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GIVEN:

the altitude of the parallelogram = 100 cm

FORMULA USED:

Area of the parallelogram = (Base × Altitude) sq. unit and Area of triangle = (1/2 × Base × Altitude) sq. unit

CALCULATION:

Let the altitude of the triangle be L₁ cm and the triangle and a parallelogram are constructed on the same base 

⇒ Area of the parallelogram = Area of triangle

⇒ (Base × Altitude) = (1/2 × Base × Altitude)

⇒ 100 = 1/2 × L₁

⇒ L₁ = 200 cm

∴ The altitude of the triangle is 200 cm

Given:

Length of pipe(h) = 10 meter = 1000 cm

Internal radius(r) = 3 cm

External radius(R) = Internal radius + thickness

= 3 cm + 0.7 cm

= 3.7 cm

Formula used:

Volume of hollow pipe = π(R² – r²)h

a² – b² = (a + b)(a – b)

Calculation:

∵ Volume of hollow pipe = π(R² – r²)h

⇒ (22/7) × 1000 × [(3.7)² – (3)²]

⇒ (22/7) × 1000 × (13.69 – 9)

⇒ (22/7) × 1000 × (4.69)

⇒ 14740 cm3

Given:

Circumference of the circle  = Area of the circle

Formula Used:

Circumference of the circle = 2πr

Area of the circle = πr2

Diameter = Radius × 2

Calculation:

According to question

2πr = πr²

⇒ 2 = r

⇒ Radius = 2 units

Diameter = Radius × 2

⇒ 2 × 2 = 4 units

∴ Diameter is 4 units

The correct option is 1 i.e. 4 units

Given:

Diagonal = 29 cm

Length = 21 cm

Formula Used:

Area of rectangle = Length × Breadth

Calculation:

Using Pythagoras theorem

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (29)² = (21)² + (Perpendicular)²

⇒ 841 – 441 = (Perpendicular)²

⇒ √400 = Perpendicular

⇒ 20 cm = Perpendicular

⇒ 20 cm = Breadth

Area of rectangle = Length × Breadth

⇒ 21 cm × 20 cm

⇒ 420 cm²

∴ The area of the rectangle is 420 cm²

The correct option is 2 i.e. 420 cm2

Given:

Radius of well = 2.8 m

Height of well = 15 m

Width of platform = 7 m

Concept:

The volume of the platform will be equal to the volume of the earth taken out, i.e. to the volume of cylindrical shaped well. The platform is also is in the shape of cylinder whose inner radius is the radius of the well and the outer radius will be the sum of radius of well and the thickness of the platform.

Formula used:

Volume of cylinder = πr²h

Volume of Hollow cylinder = π(R² – r²)h

Where, ‘R’ is the outer radius and ‘r’ is inner radius.

Calculation:

Let, the height of the platform = h

‘r’ = radius of well

‘R’ = radius (Platform + well) = 7 + 2.8 = 9.8 m

∵ Volume of the well = Volume of the Cylindrical platform

∴ πr²h = π(R² – r²)h

⇒ 2.8 × 2.8 × 15 = [(9.8)² – (2.8)²] × h

⇒ 117.6 = (96.04 – 7.84) × h

⇒ 117.6 = 88.2 × h

⇒ h = 117.6/88.2

⇒ h = 4/3 = 1.33m

GIVEN:

 length of a rectangle is decreased by 4 cm and the width is increased by 3 cm

FORMULA USED:

Area of rectangle = (l × b) sq. unit and Area of square = (side)² sq. unit 

Perimeter of rectangle = 2(l + b) units

CALCULATION:

Let the length and breadth of the rectangle be x and y 

⇒ Area of rectangle = (l × b) sq. unit

⇒ xy cm²

⇒ new length and breadth = (x - 4) and (y + 3)

⇒ (x - 4) = (y + 3)

⇒ x - y = 7 .....(1)

⇒ Area of square = (x - 4) × (y + 3)

⇒ According to question

⇒ Area of rectangle = Area of square 

⇒ xy = (x - 4) × (y + 3)

⇒ 3x - 4y = 12 .....(2)

⇒ On solving these equation 

⇒ x = 16 and y = 9

⇒ Perimeter of rectangle = 2(l + b) units

⇒ 2(16 + 9)

⇒ 2 × 25

⇒ 50 cm

∴ perimeter of the original rectangle is 50 cm

Given:

Radius of spherical ball(R) = 2.1 cm

Radius of cylinder(r) = 7 cm

Concept:

The volume of the ball will the volume of the water displaced. So, the volume of the cylindrical part with rise in water level will be equal to the volume of the ball.

Formula used:

Volume of sphere = (4/3)πr³

Volume of cylinder = πr²h

Calculation: 

Let, the rise in water level = h cm

∵ Volume of sphere = Volume of cylinder

⇒ (4/3)πR³ = πr²h

⇒ (4/3) × 2.1 × 2.1 × 2.1 = 7 × 7 × h

⇒ h = (4 × 2.1 × 2.1 × 2.1)/(3 × 7 × 7)

⇒ h = 0.252 cm

Given :

Area of the square is 8 times the area of the rectangle

Length of rectangle is half the length of the side of the square

Formula Used :

Area of the square = side²

Area of the rectangle = length × breadth

Calculations :

Let the side of the square be ‘s’

Let the length of the rectangle be ‘l’ and breadth be ‘b’

According to the question

S² = 8 × (I × b)

(2l)² = 8 × (l × b)      (l = (s/2) or s = 2 l)

4 l² = 8 × l × b

⇒ b = l/2

Now,

Diagonal of the square = √2 × side = √2s = 2√2 l     (s = 2l)

Diagonal of the rectangle = √[(l)² + (b)²] = √[(l)² + (l/2)²]

⇒ √[(5/4)(l)²]

⇒ (l/2)√5    

Diagonal of square : Diagonal of rectangle = 2√2l : (l/2)√5

⇒ 4√2 : √5

∴ The ratio of the diagonals is 4√2 : √5 

Given :

The volume of cube increased by 72.8%

The original length of the cube was 10 cm

Formula Used :

The volume of cube = a³     (where a is the side of the cube)

Calculations :

Side of the original cube = 10 cm

The volume of the original cube = 10³

⇒ 1000 cm³

The volume of the cube is increased by 72.8%

Volume of new cube = 1000 + 72.8% of 1000

⇒ 1000 + 728 = 1728 cm³

Let the length of new cube ‘x’

1728 = (x)³

⇒ x = 12 cm

∴ The length of the side of the new cube is 12 cm

Given:         

Speed on 1st side = 3 m/s

Speed on 2^nd side = 4 m/s

Speed on 3^rd side = 5m/s

Speed on 4^th side = 6m/s

Total time taken to complete 1 round = 57 seconds

Concept:

Since the distance covered on each side is same, so the time taken to cover the distance of each side is added which is the overall time to complete 1 round. So, by using the time distance speed formula, calculate the distance of each side and then calculate the area of the square.

Formula used:

Distance/Speed = time

Area of square = side × Side

Calculation:

Let, side of the square be ‘D’

∴ D/3 + D/4 + D/5 + D/6 = 57

⇒ (20D + 15D + 12D + 10D)/60 = 57

⇒ 57D = 57 × 60

⇒ D = 60 m

Area of the square of side 60 m

⇒ 60 × 60 = 3600 m²

∴ Area of the square is 3600 m²

Concept

Volume of cuboid = length × breadth × height

Volume of cube = side³

Number of cubical boxes that can fit inside cuboidal/cubical store = Volume of cuboidal/cubical store ÷ Volume of one cubical box

Calculation

Length and breadth of cuboidal store = 16 m and 13 m respectively

Boxes are to be placed to a height of 7 m 

so, height = 7 m

Volume of cuboidal store = 16 × 13 × 7 = 1456 m³

Similarly, Length and breadth of cubical box = 8 m each and height = 7 m

Volume of cubical store = 8 × 8 × 7 = 448 m³

Volume cubical box of side 1 m = 1 × 1 × 1 = 1 m³

Number of boxes to be placed = (1456 ÷ 1) + (448 ÷ 1) = 1904

So, the number of boxes = 1904.

Given:

Radius of cone = 6 cm

Radius of hemisphere = 6 cm

Height of cone = 9 cm

Formula:

Volume of cone = (1 / 3) × πr²h

Volume of hemisphere = (2 / 3) × πr³

Calculation:

Total volume of shape = Volume of cone + volume of hemisphere

⇒ (1 / 3) × π × 6² × 9 + (2 / 3) × π × 6³

⇒ (1 / 3) × π × 36 × (9 + 12)

⇒ 12π × 21

Given:

Dimension of cuboid = 4 cm × 2 cm × 1 cm

Formula:

Volume of cuboid = lbh

Total surface are of cuboid = 2 (lb + bh + hl)

Volume of cube = a³

Total surface area of cube = 6a²

Calculation:

Volume of cuboid = 4 × 2 × 1 = 8 cm³

Volume of each cube = 8 / 8 = 1 cm³

Side of each cube = ∛1 = 1 cm

Total surface area of cuboid = 2 (4 × 2 + 2 × 1 + 1 × 4)

⇒ 2 × (8 + 2 + 4)

⇒ 2 × 14

⇒ 28 cm²

Total surface area of all 8 cubes = 8 × 6 × 1²

⇒ 48 cm²

∴ Required ratio = 28 : 48 = 7 : 12

Given:

Radius of the cone r = 9 cm

Height of cone h = 9 cm

Radius of cylinder R = 9 cm

Height of cylinder H = 7 cm

Formula:

Volume of cone = (1 / 3) × πr²h

Volume of cone = πR²H

Calculation:

Volume of shape = (1 / 3) × πr²h + πR²H

⇒ (1 / 3) × π × 9 × 9 × 9 + π × 9 × 9 × 7

⇒ 81π × (3 + 7)

⇒ 81π × 10

⇒ 810π

Given:

Diameter of sphere = 42 cm

Then, Radius = 21

Formula:

Volume of sphere = 4/3 × π × R³

Calculation:

We know that –

Volume of sphere = 4/3 × π × R³   …….. (1)

Now,

Put all the given values in equation (1) then we get

Volume of sphere = 4/3 × 22/7 × (42/2)³ 

⇒ 4/3 × 22/7 × 21 × 21 × 21

⇒ 4 × 22 × 21 × 21

⇒ 38808

∴ The Volume of a sphere will be 38808 cm³

Quantity II:

Given:

Side of cube = 27 cm

Formula:

Volume of cube = (side)³

Calculation:

We know that –

Volume of cube = (side)³   …….. (1)

Now,

Put the given value in equation (1) then we get

Volume of cube = (27)³

⇒ 27 × 27 × 27

⇒ 19683

∴ The Volume of a cube will be 19683 cm³

Comparison of both the result:

∴ Quantity I > Quantity II

Given:

Ratio of area of two square = 144 : 1

Formula used:

Area of square = (side)²

Ratio of perimeter = 4 × side

Calculation:

(Area₁)²/(Area₂)²= 144/1

⇒ (side₁)²/ (side₂)²=144/1

 ⇒ Side₁/Side₂ = 12/1

Ratio of perimeter = (4 × side₁)/(4 × side₂)

⇒ (4 × 12)/(4 × 1)

⇒ 12 : 1

Given:

The circumference of a circle of radius 7 cm 

Concept used:

Circumference of circle = 2πr

Calculation:

Circumference of circle = 2πr 

Circumference of circle = 2 × 22/7 × 7

Circumference of circle = 44 cm 

Concept: Circumference of a circle = 2πr, where 'r' is radius.

Solution: 

Given: Diametre = 21 cm

So, Radius = 21/2

Using above formula, we get, 

2 × 22/7 × 21/2 = 66 cm

Hence, we conclude that the circumference is 66 cm.

Given:

Circumference of the circle = 13.2 cm

Formula used:

Circumference of circle = 2πr

Where r represents radius of the circle

Calculation:

13.2 = 2(π) r

⇒ r = (13.2 × 7)/ (2 × 22)

⇒ r = 2.1 cm

Given:

Circumference of garden = 44 cm

Formula used:

Area of circle = πr²

Circumference of circle = 2πr

Calculation:

⇒ 2πr = 44

⇒ r = (44 × 7)/ (2 × 22)

⇒ r = 7 cm

Area = πr²

⇒ (22/7) × 7 × 7

⇒ 154 cm²

Given:

Length × Breadth = 156 cm²

Breath × Height = 143 cm²

Height × Length = 132 cm²

Formula used:

Volume of a cuboid = length × Breadth × Height

⇒ Volume of a cuboid = √(l² × b² × h²)

= √[(l × b) × (b × h) × (h × l)]

Calculation:

Volume of the Cuboid = √(156 × 143 × 132)

= √[(13 × 12) × (12 × 11) × (11 × 13)]

= √[(13)² × (12)² × (11)²]

= 13 × 12 × 11

= 1716 cm³

Given:

Ratio of sides = 7 ∶ 5 ∶ 3

Total surface area of the tank = 142 m²

Formula used:

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

Volume of cuboid = l × b × h

1m³ = 1000 liters

Calculation:

Let the sides of the tank = 7x, 5x and 3x

∵ TSA of the tank = 2(lb + bh + hl)

⇒ 142 = 2[(7x × 5x) + (5x × 3x) + (3x × 7x)]

⇒ 142 = 2(35x² + 15x² + 21x²)

⇒ 142 = 2(71x²)

⇒ 71 = 71x²

⇒ x = √1 = 1      ------(Ignore the negative value)

∴ Sides of the tank are;

7x = 7 × 1 = 7m

5x = 5 × 1 = 5m

3x = 3 × 1 = 3m

∴ Volume of the tank in m³ = (7 × 5 × 3) m³ = 105 m³

∵ 1 m³ = 1000 liters

∴ 105 m³ = (1000 × 105) = 105000 liters = 105 kiloliters 

∴ Tank can hold 105 kiloliters of water.

Given:

Area of the rhombus is 216 cm²

Length of a diagonal = 18 cm

Concept used:

The two diagonals bisect each other at 90° in a rhombus.

Formula used:

Area of rhombus = (1/2) × Product of the two diagonals

Perimeter = 4 × (Side of the rhombus)

Calculations:

Let the second diagonal of the rhombus be x and the side of the rhombus be a

Area of rhombus = (1/2) × 18 × x

⇒ 216 = (1/2) × 18x

⇒ 18x = 432

⇒ x = 24 cm

If the two diagonals of rhombus ABCD intersect at O then,

⇒ AO² + BO² = AB²

⇒ (18/2)² + (24/2)² = a²

⇒ 81 + 144 = a²

⇒ a² = 225

⇒ a = 15 cm

Perimeter of the rhombus = 4a

⇒ 4 × 15

⇒ 60 cm

∴ The perimeter of the rhombus is 60 cm.

Given:

Perimeter of a rhombus is 100 cm

Length of diagonal = 14 cm

Concept used:

The two diagonals bisect each other at 90° in a rhombus.

Formula used:

Perimeter = 4 × (Side of the rhombus)

Calculations:

Let the side of rhombus be x and the diagonals be a and b

Perimeter = 4x

⇒ 100 = 4x

⇒ x = 25 cm

If the two diagonals of rhombus ABCD intersect at O then,

⇒ AO² + BO² = AB²

⇒ (14/2)² + (b/2)² = 25²

⇒ 49 + (b²/4) = 625

⇒ (b²/4) = 576

⇒ b² = 2304

⇒ b = 48 cm

Difference between the diagonals = b – a

⇒ 48 – 14

⇒ 34 cm

∴ The difference between the two diagonals is 34 cm

Given:

Area of the Rhombus = 240 cm2

Formula used:

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:

Let the diagonals be 8x and 15x

Area of Rhombus = (d1 × d2)/2

⇒ 240 = (8x × 15x)/2

⇒ x = 2

Then our diagonals are 16 cm and 30 cm

Now, d12 + d22 = 4 × s2

⇒ 162 + 302 = 4 × s2

⇒ s2 = 289

⇒ s = 17 cm

Given:

Area of a rhombus = 720 cm²

Length of first diagonal = 80 cm

Concept used:

The two diagonals bisect each other at 90° in a rhombus.

Formula used:

Area of rhombus = (1/2) × Product of the two diagonals

Calculations:

Let the length of second diagonal be x

Area of rhombus = (1/2) × 80 × x

⇒ 720 = (1/2) × 80x

⇒ x = 18 cm

If the two diagonals of rhombus ABCD intersect at O then,

⇒ AO² + BO² = AB²

⇒ (80/2)² + (18/2)² = AB²

⇒ AB² = 40² + 9²

⇒ AB² = 1600 + 81

⇒ AB² = 1681

⇒ AB = 41 cm

∴ The length of each side of a rhombus is 41 cm.

Given:

Length of side = 17 cm

Length of a diagonal = 16 cm

Concept used:

The two diagonals bisect each other at 90° in a rhombus.

Formula used:

Area of rhombus = (1/2) × Product of the two diagonals

Calculations:

Let AC = 16 cm and BD = x cm

If the two diagonals of rhombus ABCD intersect at O then,

⇒ AO² + BO² = AB²

⇒ (16/2)² + (x/2)² = 17²

⇒ 64 + (x²/4) = 289

⇒ (x²/4) = 225

⇒ x² = 900

⇒ x = 30 cm

Area of rhombus = (1/2) × 16 × 30

⇒ 8 × 30

⇒ 240 cm²

∴ The area of the rhombus is 240 cm²

Given:

Area of three consecutive faces of cuboid = 9 cm², 25 cm² and 36 cm²

Formula used:

Volume of cuboid = √area₁ × √area₂ × √area₃

Calculation:

Volume of cuboid = √9 × √25 × √36 cm³

⇒ 3 × 5 × 6 cm³

⇒ 90 cm³

∴ The volume of cuboid is 90 cm³

Given:

The areas of three consecutive faces of a cuboid are 12 cm2, 20 cm2, and 60 cm².

Formula used:

The volume of cuboid = l × b × h

Where l → length

b → breadth

h → height

Calculations:

Let the length, breadth, and height of the cuboid be x, y, and z respectively.

Then xy = 12 cm2, yz = 20 cm², and zx = 60 cm² 

By multiplying these three we'll get

xy × yz × zx = 12 × 20 × 60

⇒ x²y²z² = 14400

⇒ (xyz)² = 14400

⇒ xyz = 120 and –120, (Rejecting the negative term as volume cannot be negative)

Volume of cuboid = l × b × h = xyz

Thus xyz = 120 cm³

∴ The volume of the cuboid is 120 cm³.

Given:

Areas of three consecutive faces of a cuboid = 9 cm², 25 cm² and 36 cm²

Formula used:

Volume = √Area₁ × √Area₂ × √Area₃

Calculation:

Volume = √9 × √25 × √36 cm²

⇒ 3 × 5 × 6 cm²

⇒ 90 cm²

∴ The required volume of cuboid is 90 cm²

Given:

Radius of sphere = 3 cm 

Radius of cylinder = 6 cm

Concept:

Volume of cylinder = πr²H

Volume of sphere = (4/3)πR³ 

Where, 

r → Radius of the cylinder 

H → Height of the cylinder 

R → Radius of the sphere

When the sphere is submerged then,

The volume of sphere = Volume of cylinder

Calculation:

The volume of sphere = Volume of cylinder 

⇒ (4/3) × π × (3)³ = π × (6)² × H

⇒ H = (36/36)

⇒ H = 1 cm

⇒ The surface of the water is raised = 1 cm

∴ The surface of the water is raised by 1cm