Surface Areas And Volumes

Given, l = b = h = 4cm.

Formula Used:- total surface area = 2(lb + bh + lh)

Lateral surface area = 2h(l + b)

Here all dimensions are equal i.e. l = b = h = a(say)

Hence, total surface area = 6a²

= 6 × 4²

= 96cm²

Lateral surface area = 4a²

= 4 × 4²

= 64cm²

Given, length(l) = 8cm

Breath(b) = 6cm

Height(h) = 5cm

Formula Used:- total surface area = 2(lb + bh + lh)

Lateral surface area = 2h(l + b)

Hence, total surface area = 2((8 × 6) + (6 × 5) + (5 × 8))

= 236cm²

Lateral surface area = (2 × 5)(8 + 6)

= 10 × 14

= 140cm²

Given, total surface area of a cube = 1350m²

Let side of the cube be ‘a’

Formula used:- volume of cube = a³

Total surface area = 6a²

Hence, 6a² = 1350

a² =

a =

a = 15m

volume of the cube = a³

= 15³

= 3375m³

Given, length(l) = 12m

Breath(b) = 10m

Height(h) = 7.5m

We know, in case of area of four wall which is formed a cuboid, eliminate the area of base and top i.e. (l × b) Hence,

Area of four walls = 2(bh + lh)

= 2((10 × 7.5) + (12 × 7.5))

= 2(165)

= 330m²

Given, volume of a cuboid(v) = 1200cm³

Length(l) = 15cm

Breath(b) = 10cm

Volume of a cuboid = l × b × h

Hence, l × b × h = 1200

15 × 10 × h = 1200

h =

h = 8cm

(i) If each dimension is doubled then the total surface area becomes, = 2((4lb) + (4bh) + (4lh))

= 4 × [2(lb + bh + lh)]

∴ the area becomes four times.

(ii) If each dimension is tripled then the total surface area becomes, = 2((9lb) + (9bh) + (9lh))

= 9 × [2(lb + bh + lh)]

∴ the area becomes nine times.

It is clear from the above two solutions that the area of cuboid becomes n²times the previous area if each dimension raised to n times.

Given, a triangular prism with base dimensions 3cm, 4cm and 5cm. and height is 10cm

Volume of this type of prism = area of base × height

The triangle is right angled triangle. Hence, area of base is × 4 × 3.

Volume = × 4 × 3 × 10

= 60cm³

Given, A regular square pyramid with base 16m and height is 3m.

Volume of the pyramid = × area of the base × height

= × 16² × 3

= 256m³

Given, An Olympic swimming pool is in the shape of a cuboid of dimensions 50 m. long and 25 m. wide and 3m deep.

For calculating the volume of water required into pool, we need to find the volume of cuboidal pool i.e. (lbh).

Volume = l × b × h

= 50 × 25 × 3

= 3750m³

Given, radius of the base of a cylinder r = 56cm = 0.56m

And height h = 1.4m

To find the metal sheet required, we need to find the total surface area of the cylinder.

Total surface area of the cylinder = 2πr(r + h)

= 2 × π × 0.56 × (0.56 + 1.4)

= 6.89 ≃ 6.9m²

Given, volume of cylinder = 308 cm³

Height = 8cm

Volume of the cylinder = πr²h

308 = 3.14 × r² × 8

r² =

r =

r = 3.5cm

lateral surface area = 2πrh

= 2 × 3.14 × 3.5 × 8

= 175.9≃ 176cm²

Total surface area = 2πr(r + h)

= 2 × π × 3.5 × (3.5 + 8)

= 252.89 ≃ 253cm²

Given, A metal cuboid of dimension 22 cm. × 15 cm. × 7.5 cm. was melted and cast into a cylinder of height 14 cm

According to the question the volume remains same after casting the cuboid inti cylinder, i.e. lbh = πr²h.

22 × 15 × 7.5 = 3.14 × r² × 14

r² =

r =

r = 7.5cm

Given, An overhead water tanker is in the shape of a cylinder has capacity of 616 litres

As 1 litre = 0.001 m³


the volume of the cylinder = 0.616 m³.

Diameter of the cylinder = 5.6 m

Radius of the cylinder = = 2.8 m

Volume of the cylinder = πr²h.

0.616 = 3.14 × 2.8² × h

h = 0.025 m

= 2.5 cm

Given, A metal pipe is 77 cm. long

Inner Diameter = 4cm i.e. r = 2cm

Outer diameter = 4.4cm i.e. R = 2.2cm

(i) inner curved surface area = 2πrh

= 2 × 3.14 × 2 × 77

= 967.6cm²

(ii) outer curved surface area = 2πRh

= 2 × 3.14 × 2.2 × 77

= 1063.8cm²

(iii) total surface area = 2πrh + 2πrh + 2π(R² – r²)

= 967.6 + 1063.8 + 2 × 3.14 × (2.2² – 2²)

= 2036.6cm²

Given, A cylindrical pillar, diameter = 56cm = 0.56m i.e. r = 0.28cm

Height h = 35m

Number of pillars n = 16

Curved surface area of a pillar = 2πrh

= 2 × 3.14 × 0.28 × 35

= 61.54cmCurved surface area of 16 pillars = 16 × 61.54

= 984.64cm²

Cost of painting = 5.5 × 984.64

= 5415.52

Given, diameter = 84cm = 0.84m i.e r = 0.42m

Length h = 120cm = 1.20m

Number of revolutions n = 500

Area of the playground = n × curved surface area of the roller

= 500 × 2πrh

= 500 × 2 × 3.14 × 0.42 × 1.2

= 1582.56m²

Given, inner diameter of a circular well = 3.5 m i.e. r = 1.75m

Height = 10m

(i) curved surface area = 2πrh

= 2 × 3.14 × 1.75 × 10

= 109.9m²≃ 110cm²

(ii) cost of plastering = 40 × 110

= 4400

(i) Given, diameter = 4.2m

Radius = 2.1m

Height = 4.5m

Total surface area = 2πr(r + h)

= 2 × 3.14 × 2.1 × (2.1 + 4.5)

= 87m²

(ii) if of the steel was wasted in making the tank. The sheet used in making the tank = (1−) × 87

= × 87

= 79.75m²

Given, r = 28cm

h = 2.1m = 210cm

to find the capacity of water ,the volume of the cylinder = πr²h

= 3.14 × 28² × 210

= 516969.6cm³

1litere = 1000cm³

Hence, the water stored = = 516.9literes

Given, curved surface area = 1760cm²

Volume = 12320cm³

For finding the height of the cylinder,

curved surface area = 1760cm²

2πrh = 1760

rh =

rh = 280.25 --------(i)

volume = 12320cm³

πr²h = 12320cm³

r × rh =

r =

from equation (i), r =

r = 14cm

hence, rh = 280.25

h =

h = 20cm