Mensuration

Given: Length (l) = 4m

Breadth (b) = 3m

Height (h) =1.5m

To Find: Total Surface Area of a Cuboid

As we know, the total surface area includes area of all 6 faces of a cuboid

So, Total Surface Area of a Cuboid = 2 {(l × b) + (b × h) + (h × l)}

= 2 (4 × 3 + 3 × 1.5 + 1.5 × 4)

= 2 (12 + 4.5 + 6)

= 2 (22.5)

Total Surface Area of a Cuboid = 45 m²

Given: Length (l) = 3.5m

Breadth (b) = 2.5m

Height (h) = 3m

As we know, the room is in cuboid shape.

So, Area of Four walls = Lateral Surface Area of Cuboid = 2h (l + b)

= 2 × 3 (3.5 + 2.5)

= 6 (6)

= 36m²

Given: Length (l) = 8m

Breadth(b) = 5m

Height (h) = 4m

Firstly, we find the Lateral Surface Area of Cuboid i.e.

Area of Four Walls = 2h (l + b)

= 2 × 4 (8+5)

= 8 (13)

= 104 m²

Cost of distempering of 1 m² area = ₹ 40

So, Cost of distempering of 104 m² = ₹ 40 × 104 = ₹ 4160

Given: Length (l) = 4.8m

Breadth (b) = 3.6m

Height (h) = 2m

Total Surface Area of the room = 2 {(l × b) + (b × h) + (h × l)}

But here we should not consider the ceiling of the room.

So, the required area of the room = lb + 2 (lh + bh)

= (4.8 × 3.6) + 2(4.8 × 2 + 3.6 × 2)

= 17.28 + 2 (9.6 + 7.2)

= 17.28 + 33.6

= 50.88 m²

Cost of laying tiles per square meter = ₹ 100

Cost of laying tiles for 50.88 m²= ₹100 × 50.88

= ₹5088

Given: Length (l) = 40m

Breadth (b) = 50m

Height (h) = 60m

Total Surface Area of a cuboid = 2{(l × b) + (b × h) + (h × l)}

= 2 (40 × 50 + 50 × 60 + 60 × 40)

= 2 (2000 + 3000 +2400)

= 2 (7400) = 14800cm²

Given the Total surface area of a cube = 384 cm²

As we know, the Total surface area of a cube = 6 (side of the cube)² = 6a²

So, according to the question:

Total Surface Area of a Cube = 384 = 6a²

= ±8

a = 8cm (side can’t be negative)

Given: Lateral Surface Area of a Cube = 64cm²

Lateral Surface Area of a Cube = 4a²

So, According to the question

Lateral Surface Area of a Cube = 64 = 4a²

4a² = 64

a² = = 16

a = = ± 4 = 4cm (side can’t be negative)

Given Side of a cubical room = 4m

So, Lateral Surface Area of a Cube = 4(side of a cube)²

= 4 (4)²

= 64m²

Cost of white washing per square meter = ₹20

Cost of white washing for 64 m² = ₹20 × 64 = ₹1280

Cost of white washing the four walls of a cubical room is ₹1280

Given: Edge of a cubical box = 10cm

Dimensions of cuboidal box:

Length(l) = 12.5cm

Breadth(b) = 10cm

Height(h) = 8cm

(i) Total Surface Area of a cube = 6 (side of cube)²

= 6(10)²

= 6 × 10 × 10

= 600 cm²

Total Surface Area of a Cuboidal box = 2{(l × b) + (b × h) + (h × l)}

= 2 (12.5 × 10 + 10 × 8 + 8 × 12.5)

= 2 (125 + 80 + 100)

= 2 (305)

= 610 cm²

(ii) Given: If each edge of a cube is doubled then a’= 2a = 2 × 10 = 20cm

New Total Surface Area of a cube = 6(a’)²

= 6(20)²

=2400 cm²

So, new total surface area is 4 times the old total surface area.

Given: Length of a cube = 12cm

Total Surface Area of a cube = 6(side of cube)²

= 6(12)²

= 6(144)

= 864 cm²

Volume of a cube = (side of cube)³

= (12)³

= 12 × 12 × 12

= 1728 cm³

Given: Surface Area of a Cube = 486 cm²

So, firstly we find the side of a cube

Total Surface Area of a Cube = 486 = 6a²

a² = 81

a = = ± 9 = 9cm (side can’t be negative)

Therefore, Side of a cube = 9 cm

Now, Volume of a Cube = a³ = (9)³ = 9 × 9 × 9 = 729 cm³

Given: Volume of a cuboidal tank = 6.4 m³

Length (l) = 2 m

Breadth (b) = 1.6 m

Depth = ?

Volume of a cuboidal tank = lbh

6.4 = 2 × 1.6 × h

6.4 = 3.2 × h

h =

= 2 m

Therefore, the depth of a tank is 2 m

Given: Depth of a well = 28 m

Length = 10 m

Width = 8 m

Volume of a well = l × b × d

= 10 × 8 × 28

= 80 × 28

= 2240 m³

2240 m³ of soil has to be excavated from a rectangular well.

So, to calculate the cost of plastering.

Firstly we have to find the lateral surface of a rectangular well

Lateral Surface Area = 2h (l+b)

= 2 × 28 (10+8)

= 56 × 18

= 1008 m²

Cost of plastering per square metre = ₹ 15

Cost of plastering for 1008 m² = ₹ 15 × 1008 = ₹15120

So, the cost of plastering the vertical walls of a rectangular well is ₹ 15120

Let the side of a cubical box = ‘a’ m

Therefore, Volume of a cubical box = a³ m³

Now, Cost of 1 m³ = ₹500

Therefore, Cost of a³ m³ = ₹256

a = 0.8 m or 80 cm

Length of each side = 0.8 m

volume of a cubical box = a³ = (0.8)³ = 0.512 m³

Given, three metal cubes, V1, V2 and V3 with sides 3 cm, 4 cm and 5 cm respectively.

(i) They are melted to form a single cube and let its length be x.

We know, Volume of a cube = (side)³.

Thus,

Volume of new cube = Volume of V1+Volume of V2+Volume of V3

x³= 3×3×3 + 4×4×4 + 5×5×5

x³= 27+64+125

x³= 216

x= 6 cm

(ii) We know, Surface area of a cube = 6×(side)².

Surface Area of new cube = 6×(side)²

= 6 × (6)²

= 6×36

=216 cm²

Sum of total surface areas of the original three cubes = Surface Area of V1+Surface Area of V2+Surface Area of V3

Sum of total surface areas of the original three cubes = 6 × (3)² + 6 × (4)² + 6 × (5)²

= 6 × 9 + 6 × 16 + 6 × 25

= 54 + 96+ 150

=300

∴ Difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes= 300-216

= 84 cm²

Given, Two cubes of volume 512 cm³each and they are joined end-to-end.

The side of each cube=?

We know, Volume of a cube = (side)³.

So, Volume = (side)³

512 = (side)³

Side = 8 cm

We know, Lateral Surface Area of cuboid = 2h(l+b)

Lateral Surface Area = 2h(l+b)

= 2×8×(16+8)

= 16×24

= 384 cm²

We know, Surface area of a cuboid= 2 × (lb + bh+ hl).

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

= 2×(16×8 + 8×8 + 8×16)

= 2×( 128 + 64 + 128 )

= 2× 300

=600 cm²

Given, length, breadth, and height of a cuboid are in the ratio 6:5:3 and the total surface area is 504 cm².

Let length, breadth, and height of a cuboid be 6x, 5x and 3x.

We know, Total Surface area of a cuboid= 2 × (lb + bh+ hl).

Total Surface area of a cuboid= 504

2 × ( 6x× 5x + 5x × 3x + 3x × 6x) = 504

30x² + 15x² + 18x² = 252

63x²=252

X² =4

X=2

∴ Length = 6x = 6 × 2 =12 cm

Breadth = 5x = 5 × 2 =10 cm

Height = 3x = 3 × 2 =6 cm

We know, Volume of a cuboid= l×b×h

∴ Volume = 12×10×6

= 720 cm³

Given, a room with length, breadth, and height as 8 m, 5 m, and 3 m respectively and the cost of

whitewashing the walls per m² is Rs. 15.

We know, Area of four walls = 2×(l+b)×h

∴ Area of four walls = 2×(l+b)×h

= 2×(8+5)×3

= 2×13×3

= 78 m²

∴ Cost of white washing the walls at the rate of Rs. 15/m²= 15×78

= Rs. 1170

Given, a room with length, breadth, and height as 6 m, 4 m, and 3 m respectively and the cost of

laying tiles on its floor and four walls per m² is Rs. 80.

We know, Area of four walls = 2×(l+b)×h

∴ Area of four walls = 2×(l+b)×h

= 2×(6+4)×3

= 2×10×3

= 60 m²

Area of floor = length × breadth

= 6 × 4

= 24 m²

∴ Cost of laying tiles on its floor and four walls at the rate of Rs. 80/m²= 80×(60+24)

= Rs. 80×84

= Rs. 6720

Given, length, breadth, and height of a cuboid are in the ratio 5:3:2 and its volume is 35.937 m³.

Let length, breadth, and height be 5x, 3x and 2x.

We know, Volume of a cuboid= l×b×h

∴ Volume = l×b×h

35.937 = 5x × 3x × 2x

35.937 = 30x³

X³=1.1979

X= 1.06 m

∴ Length = 5x = 5 × 1.06 = 5.3 m

Breadth = 3x = 3 × 1.06 =3.18 m

Height = 2x = 2 × 1.06 = 2.12 m

We know, Total Surface area of a cuboid= 2 × (lb + bh+ hl).

∴ Total Surface area of a cuboid= 2 × (lb + bh+ hl)

= 2 × (5.3×3.18 + 3.18×2.12 + 2.12×5.3)

= 2 × (16.854 + 6.7416 + 11.236)

= 2 × 34.8316

= 69.66 m²

Given, the perimeter of one face of a cube is 24 cm.

We know, Perimeter of one face of a cube = 4 × side

∴ Perimeter of one face of a cube = 4 × side

24 = 4 × side

Side= 6 cm

We know, Volume of a cube = (side)³

∴ Volume= (6)³

= 6×6×6

= 216 cm²

Given, a wooden box has inner dimensions l =6 m, b =8 m and h =9 m and a uniform thickness of

10 cm.

∴ Length = 6 + 0.20

= 6.2 m

∴ Breadth = 8 + 0.20

= 8.2 m

∴ Height = 9 + 0.20

= 9.2 m

We know, Lateral Surface Area of cuboid = 2h(l+b)

Lateral Surface Area = 2h(l+b)

= 2×9.2×(6.2+8.2)

= 18.4×14.4

= 264.96 m²

The rate of painting lateral surface of the outer side = Rs. 50/m²

∴ Cost of painting = Rs. 50 × 264.96

= Rs. 13248

Let the length of a side of a cube be x.

We know, Volume of a cube = (side)³

∴ Volume of a cube = x³

Each edge of a cube is increased by 20%.

∴ New Side= x + 20% of x

= 1.2x

∴ New Volume of cube = (1.2x)³

= 1.728 x³

Percentage increase in the volume =

=0.728×100

=72.8%

Let side-length of the cube be x.

Then,

Volume of cube = (side)³

= x³

Total Surface Area of cube = 6(side)²

= 6x²

Given, the length of a cube is increased by 10% and its breadth is decreased by 10%.

∴ New length = x + 10% of x

= 1.1x

And, New Breadth = x - 10% of x

= 0.9x

∴ Volume of the new cuboid = l×b×h

= 1.1x× 0.9x × x

= 0.99 x³

∴ Surface Area of the new cuboid = 2 × (lb + bh+ hl)

= 2 × (1.1x × 0.9x + 0.9x× x + x × 1.1x)

= 2 × (0.99x² + 0.9x² + 1.1x²)

= 2 × 2.99x²

= 5.98x² m²

Percentage change in the volume =

= -0.01×100

= -1% (decrease)

Percentage change in the Surface Area

= -0.02×100

= -2% (decrease)

Hence, Volume decreases by 1% and Surface Area decreases by 2%.