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Given:

The cost of painting of cylindrical drum at the rate of Rs 5 per m2 is Rs 8360.

The radius of cylindrical drum which is opened at the top is 14 m.

Formula Used:

Curved surface area of cylinder = 2πrh

where r = radius and h = height

Calculation:

The cost of painting of cylindrical drum at the rate of Rs 5 per m2 is Rs 8360.

Then, area of the cylindrical drum which is opened at the top = 8360/5 =1672 m²

Total surface area of the cylindrical drum which is opened at the top = 2πrh + πr²

⇒ 2πrh + πr2  = 1372

⇒ 2 × 22/7 × 14 × h + 22/7 × 14 × 14 = 1672

⇒ 88h = 1672

⇒ h = 12

Therefore, the height of the cylindrical drum is 12 m.   

Given-  

External diameter = 42 cm

Internal diameter = 28 cm

Cost of painting = Rs 0.5/cm2

Height of cylinder = 5 cm

Concept Used-

Total Surface Area of a hollow cylinder = 2πh(R + r) + 2π(R² - r²)      [where R = External Radius, r = Internal Radius]

Calculation- 

External Radius(R) = 42/2

⇒ 21 cm

Internal Radius(r) = 28/2

⇒ 14

Total Surface area = 2π × 5 × (21 + 14) + 2π(21² - 14²)

⇒ 44/7 × 5 × 35 + 44/7 × (245)

⇒ 1100 + 1540

⇒ 2640 cm²

∴ Total cost of painting the cylinder = 2640 × 0.5

⇒ Rs 1320

Given :-

Spherical ball of radius 6 cm

Recast into three spherical ball having radius = 3 cm and 4 cm 

Concept :-

Volume of sphere = (4/3)πR³

Volume of spherical ball = Summation of volume of small spherical balls

Calculation :- 

Let radius of small spheres are r₁ , r₂ and r₃

r₁ = 3 cm and r₂ = 4 cm

Volume of sphere ball = Volume of first sphere + Volume of second sphere + Volume of third sphere

⇒ (4/3) × π × 6³ = (4/3) × π × 3³ + (4/3) × π × 4³ + (4/3) × π × (r)³

⇒ (4/3) × π × 216 = (4/3) × π [27 + 64 + r³]

⇒ 216 = 91 + r³

⇒ 216 - 91 = r³

⇒ r³ = 125 

⇒ r = ∛125

⇒ r = 5 

∴ Radius of third small sphere is 5 cm

Given:

Radius of circle, r = 21 m

Sector angle = 60°

Formula used:

Perimeter of sector = θ/360° × 2πr + 2r

Calculation:

Perimeter of sector = θ/360° × 2πr + 2r

⇒ Perimeter of sector = (60°/360°) × 2 × 21 × 22/7 + 2(21)

⇒Perimeter of sector = (60 × 2 × 21 × 22)/(360 × 7) + 42

⇒ Perimeter of sector = 22 + 42 = 64 m

∴ Perimeter of sector is 64 m.

Given:

Radius of an original sphere = 16 cm

Radius of a recast sphere = 4 cm

Formula Used:

Volume of a sphere = (4/3)πr³ 

Number of recast spheres =  Volume of the original sphere/Volume of the recast sphere

Calculation:

Volume of a original sphere = 4/3 π (16)³

Volume of the recast sphere = 4/3 π (4)3

Hence, number of recast spheres = Volume of the original sphere/Volume of the recast sphere

⇒ [(4/3) π (16)³]/[(4/3) π (4)³] = 64

∴ The number of recast spheres is 64.

GIVEN:

Radius of big sphere = 6 m

Radius of small sphere’s = 1 m, 2 m & 3 m

CONCEPT:

“Since the given sphere is a solid sphere which means that it is also filled inside. so we can say that the volume of the bigger sphere will be equal to the volume of the small sphere.”

CALCULATION:

So we can write it as,

⇒ 4/3πR³ = N [4/3πr₁³+ 4/3πr₂³+ 4/3πr₃³]      [N = number of spheres]

⇒ 4/3 πR³ = N 4/3π [r₁³ + r₂³ + r₃³]

⇒ Here R = 6m, r₁ = 1m,  r₂ = 2m  , r₃ = 3m

⇒ R³ = N [r13 + r23 + r33]

⇒ 216 = N [1 + 8 + 27]

⇒ N = 216/36 = 6

Given:

⇒ Outer diameter of cylinder = 16cm

⇒ Outer radius of cylinder = 8cm

⇒ Inner diameter of cylinder = 12cm

⇒ Inner radius of cylinder = 6cm

⇒ Height of cylinder = 42cm

Formula:

⇒ Volume of hollow cylinder = πh[(outer radius)² - (inner radius)²]

Calculation:

⇒ Volume of hollow cylinder = 22/7 × 42 × (8² - 6²) = 3696 cm³

⇒ Weight of hollow cylinder = 3696 × 10 = 36960 g

∴ Weight of cylinder = 36960 /1000 = 36.960 kg

Given:

Outer Radius of hollow cylinder is 4 cm.

Height is 2 cm.

Thick is 1 cm.

Formula used:

Volume of hollow cylinder = Outer Volume – Inner Volume 

= πR²h – πr²h 

Calculation:

Outer Radius(R) = 4 cm

Inner Radius(r) = (4 – 1) cm = 3 cm.

According to the question:

πR2h – πr2h 

⇒ πh(R² – r²)

⇒ (22/7) × (4² – 3²) × 2

⇒ (22/7) × 7 × 2

⇒ 44 cm³

 ∴ The volume of metal used is 44 cm³.

Given:

Radius of circle = 7 cm

Formula:

Area of circle = πr²

Area of triangle = √3 / 4 × a²

In radius of circle = a / 2√3

Calculation:

In radius of circle = a / 2√3

⇒ 7 = a / 2√3

⇒ a = 14√3

Area of triangle = √3 / 4 × (14√3)²

⇒ √3 / 4 × 14 × 14 × 3

⇒ 147√3 cm 

Given:

When each side of a cube increased by 'x' cm then the total surface area is increased by 69%.

Concept:

Total surface area of the cube = 6 × a2

Volume of the cube = a3

Where a = side

Calculation:

TSA1/TSA2 = (6 × a2)/[6 × (a + x)2] 

According to the question,

a2/(a + x)2 = 100/169

⇒ a/(a + x) = 10/13

⇒ 13a = 10a + 10x

⇒ 3a = 10x

⇒ a/x = 10/3

V1/V2 = a3/(a + x)3

⇒ 103/133 = 1000/2197

% increase = (1197/1000) × 100 = 119.7%

∴ % increase of the volume is 119.7%.

Shortcut:

Given:

When each side of a cube increased 'x' cm, the total surface area is increased by 69%.

Concept:

The total surface area of the cube = 6 × a2

The volume of the cube = a3

Where a = side

Explanation:

TSA → 100 : 169

Side → 10 : 13

Volume → 1000 : 2197

% increase = (1197/1000) × 100 = 119.7%

Given:

The length of the diagonal of the square is 4√2 cm

Concept Used:

If the length of one side of a square is 'a' unit then the length of diagonal of the square is a√2 unit.

And the arear of the square is a² square unit.

Calculation:

Let, the length of one side of the square is 'a' unit

⇒ Length of the diagonal of the square is a√2 unit

Accordingly,

a√2 = 4√2

⇒ a = 4

Area of the square is 4² = 16 cm²

Accordingly, the area of the 2nd square is 2 × 16 = 32 cm²

Let, the length of the side of the 2nd square is 'b'

Accordingly, 

b² = 32

⇒ b = √32

⇒ b = 4√2

Length of the diagonal of the 2nd square is (4√2) × √2

⇒ 8 cm

∴ The diagonal of another square whose area is double that of the first square is 8 cm.

Formula used:

Volume of cone = 1/3 πr²h

Area of circular base = πr²

Calculation:

Height = H

V = 1/3 πr²H

A = πr²

\({AH\over V} =\ { \pi r^2 \times H \over 1/3 \pi r^2 H} \)

⇒ AH/V = 3

∴ The value of \(\frac{{AH}}{V}\) is 3.

Calculation 

Incenter = Point of intersection of angle bisector 

Circumcenter = Point of intersection of perpendicular bisector 

Orthocenter = Point of intersection of altitude

and they meet at same line in equilateral triangle so, all the points are collinear in equilateral triangle 

∴ The required answer is collinear 

Given:

A square field of side 60 m and a rectangular field of length 80 m have the same perimeter.

Formula Used:

Area of the square field = (side)²

Area of the rectangular field = Length of field × Breadth of field

Perimeter of square field = 4 × side

Perimeter of rectangular field = 2 × (Length + Breadth)

Calculation:

⇒ Area of the square field = (60)² = 3600 m²

⇒ Perimeter of square field = 4 × 60 = 240 m 

⇒ Perimeter of rectangular field = 2 × (80 + Breadth) = 240

⇒ Breadth of rectangular field = 40 m

⇒ Area of the rectangular field = 80 × 40 = 3200 m2

∴ Area of the square field > Area of the rectangular field 

21 cm

Given:

The ratio of height and diameter of the right circular cone = 3 : 2

The volume of cone =  1078 cm³

Formula used:

The volume of right circular cylinder = 1/3 × π × radius² × height

Calculation:

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

r = diameter/2 = 2x/2 = x

h = 3x

The volume of right circular cylinder = 1/3 × π × r² × h

⇒ 1/3 × π × r² × h = 1078

⇒ 1/3 × π × x² × 3x = 1078

⇒ x³ = 49 × 7

⇒ x³ = 7³

⇒ x = 7

 ⇒ height of cone = 3x = 3 × 7 = 21 cm.

∴ the height of the right circular cone is 21 cm.

Given:

Length, breadth and height of tank is 3m, 2m and 1m respectively

Formula Used:

Volume = Length × breadth × height

Calculation:

Volume of tank = (length × breath × height)

Volume of tank = (3 × 2 × 1) = 6 m3

Now,

1m3 = 1000 litres

∴ 6m3 = 6000 litres

Given

Side of an equilateral triangle = 11 cm

Concept

Radius of incircle of equilateral triangle = Side/2√3

Calculation

Radius of circle = 11/2√3 cm

Radius of circle = 5.5/√3 cm

Given: 

The cost of fencing of the square plot is 156.4 rupees at the rate of 2.3 rupees per meter.  

Concept: 

Perimeter of square = 4 × side 

Area of square = (Side)²

Calculation: 

Here, the perimeter of the square plot is 

⇒ 156.4/2.3

⇒ 68 m

Now, The side of the square plot is

⇒ 68/4

⇒ 17 cm

The area of the square plot is 

⇒ (17)²

⇒ 289 cm²

∴ The required area of the square plot is 289 cm².

Given:

Diameter of Pizza = 24 cm

Radius of pizza = 12 cm

Concept:

When the pizza is divided into 8 parts, the central angle of 360° will also be divided into 8 equal parts forming 8 sectors of equal area.

Formula used:

Area of sector = (θ/360°) × πr²

Calculation:

Central angle of each part = 360°/8 = 45°

Area of each piece = (45°/360°) × π × 12 × 12

⇒ (1/8) × π × 144

⇒ 18π cm²

∴ Area of each piece is 18π cm²

Given - 

area of rectangle = 812 cm², area of semicircle = 98π cm²

Formula used - 

area of rectangle = length × breadth

area of semicircle = (1/2) × π × radius²

Solution - 

Let the radius of the semi - circle is r cm.

⇒ (1/2) × π r² = 98π  

⇒ r² = 196

⇒ r = 14 cm

⇒ diameter of the semi - circle = 28 cm

⇒ diameter of semi - circle = side of rectangle CD = 28 cm

⇒ area of rectangle = 812 cm²

⇒ BC × CD = 812

⇒ BC × 28 = 812

⇒ BC = 29 cm

⇒ We have to find perimeter of remaining figure so,

⇒ perimeter = AB + BC + AD + (1/2) × perimeter of the circle

⇒ perimeter = 28 + 29 + 29 + (1/2) × (2πr)

⇒ perimeter = 58 + 28 + (22/7) × 14

⇒ perimeter = 130 cm

∴ perimeter of remaining figure = 130 cm.

Given:

Side of equilateral triangle = 21 cm

Formula used:

Radius of circumcircle of an equilateral triangle = side/√3

Area of circle = πr2 where, r = Radius

Calculation:

The radius of circumcircle = 21/√3

⇒ 7 × √3 

⇒ 7√3 cm

Area of the circumcircle = 22/7 × 7√3 × 7√3

⇒ 22 × 7 × 3 cm²

⇒ 22 × 21 cm2

⇒ 462 cm2

GIVEN:

Side of the cube = 2√3 cm

FORMULA USED:

Total surface area of the cube = 6 × side²

CALCULATION:

Side of the cube = 2√3 cm

Total surface area of the cube = 6 × (2√3)²

⇒ 6 × 12

⇒ 72 cm²

∴ The total surface area of the cube is 72 cm2.

Given:

As the circumference of the inner circle = 22 cm

And of the outer circle = 44 cm

Formula Used:

Circumference of a circle = 2× π× R

Where R = Radius of the given circle, where π = 22/7

Calculation:

Let Us assume that the radius of the bigger circle be X unit

And the radius of the smaller circle be Y unit

By using the above formula

⇒ Circumference of the inner circle is 2× π×Y = 22 (∵ It is given in the question )

⇒ Circumference of the outer circle is 2× π×X = 44 (∵ It is given in the question )

⇒ 2× (22/7)× Y = 22 and 2× (22/7)× X = 44

⇒ Y = 3.5 cm and X = 7cm

∴ The width of the ring is X – Y = 7 – 3.5 = 3.5 cm

Given:

Diameter of the circle = 12 inches

Concept used:

1 inch = 2.54 cm

Formula used:

Circumference of circle = 2 π r

r = d/2

Calculation : 

r = 12/2

⇒ r = 6 inches  = 15.24cm

⇒ Circumference of circle = 2π × 15.24

⇒ 95.7072cm

Given:

Big cube side is  8 inches and small cube side is 2 inches

Formula Used: 

Volume of cube = \({\left( {side} \right)^3}\)

Total surface area of cube = \(6{\rm{ × }}{(side)^2}\)

Calculation:

Volume of big cube = 8× 8 × 8

 ⇒ 512

Volume of small cube = 2 × 2 × 2 

⇒ 8

Number of small cube =(volume of big cube)/ (volume of small cube)

⇒ No. of cube = 512/8

⇒ No. of cube = 64

According to question,

Total surface area of big cube = \(6 × {(8)^2}\)

⇒ 6 × 64

Total surface area of small cube = 6 × \({(2)^2}\) × 64

⇒ 6 × 4 × 64

The ratio of total surface area of sum of small cubes and big cube

⇒ 6 × 4 × 64 ∶ 6 × 64

Required ratio = 4 ∶ 1

∴ Ratio of cube is 4 : 1

Given:

Total surface area of a cuboid =126 sq. cm

Length and breadth of the cuboid is 6 cm and 5 cm respectively

Concept used:

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

⇒ 2(30 + 5h + 6h) =126

⇒ 60 + 10h + 12h =126

⇒ 22h =66

⇒ h =3 cm

∴ The height of the cuboid is 3 cm 

Given

Each side of a square field measures 10 m

Formula used

Perimeter of square = 4 × side of square

Calculation 

Perimeter of square = 4 × 10

⇒ 40 m

Given:

The perimeter of two square is 12 cm and 16 cm

Formula used:

The perimeter of square = 4 × (side)

Area of the square = (Side)²

Calculations:

The perimeter of square = 4 × (side) 

Side of first square = 12/4 = 3 cm

Side of second square = 16/4 = 4 cm

Area of the 3rd square = (3² + 4²)

⇒ (9 + 16) = 25 cm²

Side of third square = √25 = 5 cm

Perimeter of 3rd square = 4 × 5 = 20 cm

∴ Required perimeter of the third square is 20 cm

Given:

Ratio of perimeter of a rectangle and perimeter of a square = 7 : 6

Breadth of rectangle = Side of square

Length of rectangle = 40 cm

Formula used:

(i) Perimeter of rectangle  = 2(l + b)

(ii) Area of rectangle = l × b

Where,

l = length of rectangle

b = breadth of rectangle

(iii) Area of square = (a)²

(iv) Perimeter of square = 4a

Where,

a = side of square

Calculation:

Ratio of perimeter of a rectangle and perimeter of a square = 7 : 6

{2(l + b)}/4a = 7/6

⇒ 12(l + b) = 28a

⇒ 12(40 + a) = 28a

⇒ 480 + 12a = 28a

⇒ 28a – 12a = 480

⇒ 16a = 480

⇒ a = 30 cm

Area of square = (30)²

⇒ 900 cm²

Area of rectangle = l × b

⇒ 40 × 30

⇒ 1200 cm²

The ratio of the area of the square to the area of rectangle = 900 : 1200

⇒ 3 : 4

∴ The ratio of the area of the square to the area of the rectangle is 3 : 4

Given:

The total area of a circle and a square = 2611 cm² 

The diameter of the circle = 42 cm

Formula used:

If the diameter of a circle is d, then the area of a circle = (π/4) × d² 

and circumference of a circle = π × d

If the side of a square = a, then an area of a square = a²

and perimeter = 4 × a

Calculation:

The total area of a circle and a square = 2611 cm2 

⇒ (π/4) × d2 + a2 = 2611

⇒ (22/7)× (1/4) × 42² + a² = 2611

⇒ (38808/28) + a² = 2611

⇒ 1386 + a² = 2611

⇒ a² = 2611 - 1386

⇒ a² = 1225

⇒ a = 35 cm

Now, the sum of the circumference of the circle and the perimeter of the square =  perimeter of a square + circumference of a circle

⇒ 4 × 35 + π × 42

⇒ 140 + (22/7) × 42

⇒ 140 + 132

⇒ 272 cm

∴ The sum of the circumference of the circle and the perimeter of the square is 272 cm.

Given:

Product of base and height of triangle = 5/14 of radius² 

Formula used:

Area of circle = πr² 

Area of triangle = (Base × Height)/2

Calculation:

Let the base of the triangle be B and height of the triangle be H and the radius of the circle be r

According to question

B × H = 5/14 of r² 

Area of circle ∶ Area of triangle = πr2 ∶ (BH)/2

⇒ Area of circle ∶ Area of triangle = \(\frac{{\frac{{22}}{7}{r^2}}}{{\frac{1}{2}BH}}\)

⇒ Area of circle ∶ Area of triangle = \(\frac{{\frac{{22}}{7}{r^2}}}{{\frac{1}{2} \times \frac{5}{{14}}{r^2}}}\)

⇒ Area of circle ∶ Area of triangle = 88/5

∴ Area of circle ∶ Area of the triangle is 88 ∶ 5

Given:

Radius of circle = 8 cm

An area of a circle = An area of a triangle

The base of a triangle = 8 cm 

Formula used:

If the diameter of a circle is r, then the area of a circle = π × r2 

If in a triangle, the base is b and altitude is h, then an area of triangle = (1/2) × b × h

Calculation:

An area of a circle = An area of a triangle

⇒ π × r2 = (1/2) × b × h

⇒ π × 8² = (1/2) × 8 × h

⇒ h = (π × 64 × 2)/8

⇒ h = 128 π/8

⇒ h = 16π cm

∴ The length of the corresponding altitude of the triangle is 16π​​ cm.

Given:

Circumference of circle = 18π cm

Formula used:

Circumference of circle = 2πr

Area of circle = πr²

Calculation:

Circumference of circle = 2πr

⇒ 18π = 2πr 

⇒ r = 9 cm

Area of circle = πr²

⇒ Area = 9² × π 

⇒ Area = 81π sq. cm

∴ The area of circle is 81π square cm.

Given:

Radius of original cylinder is 3 times the height.

Two cuts made parallel to base of equal heights.

Formula Used:

Total Surface Area = 2πr(h + r), where r is radius of base and h is height of cylinder

Concept Used:

Since Base of a cylinder is circle, so after every cut parallel to base, there will be an increase of two new circles with area πr² each.

Calculation:

Let the height of original cylinder be 3x then radius will be 9x units

Total Surface Area of original Cylinder = (2π × 9x)(3x + 9x)

⇒ T.S.A. = 216πx²

Using the Statement of concept used, 

One cut will produce 2 new circles then 2 cuts will form 4 new circles.

Increased Surface Area = 4π × (9x)²

⇒ Increased Surface Area = 324πx²

 Increased surface area/Original surface area = 324πx²/216πx²

⇒ Increased surface area/Original surface area = 3/2

∴ The ratio of Increased surface area and original Surface area is 3 : 2

Given

Radius of circular park = 3 metre

Width of track = 6 metre

Concept

Area of circular track = Area of outer circle – Area of inner circle

Formula used

Area of circle = πr²

Explanation

Area of track =πr₁²- πr₂ ²

r₁ and r₂ represents radius of outer and inner circle

Area of track= π(r₁ – r₂) (r₁ + r₂)

⇒ π( 9 – 3) (9 + 3)

Given:

Total surface area of a hemisphere = 41.58 cm2

Radius of hemisphere = Radius of sphere.

Formula Used:

Total surface area of a hemisphere = 3πr²

Total surface area of a sphere = 4πr² 

Calculation:

Total surface area of a hemisphere is 41.58 cm2

⇒ 3πr2 = 41.58

⇒ r² = 41.58/3π 

As the radius of a sphere is the same as that of a hemisphere.

⇒ Total surface area of a sphere = 4πr2 

Using the above value of r²

⇒ Total surface area of a sphere = 4π × (41.58/3π)

⇒ Total surface area of a sphere = 4 × 13.86

⇒ Total surface area of a sphere = 55.44 

∴ Total surface area of a sphere is 55.44 cm²

Given:

For Cylinder,

Radius (r) = 7 cm

Height (h) = 4 cm

For Cone,

Radius (R) = 21 cm

Height (H) = 15 cm

Formula Used:

Volume of cone = (1/3)πR²H

Volume of cylinder = πr²h

Concept Used;

Volume of cone = n × Volume of cylinder, where n is the number of cylindrical glasses

Calculation:

Let the number of cylindrical glasses be n

Using concept,

(1/3)πR²H = n × πr²h

(1/3) × (21)² × 15 = n × (7)² × 4

⇒ n = (21 × 21 × 15)/(7 × 7 × 4 × 3)

⇒ n = 11.25

We have to take integer as our answer

∴ The number of glasses filled with wine is 11.

Given:

The length of a room floor exceeds its breadth by 20 m. The area of the floor remains unaltered when the length is decreased by 10 m and the breadth is increased by 5 m

Concept used:

Area of Rectangle = Length × Breadth

Calculation:

Let the breadth of the floor be x m.

Length = (x + 20) meter

The area of the floor = Length × Breadth 

Area of the floor = (x + 20) x m²

According to the question

Length is decreased by 10 m = (x + 20 - 10) = (x + 10) m

Breadth is increased by 5 m = (x + 5) m

⇒ (x + 10) (x + 5) = x (x + 20)

⇒ x² + 15x + 50 = x² + 20x

⇒ 5x = 50

⇒ x = 10 meter

Length of the floor = x + 20 = 30 m

Area of the floor = 30 × 10

∴ Area of the floor = 300 m²