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Given:

The dimensions of the rectangle are 24 cm and 16 cm.

The circumference of the circle = 110% of the perimeter of rectangle.

Formula Used:

Perimeter of the rectangle = 2 × (length + breadth)

Circumference of the circle = 2πr

Area of the circle = πr²

Where r = radius of the circle

Calculation:

Circumference of the circle = 110% of Perimeter of the rectangle

⇒ 2πr = (110/100) × 2 × (24 + 16)

⇒ 2 × (22/7) × r = (11/10) × 2 × 40

⇒ r = (11 × 2 × 40 × 7)/(10 × 2 × 22)

⇒ r = 14

Area of the circle = πr²

⇒ (22/7) × 14²

⇒ 22 × 2 × 14

⇒ 616

∴ Area of the circle is 616 cm².     

Given:

Number of cubes = 42

Side of cube = 1 cm

Perimeter of base of cuboid = 18 cm

Concept used:

The cubes that are at the 4 corners contribute 2 cm to the perimeter, and the rest cubes contribute 1 cm to the perimeter

Formula used:

Height of the cuboid = (Total number of cubes/Number of cubes in the base) × Side of the cube

Calculations:

Number of cubes in the base of the cuboid

⇒ 18 - 4 

⇒ 14 

Height of cuboid = (42/14) × 1

⇒ 3 cm

∴ The height of the cuboid is 3 cm

Given:

Area of right angled isosceles triangle = 676 sq.cm

Formula:

Area of right angled isosceles triangle = 1/2 × a × a

a = equal sides of triangle

Calculation:

Let height or base of right angled triangle be a cm.

⇒ 676 = 1/2 × a²

⇒ a = 26√2 cm

Height and base of isosceles triangle = 26√2 cm

In right angled triangle,

Using pythagoras theorem,

hypotenuse² = a² + a²

hypotenuse = 52

∴ Length of its hypotenuse is 52cm.

Given:

Diameter of the each sphere = 3 cm

Diameter of the solid metal cylinder = 4 cm

Height of the solid metal cylinder = 54 cm

Concept used: 

Volume of spherical ball = 4/3 × π × r3

Volume of cylinder = π × r² × h

Number of small balls (n) = (Volume of cylinder)/(Volume of sphere)

Calculation:

Let the number of solid spheres be n.

Radius of the each sphere = 3/2 cm

Radius of the solid metal cylinder = 4/2 = 2 cm

Volume of each sphere = (4/3) × π × (3/2)3 

⇒  π × (4/3) × (27/8)

⇒ 9π/2 cm³

Volume the solid metal cylinder = π × (2)² × 54 = 216π cm³

n = 216π/(9π/2)

⇒ n = 48

∴  The number of solid spheres is 48.

Given:

Radius of a solid sphere = 10 cm

Number of solid spherical balls = 8

Concept Used:

For number of balls we have to equate the volume of all the balls and the solid sphere volume

Formula Used:

Volume of a sphere = (4/3)πr³, here r = Radius of the sphere

Calculation:

Let radius of a solid spherical ball be R cm

(4/3)π(10 cm)³ = 8 × (4/3)πR³ 

⇒ R³ = 1000/8

⇒ R3 = 125

⇒ R = 5

Radius of a spherical ball = 5 cm

∴ The radius of each small spherical ball is 5 cm

Given:

The ratio of length and breadth = 6 : 5

Difference between the length and breadth of a rectangular plot = 34 meters

Formula Used:

Area of the rectangle = l × b square units

The perimeter of the rectangle = 2 × (l + b)

Calculation:

let us take the  length and breadth  of the rectangle be 6x, 5x respectively

According to the question,

The difference between the length and breadth of plot = Difference in the ratio of length and breadth of the plot.

⇒ 34 = 6x - 5x

⇒ x = 34 

Lenght of the rectangle = 6 × 34 = 204 meters

Breadth of the rectangle = 5 × 34 = 170 meters

The perimeter of the rectangle = 2 × ( 204 + 174) = 748 meters 

∴ The perimeter of the rectangle is 748 meters

Given:

Length = 30 m

Breadth = 25 m

Depth = 4.5 m

Side of the square shape tiles = 40 cm

Formula used:

Lateral surface area of a cuboid (L.S.A) = 2 × (l + b) × h

Area of a rectangle = (l × b)

l = length 

b = breadth

h = height

area of a square = a²

a = side of the square

Calculation:

Area of the base of the pool = (30 × 25)m2

⇒ 750 m²

Area of the four walls of the pool = 2 × (30 + 25)m × 4.5 m 

⇒ 2 × (30 + 25)m × 4.5 m

⇒ 495 m²

Total area = (750 + 495)m²

⇒ 1245 m²

Area of the square tiles = (0.4)²

⇒ 0.16 m²

No of tiles required = 1245/0.16

⇒ 7781.25 ≈ 7781

∴ Required titles is 7781 (approx)

Given:

⇒ Length of rectangle = 98/2 - 16 = 33 cm

⇒ Area of rectangle = 33 × 16 = 528 sq.cm

⇒ Curved surfece area of cylinder = 528 = 2πrh

⇒ rh = 84

Then,

⇒ h - r = 5

Solving,

⇒ r = 7 cm and h = 12cm

⇒ Volume of cylinder = πr²h

= 22/7 × 7 × 7 × 12

= 1848 cubic.cm

∴ Volume of cylinder = 1848 cubic.cm

Given:

Length of the hall = 10 m

The breadth of the hall = 6 m

Height of the hall = 4 m

Concept used:

The longest rod that can be fit into a cuboidal hall = diagonal of the cuboid

Diagonal of the cuboid = √(l² + b² + h²)

Calculation:

Longest rod that can be placed in the hall = √(10² + 6² + 4²) = √(100 + 36 + 16)

The longest rod that can be placed in the hall = 2√38 m

∴ The longest rod that can be placed in the hall 2√38 m.

Given:

⇒ Perimeter of an equilateral triangle = 15 × 3 = 45cm

⇒ Perimeter of an isosceles triangle = 45cm

⇒ The unequal side of an isosceles triangle = 45 - 2 × 14 = 17cm

∴ Required ratio = 14 : 14 : 17

Given:

Using pythagoras theorem,

⇒ (slant height)² = 21² + 20²

⇒ slant height = 29 cm

⇒ Radius of cylinder = 21 - 7 = 14cm

⇒ Volume of cylinder = 6160 = πr²h

⇒ h = 10cm

⇒ Curved surface area of cone = πrl = π × 21 × 29 = 609π sq.cm

⇒ Curved surface area of cylinder = 2πrh = 2π × 14 × 10 = 280π sq.cm

∴ Required ratio = 280π : 609π = 40 : 87

Given:

Diagonal of square = 14√2 cm

Formula used:

Diagonal of square = √2 × side

Area of square = (side)²

Calculation:

Diagonal of square = √2 × side

⇒ 14√2 = √2 × side

⇒ side = 14 cm

Area of square 

⇒ 14 × 14

⇒ 196 cm²

∴ The area of square is 196 cm².

Given:

Ratio of Parallel sides = 3 ∶ 5

Area of trapezium = 240 cm²

Height of trapezium = 12 cm

Concept:

Replace the ratio with a constant and using the formula for the area of the trapezium, calculate the sides of the trapezium.

Formula used:

Area of trapezium = 1/2 × (Sum of parallel sides) × height

Calculation:

Let the parallel sides be 3x and 5x

Area of trapezium = 1/2 × (Sum of parallel sides) × height

⇒ 240 cm² = 1/2 × (3x + 5x) × 12 cm

⇒ 240 cm² = (1/2) × (8x) × 12 cm

⇒ 8x = (240 × 2)/12 cm

⇒ x = 40/8 cm

⇒ x = 5 cm

Shorter parallel side of the trapezium = 3x = 3 × 5 cm = 15 cm.

∴ Shorter parallel side of the trapezium is 15 cm.

Given:

The sides of a triangle are in the ratio 4 ∶ 3 ∶ 2.

Formula used:

Semi perimeter of a Δ = (a + b + c)/2

Area of Δ = √{s(s - a)(s - b)(s - c)}

Where a, b, c are sides of the Δ.

s → semiperimeter

Calculations:

Let the ratio of sides be 2x, 3x, and 4x.

Then 4x + 3x + 2x = 54

⇒ 9x = 54

⇒ x = 6

So the sides of the Δ are 24 cm, 18 cm, and 12 cm.

Semi perimeter of the Δ = (a + b + c)/2 = (24 + 18 +12)/2 = 27 cm

Area of the Δ = √{s(s - a)(s - b)(s - c)}

⇒ √{s(s - a)(s - b)(s - c)} = √{27(27 - 24)(27 - 18)(27- 12)}

⇒ √(27 × 3 × 9 × 15) = 27√15 cm²

∴ The area of the triangle is 27√15 cm².

Given:

Length of rectangular region = 30 m

Width of rectangular region = 20 m

Formula used:

Perimeter = 2 × (Length + Breadth)

Calculation:

Perimeter = 2 × (30 + 20) m

⇒ 2 × 50 m

⇒ 100 m

∴ The perimeter of rectangular region is 100 m

Given: 

Quantity I:

Radius and height of the cone = 6 cm and 8 cm

The total surface area of sphere = 3 × curved surface area of the cone

Quantity II:

The ratio of length, height, and breadth of cuboid = 5 ∶ 2 ∶ 3

Difference between length and height = 5 cm

Concept used:

The volume of sphere = 4/3 × π × r³

The total surface area of sphere = 4πr²

The curved surface area of cone = π × r × l

Where l = √(r² + h²)

Total surface area of the cuboid = 2 × (lb + bh + hl)

Where, l, b, and h are the length, breadth, and height.

Calculation:

Quantity I:

l = √(6² + 8²) = 10 cm

curved surface area of cone = π × 6 × 10 = 60π

total surface area of sphere = 4πr² = 3 × 60π

⇒ r² = 45 cm²

Total surface area of the hemisphere = 3πr² = 3π × 45

Total surface area of the hemisphere = 135 × 3.14

Quantity I = 135 × 3.14 = 423.9 cm²

Quantity II:

Ratio of length, height and breadth of cuboid = 5 ∶ 2 ∶ 3

Then, length, breadth and height of cuboid will be = 5x, 3x and 2x

Difference between length and breadth = 5x – 3x = 2x

⇒ 2x = 5

Or, x = 5/2 cm

Total surface area of the cuboid = 2 × (15x² + 6x² + 10x²) = 62 × (5/2)²

Quantity II = 62 × (5/2)² = 387.5 cm²

∴ From the table, we can see that Quantity I > Quantity II.

Given

Radius of the cylindrical wire = 6 cm

Radius of the sphere = 21cm

Formula

Volume of cylinder = πr²h

Volume of sphere = (4/3)πr³

Calculation

πr2h = (4/3)πR3

⇒ h = (4/3)R³/r² 

⇒ (4/3) × (21³/6²)

⇒ 343 cm

∴ The length of the cylinderical wire is 343 cm.

Given:

Two adjacent sides of a parallelogram are 12 cm and 9 cm.

And one of the angle is 30°

Formula used:

Area of parallelogram = ab × sinθ  

Where, a and b are two adjacent sides.

Calculation:

a = 12 cm

b = 9 cm 

θ = 30° 

Area of parallelogram = ab × sinθ 

⇒ 12 × 9 × sin30° = 12 × 9 × (1/2)

⇒ 6 × 9 =54 cm²

∴ The area of parallelogram is 54 cm².

Given:

Area of parallelogram = 960 cm²

Ratio of adjacent sides = 5 ∶ 8

Length of Altitude (L₁) = 20 cm

Formula used:

Area of parallelogram = Base × Height

Calculation:

Let the other altitude be ‘y’

Let the adjacent sides be 5x and 8x

Area of parallelogram = Base × Height

⇒ 960 = (8x) × 20

⇒ 8x = 960/20

⇒ x = 6

So, Sides are;

5x = 5 × 6 = 30 cm

8x = 8 × 6 = 48 cm

Area of parallelogram = 30 × y

⇒ 48 × 20 = 30 × y

⇒ y = 32 cm

∴ The altitude on smaller side is 32 cm

Given:

DEF is a right triangle.

∠E = 90° 

DE = 5 cm

EF = 12 cm

DF = 13 cm

Formula used:

Radius of incircle = area of triangle/semi perimeter of triangle

Calculation:

Radius of incircle = [(1/2) × 12 × 5]/ [(13 + 12 + 5)/2]

⇒ 2 cm

Given:

Length of pipe = 2 m

Breadth of pipe = 10 m

Speed of water = 10 km/hr

Time = 15 minutes

Calculation:

Speed = 10 km/hr = 10 × (5/18) m/sec = 25/9 m/s

Time = 15 minutes = (15 × 60) sec = 900 sec

Length (H) = (25/9) × 900 = 2500

Volume = L × B × H

= 2 × 10 × 2500

= 50,000 m³

∴ Volume of water collected in 15 minutes is 50,000 m³.

Given:

The heights of the two prisms are 14 cm and 10 cm

The ratio of the area of bases of two square prisms are 7 ∶ 9

The volume of first prism = 882 cm3

Formula used:

The volume of the square prism = Area of the base × height

The total surface area of the square prism = 2 × Area of base + 4 × Area of lateral sides

Calculation:

Let the area of the base of the first and second prism be 7x and 9x

The volume of the first prism = Area of the base × height

⇒ 7x × 14 = 882

⇒ x = 9

So, Area of the base of the second prism = 9 × 9 = 81

So, The length of the side of the second prism = √81 = 9 cm

Now, Total surface area of the second prism = 2 × area of base + 4 × area of the lateral side

⇒ 2 × 92  + 4 × 9 × 10 = 522 cm2

∴ The Total surface area of the second square prism is 522 cm2

Given

Dimensions of cuboid = a × b × c

Volume of cuboid = V

Surface area of cuboid = S

Formula used

For a cuboid with length, breadth and height x, y and z respectively, volume = xyz  and surface area = 2xy + 2xz + 2yz

Calculation

V = abc

S = 2(ab + bc + ca)

On dividing S by V we get,

\({S \over V} = {2(ab + bc + ca) \over abc}\)

\(\Rightarrow {S \over V} = 2( {ab \over abc}+ {bc \over abc} + {ca \over abc})\)

\(\Rightarrow {S \over V} = 2( {1 \over c}+ {1 \over a} + {1 \over b})\)

\(\Rightarrow {1 \over V} = {2\over S}( {1 \over a}+ {1 \over b} + {1 \over c})\)

Given:

The angle of the triangle are in the ratio = 4 ∶ 5 ∶ 6

Concept used:

The Sum of all the angles of the triangle is 180° 

An acute-angled triangle is a type of triangle in which all the three internal angles of the triangle are acute, i.e less than 90°

An obtuse-angled triangle is a triangle in which one of the interior angles measures more than 90° 

A right triangle or right-angled triangle is a triangle in which one angle is a right angle

An isosceles triangle is a triangle that has two sides of equal length

Calculations:

The sum of angles of triangle = 180°

The sum of the angles = (4x + 5x + 6x) 

⇒ 15x 

15x = 180° 

⇒ x = 180°/15

⇒ x = 12

Hence angles in the triangle are 4x, 5x, 6x = 48, 60, and 72 

In a triangle, If all the three angles are less than 90° then, the triangle is known as Acute angle triangle

∴ The triangle is an Acute triangle.

Given

Speed of bicycle is  14.4 km/h

Formula used

Area of circle = πr2

Circumference of circle = 2πr

Calculation

Speed of the rider = 14.4 km/h

⇒ 14.4 × 5/18 = 4 m/sec

Total time taken by him to complete 1 round is (60 + 28)sec

⇒ 88 sec

Total distance = circumference = speed × time

⇒ 4 × 88

⇒ 352 m

⇒ 2πr = 352

⇒ r = 56 m

Area of circle =  πr2

⇒ 22/7(56 × 56)

⇒ 9856 m2

Given:

Total distance traveled = 4.4 km

Number of revolution = 2000

Concept used:

Total distance traveled = (Number of revolution) × (Circumference of the circle)

Calculation:

Let the radius of the circle be r cm.

Circumference of circle = 2πr

⇒ Circumference = 2 × 22/7 × r

Total distance traveled = (Number of revolution) × (Circumference of the circle)

⇒ 4.4 × 100000 cm = 2000 × 2 × 22/7 × r

⇒ r = (4.4 × 100000 × 7)/(2000 × 2 × 22)

⇒ r = 35 cm

Diameter = 2 × Radius

⇒ Diameter = 2 × 35 cm

⇒ Diameter = 70 cm

∴ The diameter of wheel is 70 cm.

Given:

Length = 12 m

Breadth = 8 m

Height = 10 m

Cost of 1 m²  wall painting = Rs. 25 

Formula used:

Area of four walls = 2(l + b) × h

Where,

Length = l 

Breadth = b 

Height = h

Calculation:

Area of four walls = 2(l + b) × h

⇒ 2(12 + 8) × 10

⇒ 2 × 20 × 10

⇒ 400 m²

Cost of 1 m2 wall painting = Rs. 25 

Cost of 400 m2 wall painting = Rs. 25 × 400 = 10000

∴ Total cost (in Rs.) of whitewashing only all four walls of the room is Rs. 10000.

Given:

Area = 400 hectares

Length = 500 m

Formula used:

Area of rectangle = Length × Width

1 Hectare = 10000 m²

1 m = 100 cm

1 km = 1000 m

Calculations:

⇒ 400 × 10000 = 500 × Width

⇒ Width = 8000 m

⇒ Width = 8 km

∴ The width of the rectangular field is 8 km

Given :

Circumference of base of hemisphere = 88 m

 Radius of hemisphere : height of cylinder = 1 : 4

 Cost of painting = Rs. 2 per meter sq.

Formula Used:

 Area of paint = curved surface area of hemisphere + curved surface area of cylinder

 Cost of painting = total curved surface area × cost of painting

Calculations:

Let the radius of hemisphere and height of cylinder be r and h respectively

⇒ Circumference of base of hemisphere = 2πr = 88 m

⇒ r = 14 m

⇒Height of cylinder =  14 × 4

⇒ h =  56 m

⇒CSA of hemisphere = 2πr² = 2 π × 14 ×14

⇒CSA of hemisphere = 1232 m²

⇒ CSA of cylinder = 2πrh= 2π × 14× 56

⇒ CSA of cylinder = 4928 m²

⇒ Total curved surface area = (4928 + 1232) m²

 ⇒ Total curved surface area =  6160 m²

⇒Cost of painting = Rs. 6160 × 0.1

⇒ Cost of painting = Rs. 616

Given:

Radius of cone = 7 cm

Curved surface area of cone = 220 cm²  

Formula used:

Curved surface area of cone = πrl

Where, r = radius of cone

l = slant height of cone

h = height of cone

l² = r² + h²

Calculation:

220 = π × 7 × l

⇒ l = 220/22 = 10 cm

Also,

10² = 7² + h²

⇒ h² = 100 – 49 = 51

⇒ h = √51 cm

∴ The height of the cone is √51 cm. 

Given:

Height of the cone = 21 cm

Formula used:

Curved surface area of a right circular cone = πrl

Volume of a cone = \(\left( {\frac{1}{3}} \right) \times {\rm{\pi }} \times {{\rm{r}}^2} \times {\rm{h}}\)

Area of circle = πr²

\({\rm{h}} = {\rm{\;}}\sqrt {{{\rm{l}}^{2{\rm{\;}}}} - {\rm{\;}}{{\rm{r}}^2}} \)

Where r = radius, l = slant height, h = height of a cone

Calculation:

According to the question,

πrl = 3πr²

⇒ l = 3r

⇒ l/r = 3/1

\({\rm{h}} = {\rm{\;}}\sqrt {{{\rm{l}}^{2{\rm{\;}}}} - {\rm{\;}}{{\rm{r}}^2}} \)

⇒ \({\rm{h}} = {\rm{\;}}\sqrt {{{\rm{3}}^{2{\rm{\;}}}} - {\rm{\;}}{{\rm{1}}^2}} \)

⇒ \({\rm{h}} = {\rm{\;}}\sqrt {{{\rm{9}{\rm{\;}}}} - {\rm{\;}}{{\rm{1}}}} \)

⇒ h = √8

⇒ h = 2√2

⇒ 21 = 2√2

⇒ r = 21/2√2

\(\left( {\frac{1}{3}} \right) \times {\rm{\pi }} \times {{\rm{r}}^2} \times {\rm{h}}\) = \(\left( {\frac{1}{3}} \right) \times {\frac{22}{7}} \times ({{\frac{21}{2\sqrt 2}})^2} \times {\rm{21}}\) 

⇒\(\left( {\frac{1}{3}} \right) \times {\frac{22}{7}} \times ({{\frac{441}{8}})} \times {\rm{21}}\)

⇒ 22 × 63/8 × 7

⇒ 1212.75

⇒ Volume = 1212.75 cm³ ≈ 1213 cm³

∴ Volume of the cone is 1213 cm³

Given:

The height and radius of cone is 8 cm and 6 cm respectively

Formula used:

Curved surface area of cone = π r l

Slant height = √ (h)² + (r)²

Where r is radius, h is height and l is the slant height of cone

Calculation:

To calculate the curved surface area first we have to calculate the slant height

∴ Slant height = √ (8)² + (6)² = √100 = 10 cm

Now, curved surface area of cone = π r l = (22/7) × 6 × 10 = 1320/7 square cm

Hence, option (2) is correct

Given:

Volume = 15400 cm3

Height, (h) = 25 cm

Formula used:

The volume of the cylinder = π r2 h

The curved surface area of the cylinder = 2 π r h

Here, h and r is the height and radius respectively

Calculation:

Let the radius of the cylinder be r

The volume of the cylinder = π r2 h

⇒ 15400 = 22/7 × r2 × 25

⇒ r2 = 196

⇒ r = 14 cm

Now, Curved surface area of the cylinder = 2 π r h

Curved surface area of the cylinder = 2 × 22/7 × 14 × 25 = 2200

∴ The curved surface area of the cylinder is 2200 cm2

Given:

Area of square = 484 sq. cm

Square wire is converted into circle

Formula used:

Perimeter of circle = 2π r, where r = radius of circle

Area of circle = π r²

Perimeter of square = 4a, where a = side of the square

Concept used:

When square shape bent into circle, perimeter of square becomes equal to perimeter of circle

Area of square = a²

∴ a = √area = √484 = 22 cm

∴ perimeter of square = 4a = 4 × 22 = 88 cm

Now, perimeter of circle = perimeter of square = 88 cm

⇒ 2π r = 88

∴ r = 88 / 2π = 88 × 7 / 22 × 1 / 2 = 14 cm

∴ Area of circle = π r² = 22 / 7 × 14² = 22 / 7 × 196 = 616 sq.cm

Given:
Ratio of the radius and the height = 4 : 7
Volume of a cylinder = 2816 cm³

Concept used:
Volume of a cylinder = π × r² × h

Calculation:
Let the radius and the height of a cylinder be 4x and 7x respectively.
⇒ 22/7 × (4x)² × 7x = 2816
⇒ 16x² × 7x = (2816 × 7)/22
⇒ (16 × 7)x³ = 128 × 7
⇒ x³ = (128 × 7)/(16 × 7)
⇒ x³ = 8
⇒ x = 2
Radius = 4 × 2 = 8 cm
∴ The radius of a cylinder is 8 cm.

Given:

The curved surface area of a cylinder is five times the area of its base.

Concept used:

The curved surface area of a cylinder = 2πrh

The area of base of a cylinder = πr²

Calculation:

According to the question,

2πrh/πr² = 5/1

⇒ 2h/r = 5/1

⇒ r : h = 2 : 5

The ratio of the radius and height of the cylinder is 2 : 5.

Given:

Speed of water flowing through the pipe = 2.5 km/h = 2500 m/h

Radius of pipe = 3.5 cm = 0.035 m

Length of tank = 10 m

Breadth of tank = 8.8 m

Height upto which water to be raised in the tank = 42 cm = 0.42m

Formula used:

(i) Volume flow rate of water = A × V

where,

A = Cross section area of pipe

V = Speed of water

(ii) Volume of cuboid tank = l × b × h

where,

l = length of tank

b = breadth of tank

h = height of tank

(iii) Time taken to fill the tank = (Total volume to be filled inside the tank)/(Volume flow rate of water)

Calculation:

Volume flow rate of water through pipe = 2500 × π × 0.035 × 0.035

⇒ 9.62 m³/hr

Total volume of water to be filled inside the tank = 10 × 8.8 × 0.42

⇒ 36.96 m³

Time taken to fill the tank = 36.96/9.62

⇒ 3.84 hours

∴ The time (in hours) in which the level of water in the tank will rise by 42 cm is 3.84 hours.

Mistake point:

Don't forget to make all units same.

Given:

New cylinder radius = 2 × Radius of original cylinder

New cylinder height = Original cylinder height

Formula used:

Volume of cylinder= π × (radius)² × height

Calculation:

Let the radius of original cylinder be r cm.

Radius of new cylinder = 2r cm

Height of new and original cylinder = h cm

Volume of original cylinder = πr²h cm²

Volume of new cylinder = π(2r)²h

Now, change in volume = 4πr²h – πr²h = 3πr²h

Given:

Radius = 28 cm 

Height = 5 cm

Formula used:

The curved surface area of the cylinder = 2 π r h

Calculation:

The curved surface area of the cylinder = 2 π r h

⇒ 2 × (22/7) × 28 × 5

⇒ 880 cm²

∴ The curved surface area of the cylinder is 880 cm²

Given:

The number of revolutions = 1000

And total cover distance = 440 m.

Formula used:

One revolution = total distance covered by wheel/the number of revolutions

Circumference of wheel = 2 × π × r = distance covered by a wheel in one revolution

Diameter of a wheel = 2 × r

Calculation:

Distance in one revolution = 440/1000

⇒ Distance in one revolution = 0.44 m

⇒ 2 × π × r = 0.44

⇒ r = 0.44/(2 × π), where π = 22/7

⇒ r = 0.07 m

Diameter = 2 × 0.07

⇒ Diameter = 0.14 m

∴ The diameter (in meter) of the wheel is 0.14 m.

Given:

Area of wire bent in the form of square = 121 cm² 

Concepts used:

Area of square = (Side)²

The perimeter of wire bent in the form of square = Perimeter of wire bent in the form of the circle

The perimeter of square = 4 × side of the square

Perimeter/Circumference of circle = 2πr

Area of circle = πr²

Where r → Radius

Calculation:

Area of square = (Side)2 

⇒ 121 cm² = (Side)²

⇒ Side = √121 cm

⇒ Side of square = 11 cm

The perimeter of wire bent in the form of square = 4 × side of the square

⇒ 4 × 11 cm

⇒ 44 cm

Let the radius of the circle be r cm.

Perimeter of wire bent in the form of square = Perimeter of wire bent in the form of circle

44 cm = 2πr

⇒ 44 cm = 2 × (22/7) × r

⇒ (44 × 7)/44 cm = r

⇒ r = 7 cm

Area of circle = πr² 

⇒ 22/7 × (7)² cm² 

⇒ (22/7) × 49 cm² 

⇒ 154 cm² 

∴ The wire encloses an area of 154 cm² when bent in the form of circle.

Given :- 

Diameter = 14√3

Concept :- 

Radius = Diameter/2

Total surface area of sphere = 3πR²

Where, π = (22/7)

Calculation :-

⇒ Radius = (14√3)/2 = 7√3

⇒ Total surface area = 3 × (22/7) × (7√3)²

⇒ Total surface area = 1386 cm²

∴ Total surface area = 1386 cm²

Formula used :

Area of the circle = π.r² (where r is the radius of the circle)

Diameter of a circle = 2 × the radius of the circle  

Calculations :

Let the diameter be x 

So the radius will be 2x

⇒ Area of the original circle = π (2x)² = 4.π.x²

Now the diameter has been doubled 

The new diameter will be 2x

The new radius will be 4x 

⇒ Area of the circle with new circle = π.(4x)² = 16.π.x² 

⇒ Area of the new circle : Area of the original circle = 16 : 4 

⇒ 4 : 1

∴ the area of the new circle will be 4 times the original circle

Given:

A metal wire when bent in the form of a square encloses an area 121 cm2.

Formula used:

Area of circle = \(\pi \) × r²

Radius = r , \(\pi \) = 22/7

Calculation: 

Side of square = √ area

⇒ √ 121

⇒ 11 cm

Length of the wire or circumference of the wire = 4 × side

⇒ 4 × 11

⇒ 44 cm

Circumference of the circle = 2\(\pi \)r

⇒  2\(\pi \)r = 44

⇒ r = (44 × 7)/(22 × 2)

⇒ r = 7 cm

Area of the circle =  \(\pi \) × r2

⇒ (22/7) × 7 × 7

⇒ 154 cm²

∴ Area of the circle is 154 cm².

Given:

Height = 25 cm

Area = 200 cm2

Calculations:

We know that,

Area of a rectangle = length × breadth

So, the breadth of the rectangle = 200/25 = 8 cm

∴ The breadth of the given rectangle is 8 cm.

Given:

Ratio of length and breadth of rectangle = 6 : 5

Area of rectangle = 6750 cm²

Formula used:

Area of rectangle = Length × Breadth

Calculation:

Let the length and breadth be 6x and 5x respectively

According to question,

6x × 5x = 6750 cm²

⇒ 30x² = 6750 cm²

⇒ x² = 225 cm²

⇒ x = 15 cm

Breadth = 5 × 15 cm

⇒ 75 cm

Required ratio = 75 cm : 6750 cm²

⇒ 1 : 90

∴ The ratio of its breadth and area is 1 : 90