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Given:

Perimeter of rectangle = 420 m

Length = Breadth + 30 m

Rate of speed = 10 m/s

Formula used:

Perimeter of rectangle = 2 × (Length + Breadth)

Length of diagonal = √(Length² + Breadth²)

Time taken = Distance/Speed

Calculation:

Let the breadth of rectangle be x meter

Then, length of rectangle = (x + 30) m

Now, according to question,

2 × (x + x + 30) = 420 m

⇒ 2 × (2x + 30)  = 420 m

⇒ 4x + 60 = 420 m

⇒ 4x = 360

⇒ x = 90

Breadth = 90 m

Length = (90 + 30) m

⇒ 120 m

Length of diagonal = √(120² + 90²) m

⇒ √(14400 + 8100) m

⇒ √22500 m

⇒ 150 m

So, Time taken = 150/10 seconds

⇒ 15 seconds

∴ The time taken to cross diagonally is 15 seconds

Given:

Radius of circle =  Diagonal of square

Perimeter of Square = √3 × area of equilateral triangle 

Circumradius of equilateral triangle = 4√3 cm

Formulas Used:

CircumRadius (R) = Side of equilateral triangle/√3

Area of Equilateral Triangle = (√3/4) × (Side)²

Perimeter of Square = 4 × Side of square(a)

Diagonal of Square = √2 × Side of Square

Area of circle = π × (Radius)²

Calculation:

Side of equilateral triangle = √3 × 4√3 = 12 cm

Area of Equilateral Triangle = (√3/4) × (12)² = 36√3 cm²

Here it is given that perimeter of square is √3 times the Area of equilateral triangle

Perimeter of Square = 4 × a = √3 × 36√3 = 108 cm

Side of Square (a) = 108/4 = 27 cm

Diagonal of Square = √2 × a = √2 × 27 = 27√2 cm

Radius of Circle (r) = Diagonal of square

⇒ r = 27√2 cm

Area of circle = π × (27√2)² = 1458π 

Double the area of circle = 2 × 1458π = 2916π 

∴ Double the area of the circle is 2916π.

Given:

Breadth of rectangle = a

Length of rectangle = 2a

Semi-circle is drawn on the diagonal of rectangle.

Formulas Used:

Area of rectangle = Length × breadth

Diagonal of rectangle = √[(Length)² + (Breadth)²]

Area of Semi-circle = [π × (radius)²]/2

Calculation:

Area of rectangle = 2a × a = 2a² 

Diagonal of rectangle = √[(2a)² + (a)²] = √5 × a

Diagonal of rectangle will be diameter of the semi-circle

So radius of semi-circle = (√5 × a)/2

Area of Semi-circle = π/2 × [(√5 × a)/2]²

⇒ Area of Semi-circle = 5a²π/8

Area of semi-circle/Area of rectangle = (5a²π/8)/2a²

⇒ Area of semi-circle/Area of rectangle = (5 × 22 × a²)/(2 × 8 × 7 × a²) = 55/56

⇒ Area of semi-circle/Area of rectangle = 55/56

∴ The Ratio between the Area of semi-circle and the Area of rectangle is 55 : 56

Given:

Length of the rectangle = 80 m

The breadth of the rectangle = 60 m

Diagonal of the rectangle = a

Side of new square = a

Formula:

(Diagonal)² = (Length)² + (Breadth)²

Area of the square = a²

Calculation:

a² = 80² + 60²

⇒ a² = 6400 + 3600

⇒ a² = 10000

⇒ a = 100 m

∴ Area of the square = 100 × 100 = 10,000 m²

Given:

The ratio of length, width and height = 3 : 2 : 1

Volume = 3072 cubic meters

Formula used:

The volume of cuboid = length × width × height

Calculation:

Let the length, width, and height be 3x, 2x, and x meters.

The volume of room = length × width × height

⇒ 3072 = 3x × 2x × x

⇒ 6x³ = 3072

⇒ x³ = 512

⇒ x = 8

Width = 2x

⇒ Width = 2 × 8

⇒ Width = 16 m

∴ The width of room is 16 meter.

Given:

The difference between length and breadth of rectangle is 3 cm and area is 28 square cm

Formula used:

Area of rectangle = Length × Breadth

Calculation:

Let the Length of the rectangle be x cm and breadth be y cm

∴ x – y = 3 cm

⇒ x = (y + 3)

Now area of rectangle is 28 square cm

∴ y × (y + 3) = 28

⇒ y² + 3y – 28 = 0

⇒ (y + 7) × (y - 4) = 0

⇒ y = 4 and -7

So, the breadth of rectangle is 4 cm and length is 7 cm

∴ Sum of length and breadth = 4 + 7 = 11 cm

Hence, option (2) is correct

Given:

The ratio of area of square is 1 ∶ 2

Formula used:

Area of square = Side × side

Length of diagonal of square = √2 × side

Calculation:

Let the area of first square is A1 and area of Second Square is A2 similarly S1 and S2 is sides and D1 and D2 are the diagonals of the square

∴ A1 ∶ A2 = 1 ∶ 2

Now, we know that

Area of square = Side × side

∴ S1 ∶ S2 = 1 ∶ √2

Now ratio of diagonals = √2 × 1 ∶ √2 × √2 = 1 ∶ √2

∴ D1 ∶ D2 = 1 ∶ √2

Hence, option (1) is correct

Given:

Diagonals of a rhombus are 24 cm and 32 cm respectively

Formula used:

Perimeter of Rhombus = 4 × side

Area of Rhombus = (d₁ × d₂)/2

Area of Rhombus = Side × Altitude

d₁² + d₂² = 4 × s²

Here, d₁, d₂, and s are diagonals and side of rhombus respectively.

Concept used:

All sides of a rhombus are equal and diagonals bisect each other at a right angle.

Calculation:

Area of Rhombus = (d₁ × d₂)/2

⇒ Area of Rhombus = (24 × 32)/2

⇒ Area of Rhombus = 384

Now, d₁² + d₂² = 4 × s²

⇒ 24² + 32² = 4 × s²

⇒ s² = 400

⇒ s = 20

Perimeter = 4 × side = 4 × 20 = 80

∴ The perimeter and area of the Rhombus are 80 cm and 384 cm²

Given:

ABCD is a trapezium, AB ∥ DC

M is mid-point of AD

N is mid-point of BC

AB = 40.25 cm

MN = 49 cm

Formula Used:

MN = 1/2 × (AB + CD)

In a trapezium AB ∥ DC

And M and N are mid-points of AD and BC respectively.

Calculation:

MN = 1/2 × (AB + CD)

⇒ 49 = 1/2 × (40.25 + CD)

⇒ 98 = 40.25 + CD

⇒ CD = 98 – 40.25

⇒ CD = 57.75 cm

∴ The length of CD is 57.75 cm.

Given:

ABCD is a trapezium, AB ∥ DC

M is mid-point of AD

N is mid-point of BC

AB = 18 cm

CD = 12 cm

Formula Used:

MN = 1/2 × (AB + CD)

In a trapezium AB ∥ DC

And M and N are mid-points of AD and BC respectively.

Calculation:

MN = 1/2 × (AB + CD)

⇒ MN = 1/2 × (18 + 12)

⇒ MN = 1/2 × 30

⇒ MN = 15

∴ The length of MN is 15 cm.

Given:

ABCD is a trapezium, AB ∥ DC

M is mid-point of AD

N is mid-point of BC

AB = 71.5 cm

CD = 52 cm

Formula Used:

MN = 1/2 × (AB + CD)

In a trapezium AB ∥ DC

And M and N are mid-points of AD and BC respectively.

Calculation:

MN = 1/2 × (AB + CD)

⇒ MN = 1/2 × (71.5 + 52)

⇒ MN = 1/2 × 123.5

⇒ MN = 61.75 cm

∴ The length of MN is 61.75 cm.

Given:

Total distance = 44km

Number of rotation = 5000

Concept used:

Total distance = Number of rotation × Circumference of wheels

Circumference of circle = 2πr

Calculation:

Total distance = Number of rotation × Circumference of wheels

⇒ 44 × 100000 cm = 5000 × 2πr

⇒ 44 × 100000 cm = 5000 × 2 × 22/7 × r

⇒ r = (44 × 100000 × 7)/(5000 × 2 × 22)

⇒ r = 140 cm

∴ The radius of wheels is 140 cm.

Given:

ABCD is a trapezium, AB ∥ DC

AB = 17 cm

CD = 13 cm

Height of the trapezium = 12 cm

Formula Used:

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

Calculation:

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

⇒ 1/2 × 12 × (17 + 13)

⇒ 6 × 30

⇒ 180

∴ The area of trapezium is 180 cm².

Given:

ABCD is a trapezium, AB ∥ DC

M is mid-point of AD

N is mid-point of BC

CD = 19 cm

MN = 15.5 cm

Formula Used:

MN = 1/2 × (AB + CD)

In a trapezium AB ∥ DC

And M and N are mid-points of AD and BC respectively.

Calculation:

MN = 1/2 × (AB + CD)

⇒ 15.5 = 1/2 × (AB + 19)

⇒ 31 = AB + 19

⇒ AB = 31 – 19

⇒ AB = 12 cm

∴ The length of AB is 12 cm.

Given:

ABCD is a trapezium, AB ∥ DC

M is mid-point of AD

N is mid-point of BC

AB = 14 cm

MN = 17 cm

Formula Used:

MN = 1/2 × (AB + CD)

In a trapezium AB ∥ DC

And M and N are mid-points of AD and BC respectively.

Calculation:

MN = 1/2 × (AB + CD)

⇒ 17 = 1/2 × (14 + CD)

⇒ 34 = 14 + CD

⇒ CD = 34 – 14

⇒ CD = 20 cm

∴ The length of CD is 20 cm.

Concept:

Let the radius of a sphere is R meters,

The volume of the sphere = \(V=\frac{4}{3}{\rm{\pi }}{{\rm{R}}^3}\)

Surface area of the sphere = S = 4πr2

Calculation:

Given that V = 792/7 cc

\(\frac{792}{7}=4\times\frac{22}{7}\times\frac{R^3}{3}\)

R³ = 27

R = 3 cm

Given:

ABCD is a trapezium, AB ∥ DC

The area of ABCD = 180 cm²

AB = 17 cm

Height of the trapezium = 12 cm

Formula Used:

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

Calculation:

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

⇒ 180 = 1/2 × 12 × (AB + CD)

⇒ 180 = 6 × (17 + CD)

⇒ CD = 30 – 17

⇒ CD = 13

GIVEN:

Three statements.

CONCEPT:

Mensuration

FORMULA USED:

Area of the floor of the room = LB

Volume of the room = LBH

Length of body (main) diagonal of the room = √(L² + B² + H²)

CALCULATION:

H = 12 cm

Let L = 6x and B = 5x

Now,

2 (12 × 6x + 12 × 5x) = 792

⇒ 264x = 792

⇒ x = 3

L = 6x = 18 cm

B = 5x = 15 cm

A:

Area of the floor of the room = LB

= 18 × 15

= 270 cm²

B:

Volume of the room = LBH

= 18 × 15 × 12

= 3240 cm³

C:

Length of body (main) diagonal of the room = √(L² + B² + H²)

= √(18² + 15² + 12²)

= √693

= 3√77 cm

Hence, only statement B is TRUE.

Given:

The length of the room is (3x + 10) m and breadth of the room is (2x + 5) m. The area of four walls of the room is (60x + 180) m². We have to find the height of the room.

Concept Used:

If the length, breadth and height of a room be l, b and h then the area of four walls of the room is [2 × (l + b) × h]

Calculation:

Let, the height of the room be h

Accordingly,

2 × {(3x + 10) + (2x + 5)} × h = (60x + 180)

⇒ (5x + 15)h = 30x + 90

⇒ (5x + 15)h = 6(5x + 15)

⇒ h = 6

∴ The height of the room is 6 m.

Given:

A sphere of maximum volume is cut out from a solid hemisphere of radius r.

Concept used:

Volume of hemisphere = (2/3)πr³

Calculation:

A sphere of maximum volume is cut.

Hence the diameter of sphere = r

Volume = \(\frac{4}{3} \times \pi \times {\left( {\frac{r}{2}} \right)^3}\)

⇒ \(\frac{4}{3} \times \pi \times \frac{{{r^3}}}{8}\)

⇒ \(\frac{{\pi {r^3}}}{6}\)

Required Ratio of the volume of the hemisphere to that of the sphere

⇒ \(\frac{2}{3}\pi {r^3}:\frac{1}{6}\pi {r^3}\)

∴ 4 : 1

Given:

Ratio of the radius of sphere and hemisphere = 2 : 1

Formula Used:

Total surface area of a Sphere = 4πr² 

Total surface area of a hemisphere = 3πr²

Solution:

Let the radius of a hemisphere be x

⇒ Then the radius of a sphere is 2x.

⇒ Total surface area of a Sphere = 4πr2 

⇒ Total surface area of a Sphere = 4π × (2x)²

⇒ Total surface area of a hemisphere = 3πr²

⇒ Total surface area of a hemisphere = 3π × (x)²

Now the ratio of their respective areas,

⇒ 4π(2x)2/3π(x)2

⇒ 4(4x²)/3(x²)

⇒ 16/3

∴ The ratio of their total surface area is 16 : 3 

Concept

Total surface area of sphere = 4πr²

Total surface area of hemisphere = 3πr²

Calculation

Ratio of total surface area of a hemisphere to surface area of sphere = 3πr² : 4πr²

⇒ 3 : 4

Given:

Area(A) of the equilateral triangle = 4√3 cm²

Formula used:

Area of an equilateral triangle = (√3/4)a²; a = length of the side of the triangle

Area of a triangle = bh/2; b = base(side) of the triangle, h =height of the triangle 

Area of a square = x²

Diagonal of a square = √2x 

x = length of the side of the square

Calculation:

According to the question:

(√3/4)a² = 4√3

⇒ a = 4 cm = b

Also, 4√3 = bh/2

⇒ 4√3 = 4h/2

⇒ h = 2√3 = x

Diagonal of the square = √2x = √2(2√3)

∴ Diagonal of the square = 2√6 cm

Given:

Side of cube A = 4√3 cm

Diagonal of cube A = Side of cube B

Formula used:

Diagonal of cube = √3a

Volume of cube = a³

Calculation:

Let, the side of cube A = a

Side of cube B = b

∵ Diagonal of cube A = √3a

⇒ Diagonal of cube A = √3 × 4√3

= 12 cm

∵ Diagonal of cube A = Side of cube B

∴ Side of cube B = 12cm

⇒ Volume of Cube B = (12)³

= 1728 cm³

Given:

The total surface area of the room is 96 m²

Concept Used:

The total surface area of a cube of side ‘a’ unit is 6a² square unit

The maximum length of a rod can be placed in a room is as the same length of the diagonal of the room.

Length of the diagonal of a cube of side ‘a’ unit is a√3 unit

Calculation:

Let, length of a side of the cube is a unit

The total surface area of the cube is 6a² unit²

Accordingly,

6a² = 96

⇒ a² = 16

⇒ a = 4

Sides of the cube is 4 meter

Length of the diagonal of the cube is 4√3 meter

∴ The maximum length of the rod which can place in that room is 4√3 meter.

Given:

Radius of three iron balls are 5 cm, 4 cm and 3 cm.

Radius of bigger iron ball = x cm

Formula:

Volume of sphere = (4 / 3) × πr³

Calculation:

According to the question

(4 / 3) × π × x³ = (4 / 3) × π × [5³ + 4³ + 3³]

⇒ x³ = 125 + 64 + 27

⇒ x³ = 216

∴ x = 6 cm

Given:

Radius of hemisphere = Radius of sphere/2

Formula used:

Volume of sphere = 4/3 × π × (radius)³

Volume of hemisphere = 2/3 × π × (radius)³

Calculation:

Let the radius of sphere is r cm

So, radius of hemisphere = r/2 cm

Now,

Volume of sphere/Volume of hemisphere = [(4/3)πr³]/[(2/3)π(r/2)³]

⇒ 2/(1/8)

Given:

Diagonal of the cube is 4√3 m

Concept Used:

Length of the diagonal of a cube of side ‘a’ unit is a√3 unit

The volume of a cube of side ‘a’ unit is a³ cube unit

If ‘a’ be the side of a cube then total surface area of the cube is 6a²

Calculation:

Let, the length of the side of the cube be a

Accordingly,

a√3 = 4√3

⇒ a = 4

Length of the side of the cube is 4 m

The volume of the cube is 4³ = 64 m³

The total surface area of the cube is 6 × 4² = 96 m²

The ratio of the numerical value of volume and total surface area is 64 : 96

⇒ 2 : 3

∴ The ratio of the numerical value of volume and total surface area is 2 : 3

Given:

Diagonal of a cube is of length = 3 l

Concept:

Diagonal of cube = √3 × a

The total surface area of cube = 6 × a²

Where a = side

Explanation:

According to the question,

√3 × a = 3 l

⇒ a = √3 l

The total surface area of cube = 6 × (√3 l)²

⇒ The total surface area of cube = 18 l²

∴ The total surface area of the cube is 18 l².

Given:

The numerical value of total surface area is twice to the volume of the cube.

Concept:

The volume of the cube = a³

The total surface area of the cube = 6 × a²

Diagonal of the cube = √3 × a

Where a = side 

Explanation:

according to the question,

6 × a² = 2 × a³

⇒ 6 = 2 × a

⇒ a = 3 cm

Diagonal of the cube = √3 × 3 = 3√3 cm

∴ The square of the diagonal of the cube is (3√3)² is 27 cm².

Given:

Height of cone = 3 cm

Radius of cone = 7 cm

Height of cylinder = 1 cm

Formula used:

Volume of cone = (1/3)πr²h

Where, r and h represents radius and height of cone

Calculation:

Volume of cone = (1/3) × (22/7) × 7 × 7 × 3 = 154 cm²

Volume of cone = Volume of cylinder

⇒ 154 = π(radius)²height

⇒ 154 = (22/7) × (radius)² × 1

Given:

Parallel sides of a trapezium are of length 17 cm and 11 cm

Height of trapezium = 8 cm

Cost of silver plating is Rs 2 per cm² 

Formula used:

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

Calculation:

Sum of parallel sides = 17 + 11 = 28

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

⇒ Area of trapezium = (1/2) × 28 × 8

⇒ Area of trapezium is = 112 cm²

The total cost of silver plating = 2 × 112 = 224

∴ The total cost of silver plating a trapezium plate is Rs. 224

Calculation:

Circumference of circle = 1 m = 100 cm

Formula used:

Circumference of circle = 2πr

Area of circle = πr2

Value of π = 22/7

Calculation:

According to question,

2 × (22/7) × r = 100 cm

⇒ r = 700/44 cm

⇒ 175/11 cm

Now, area of circle = π × (175/11) × (175/11) cm2

⇒ (22/7) × (175/11) × (175/11) cm2

⇒ 795.45 cm2 

∴ The area of circle is 795.45 cm2

Given: 

Face diagonal of the cube is 10 cm

Concept:  

Face diagonal of cube = √2 × side

The volume of the cube = (Side)³

Calculation: 

Face diagonal of cube =  √2 × side

⇒ 10 = √2 × side

⇒ Side = 5√2 cm 

The volume of the cube is 

⇒ (5√2)³ 

⇒ 250√2 cm³

∴ The required volume of the cube is 250√2 cm³.  

Given:

Surface area of cube = 2 × Volume of a cube

Formula used:

Volume of a cube = (Side)³

Surface area of cube = 6(Side)²

Calculation:

Surface area of cube = 2 × Volume of cube

⇒ 6(Side)² = 2(Side)³

⇒ Side = 3 units

Given:

Ratio of radii of two spheres = 6 : 11

Formula used:

Volume of sphere = 4/3 × (π) × (r)³

Where, r represents radius of sphere.

Calculation:

Let r₁ and r₂ are the radii of first and second sphere.

Volume of first sphere/Volume of second sphere = [(4/3)π(r₁)³]/[(4/3)π(r₂)³]

Let radius of spheres be 6x and 11x.

⇒ (4/3) × π × (6x)³/[(4/3) × π × (11x)³]

⇒ (6 × 6 × 6)/(11 × 11 × 11)

Given:

Area of wheel = 616 cm²

Total distance = 1.76 km

Concept used:

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

Calculation:

Area = πr²

⇒ 616 = 22/7 × r²

⇒ r² = 28 × 7

⇒ r² = √196

⇒ r = 14 cm

Circumference = 2πr

⇒ Circumference = 2 × 22/7 × 14

⇒ Circumference = 88 cm

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

⇒ 1.76 km = Number of revolution × 88

⇒ Number of revolution = 1.76/88 × 100000

⇒ Number of revolution = 2000

∴ The wheel makes 2000 revolution during the journey.

Given:

The circumference of the base of the container is 88 cm and the height of the container is 10 cm.

Concept Used:

The base of a cylinder is a circle, so, the circumference of the base is 2πr where r is the radius of the base

The volume of a cylinder is πr²h where r is the radius of the base and h is the height of the cylinder

Calculation:

Let, the radius of the base be r

Accordingly,

2πr = 88

⇒ 2 × 22/7 × r = 88

⇒ r = 14

The volume of the glass is π(14)² × 10

⇒ (22/7) × 14 × 14 × 10

⇒ 6160 cm³

∴ The capacity of the glass is 6160 cm³.

Given:
Length of a room = 16 m

Breadth of a room = 12 m

Height of a room = 20 m

Rate of white washing = Rs. 5000 per cm² = Rs. 0.5 per m²

Rate of painting = Rs. 3 per m²

Concept used:

Area of four walls = 2(l + b) × h 

Area of rectangular ceiling = l × b

Cost of painting = Rate × Total area of painting

Calculation:

Area of four walls = 2(16 + 12) × 20

⇒ 2 × 28 × 20

⇒ 1120 m²

Cost of white washing the walls = 1120 × 0.5  

⇒ Rs. 560

Area of ceiling = 16 × 12

⇒ 192 m²

Cost of painting the ceiling = 192 × 3

⇒ Rs. 576

Total cost of painting = 560 + 576 = Rs. 1136

∴ The total cost of painting the room is Rs. 1136.

Given:

Curved surface area of cylinder = 3300 cm²

Height of cylinder = 25 cm

Formula used:

Curved surface area of cylinder = 2π × radius × height

Calculation:

2 × (22/7) × radius × 25 = 3300

⇒ Radius = (3300 ×7)/(2 × 22)

⇒ Radius = 21 cm

Given:

Area of base = 616 cm²

Height = 8 cm

Formula used:

Area of circle = πr²

Curved surface area (CSA) of cylinder = 2πrh

Calculation:

Area of base = Area of circle = πr²

⇒ πr² = 616

⇒ (22/7) × r² = 616

⇒ r² = (616 × 7)/22

⇒ r² = 196

⇒ r = 14 cm

CSA of cylinder = 2πrh = 2 × (22/7) × 14 × 8

⇒ CSA of cylinder = 704

∴ Curved surface area of cylinder is 704 cm²

Given:

The curved surface area of the right cylinder = 3696 cm²

Height of cylinder = 3 × radius of the cylinder

Formula used:

The curved surface area of cylinder = 2πrh

Volume of cylinder = πr²h

Where,

r = radius of cylinder

h = height of cylinder

Concept used:

1 cm³ = 0.001 litre

Calculation:

Let the radius of the cylinder be ‘r’ and the height of the cylinder be ‘h’.

Curved surface area of cylinder = 2πrh

⇒ 2πrh = 3696

⇒ 2πr × 3r = 3696      (h = 3r)

⇒ r² = (3696 × 7)/(22 × 6)

⇒ r² = 196

⇒ r = 14 cm

⇒ h = 3r = 3 × 14 = 42 cm

Volume of cylinder = πr²h

⇒ (22/7) × 14 × 14 × 42

⇒ 25872 cm³

We know, 1 cm³ = 0.001 litre

⇒ 25872 × 0.001

⇒ 25.872 litres

Given:

Areas of 3 adjacent faces of cuboidal tank = 3 m2, 12 m2, 16 m2

Formula Used:

1 m³ = 1000 Litres

If areas of 3 adjacent faces of a cuboid are x, y and z respectively, then

Volume of cuboid = √(x × y × z)

Calculations:

Areas of 3 adjacent faces of cuboidal tank = 3 m2, 12 m2, 16 m2

Volume of cuboid = √(x × y × z)

⇒ Capacity of tank = √(3 × 12 × 16) m³

⇒ Capacity of tank = 24 m³

1 m3 = 1000 Litres

⇒ Capacity of tank = 24 × 1000 lites

⇒ Capacity of tank = 24000 litres

∴ The capacity of the tank, (in litres) is 24000 litres.

Given:

The length of open cuboidal cistern is 1.6 m.

The breadth of open cuboidal cistern is 1.3 m.

The height of open cuboidal cistern is 53 cm.

The capacity of open cuboidal cistern is 900 litres.

The thickness of wall is 5 cm.

Formula Used:

Volume of Cuboidal  = length × Breadth × Height

Calculation:

The length of open cuboidal cistern = 1.6 m = 160 cm

The thickness of wall is 5 cm.

The resultant length of open cuboidal cistern = 160 - 2 × 5 = 150 cm

The breadth of open cuboidal cistern = 1.3 m = 130 cm

The thickness of wall is 5 cm.

The resultant breadth of open cuboidal cistern = 130 - 2 × 5 = 120 cm

The height of open cuboidal cistern = 53 cm 

The thickness of bottom is x cm.

The resultant height of open cuboidal cistern = (53 - x) cm

The capacity of open cuboidal cistern = 900 litres = 900 × 1000 cm³

⇒ 150 × 120 × (53 - x) = 900 × 1000 

⇒ 180 × (53 - x) = 9 × 1000 

⇒ 20 × (53 - x) = 1 × 1000 

⇒ 53 - x = 50

⇒ x = 3

∴ The thickness of bottom is 3 cm.

Given:

The radius of a circular wheel is 1.75 m. Distance covered is 11 km

Concept used:

Number of revolutions (n) = d/2πr

Calculation:

Radius of wheel (r) = 1.75

Number of revolutions 

⇒ \(\frac{{11\ \times\ 1000\ \times\ 7}}{{2\ \times\ 22\ \times\ 1.75}}\)

∴ 1000

Given:

First cube edge (a₁) = 3 cm

Second cube edge (a₂) = 4 cm

Third cube edge (a₃) = 5 cm

Formula used:

Volume of cube = a³

Calculations:

Volume of three cubs =  Volume of new cube

⇒  (a1)³ + (a2)³ + (a₃)³ = V

⇒ 3³ + 4³ + 53 = V

⇒ V = 216 cm³

⇒ Edge = 6 cm

∴ The edge of new cube is 6 cm.

Given:

The length of rectangle = 48 cm

The breadth of rectangle = 21 cm 

The side of a square =  two-thirds the length of the rectangle 

Formulae required: 

Area of rectangle = Length × breadth

Area of square = side × side

Calculations:

Area of rectangle = 48 × 21

⇒ 1008 cm²

Side of square = (2/3) × length of a rectangle

⇒ (2/3) × 48 = 32cm

Area of square = 32 × 32 

⇒ 1024 cm²

Sum of the areas = 1024 + 1008 

⇒ 2032 cm²

∴ Sum of the areas is 2032 cm²

GIVEN:

Diagonals of the Rhombus = 6 cm and 8 cm

FORMULA USED:

_d1² + d₂² = 4 × (a)² Where d₁ & d₂ are the diagonals of the Rhombus and a is the side

CALCULATION:

Diagonals of the Rhombus = 6 cm and 8 cm

⇒ d12 + d22 = 4 × (a)2

⇒ 6² + 8² = 4 × (a)²

⇒ 100 = 4 × (a)²

⇒ 25 = (a)²

⇒ a = 5

⇒ Perimeter of the Rhombus = 4 × a

⇒ 4 × 5

⇒ 20

∴ Perimeter of the Rhombus is 20 cm

Given :-

Area of equilateral triangle = 36√3 m²

Concept :-

Area of an equilateral triangle = (√3/4) × Side²

Perimeter of equilateral triangle = 3 × Side 

Calculation :-

⇒ 36√3 = (√3/4) × Side²

⇒ Side² = 36 × 4

⇒ Side = √(36 × 4)

⇒ Side = 6 × 2 = 12 m

Now,

⇒ Perimeter of equilateral triangle = 3 × 12

⇒ Perimeter of equilateral triangle = 36 m

∴ Perimeter of equilateral triangle is 36 m

Given:

Diagonals of the small tiles are 39 m and 30 m

Total number of tiles = 290

Rate of painting = Rs. 3 per m2

Formula used:

Area of Rhombus = (d1 × d2)/2

Here, d1 and d2 are diagonals of a rhombus

Concept used:

All sides of a rhombus are equal and diagonals bisect each other at a right angle.

Calculation:

Area of Rhombus = (d1 × d2)/2

⇒ Area = (39 × 30)/2

⇒ Area = 585 m2

Total area of hall’s floor = 585 × 290 = 169650 m2

Cost of painting floor = 169650 × 3 = Rs. 508950

∴ The cost of painting the floor of the hall is Rs. 508950