
= 196 - [77 + 77] cm² = (196 - 154) cm²
= 42 cm².

SOLUTION:

Soln. : Area of the circle with radius 6 cm

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


Q Each angle of an equilateral triangle = 60°
∴ ∠AOB = 60°
∴ Area of sector COD


Now, area of the shaded region,
= [Area of the circle] + [Area of the equilateral triangle] - [Area of the sector COD]

SOLUTION:

Soln. : Side of the square = 4 cm ∴ Area of the square ABCD = 4 × 4 cm²
= 16 cm²
Q 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 of the middle circle = 1 cm
∴ Area of the middle circle

Now, area of the shaded region
= [Area of the square ABCD] -
[(Area of the 4 quadrant circles)
+ (Area of the middle circle)]

SOLUTION:

Soln. : Area of the circle = πr²

'O' is the centre of the circle,
∴ AO = OB = OC = 32 cm
⇒ ∠AOB = ∠BOC = ∠AOC = 120°
Now, in ∠AOB, ∠1 + ∠2 = 60°
and OA = OB ⇒ ∠1 = ∠2
∴ ∠1 = 30°
If OM ⊥AB, then


Also,



Now, area of ∠AOB,

Since area of ∠ABC = 3 ×[area of ∠AOB]

Now, area of the design = [Area of the circle] - [Area of the equilateral triangle]

SOLUTION:

Soln. : Side of the square ABCD = 14 cm
∴ Area of the square ABCD = 14 × 14 cm²
= 196 cm².
Q Circles touch each other
Radius of the circle

Now, area of a sector of radius 7 cm and sector angle θ as 90°

Area of 4 sectors

Area of the shaded region =
[Area of the square ABCD]
- [Area of the 4 sectors]
= 196 cm² - 154 cm² = 42 cm².

SOLUTION:

Soln. : (i) Distance around the track along its inner edge

= BC + EH +




(ii) Now, area of the track = Area of the shaded region = (Area of rectangle ABCD) + (Area of rectangle EFGH) + 2 [(Area of semi-circle of radius 40 m) (Area of semi-circle of radius 30 cm)] [Q The track is 10 m wide]
⇒ Area of the track
= (106 × 10 m²) + (106 × 10 m²)

= 1060 m² + 1060 m²


[(40 + 30) × (40 - 30)] m²

= 2120 m² + 2200 m² = 4320 m².

SOLUTION:

Soln. : O is the centre of the circle, OA = 7 cm

⇒ AB = 2OA = 2 × 7 = 14 cm
OC = OA = 7 cm
Q AB and CD are perpendicular to each other
⇒ OC ⊥AB
∴ Area of ∠ABC

Again OD = OA = 7 cm
∴ Radius of the small circle

∴ Area of the small circle

Radius of the big circle
Area of the semi-circle OACB

= 11 × 7 cm² = 77 cm²
Now, Area of the shaded region
= [Area of the small circle] + [Area of the big semi-circle OABC] - [Area of ∠ABC]

SOLUTION:

Soln. : Area of ∠ABC = 17320.5 cm²
∴ ∠ABC is an equilateral triangle and area of an equilateral

∴




⇒ (side)² = 40000
⇒ (side)² = (200)² ⇒ side = 200 cm
∴ Radius of each circle
Since each angle of an equilateral triangle is 60°,
∴ ∠A = ∠B = ∠C = 60°
∴ Area of a sector having angle of sector as 60° and radius 100 cm.


Area of 3 equal sectors

Now, area of the shaded region
= [Area of the equilateral triangle ABC]
- [Area of 3 equal sectors]
= 17320.5 cm² - 15700 cm² = 1620.5 cm².

SOLUTION:

Soln. : Q The circles touch each other. ∴ The side of the square ABCD
= 3 × diameter of a circle
= 3 × (2 × radius of a circle) = 3 × (2 × 7cm)
= 42 cm
⇒ Area of the square ABCD = 42 × 42 cm²
= 1764 cm².

Now, area of one circle

Q There are 9 squares
∴ Total area of 9 circles = 154 × 9 = 1386 cm²
∴ Area of the remaining portion of the handkerchief = (1764 - 1386) cm² = 378 cm².

SOLUTION:

Soln. : (i) Here, centre of the circle is O and radius = 3.5 cm.
∴ Area of the quadrant OACB

(ii)

∴ Area of the shaded region
= (Area of the quadrant OACB)
- (Area of ∠BOD)

SOLUTION:

Soln. : OABC is a square such that its side
OA = 20 cm
∴ OB² = OA² + AB²
⇒ OB² = 20² + 20²

= 400 + 400 = 800

⇒ Radius of the circle
Now, area of the quadrant

Area of the square OABC = 20 × 20 cm²
= 400 cm²
∴ Area of the shaded region
= 628 cm² - 400 cm² = 228 cm².

SOLUTION:

Soln. : Q Radius of bigger circle R = 21 cm and sector angle θ = 30°
∴ Area of the sector OAB



Again, radius of the smaller circle, r = 7 cm
Also, the sector angle is 30°
∴ Area of the sector OCD

∴ Area of the shaded region

SOLUTION:

Soln. : Radius of the quadrant = 14 cm
Therefore, area of the quadrant ABPC


= 22 × 7 cm² = 154 cm²
Area of right ∠ABC
⇒ Area of segment BPC = 154 cm² - 98 cm²
= 56 cm²
Now, in right ∠ABC,
AC² + AB² = BC²
⇒ 14² + 14² = BC²
⇒ 196 + 196 = BC²
⇒ BC² = 392
∴ Radius of the semi-circle BQC

∴ Area of the semi-circle BQC

= 11 × 7 × 2 cm² = 154 cm²
Now, area of the shaded region
= [Area of semi-circle BQC]
- [Area of segment BPC]
= 154 cm² - 56 cm² = 98 cm².

SOLUTION:

Soln. : Q Side of the square = 8 cm
∴ Area of the square (ABCD) = 8 × 8 cm²
= 64 cm²
Now, radius of the quadrant ADQB = 8 cm
∴ Area of the quadrant ADQB


Similarly, area of the quadrant

Sum of the two quadrant


Now, area of design
= [Sum of the area of the two quadrant] -
[Area of the square ABCD]