Subject: Compulsory Maths

Solid means having three dimensions (length, breadth, and thickness), as a geometrical body or figure. Or it may also defined as relating to bodies or figures of three dimensions.

Solid means having three dimensions (length, breadth and thickness), as a geometrical body or figure. Or it may also be defined as relating to bodies or figures of three dimensions.

Prisms are the solid object that has two opposite faces congruent and parallel.

The above figures are different solid figures which have congruent opposite face.

Each congruent face of a prism is called its cross-section. There are infinite numbers of imaginary surfaces that cut the prism perpendicular to its height or length with congruent surfaces.

**Total surface area of prism**

Let's take a cartoon box. While unfolding the cartoon box, there we can find many faces of the boxes. Measure all the surfaces and find the area of each surface. If the length is “l” breadth is “b” and height is ‘h’, what will be the total surface area?

In a box, altogether there are 6 surfaces. Among them three pairs are congruent. Thus, total surface area (A) = 2(l \(\times\) b) + 2(b \(\times\)h) + 2(l \(\times\) h) square units

= 2(lb + bh + lh) square units.

In the case of irregular shape, area of surface are calculated separately and added to find out the total surface area.

**Lateral surface area of prism**

The sum of areas of four lateral surfaces of the prism is called the lateral surface area of the prism.

Lateral surface area (S) = 2(l \(\times\) h) + 2(b \(\times\) h)

= 2lh + 2bh

= h \(\times\) 2(l+b)

= h \(\times\) p

\(\therefore\) lateral surface area (S) = height \(\times\) perimeter of base

Total surface area of the prism (TSA) = Lateral surface area + 2 \(\times\) area of cross section

i.e. TSA =LSA + 2A

**Volume of prism**

let's take a cuboidal prism.

For a cuboid | For a cube |

Volume (V) = l \(\times\) b \(\times\) h V = A \(\times\) H Where, A = l \(\times\) b |
Volume (V) = l \(\times\) l \(\times\) l (V) = l (V) = A \(\times\) l |

Thus, Volume = Area of cross-section × height

S.N |
Solid Figures |
Area of base or cross section |
Lateral Surface Area |
Total surface Area |
Volume |

1. | Cuboid | A = l \(\times\) b | 2h(l + B) | 2(lb + bh + lh) | V = l\(\times\) b\(\times\) h |

2. | Cube | A = l^{2} |
4l^{2 }or 4A |
6l^{2}or 6A |
V = l^{3 } |

3. | Prism | A = base area or A = area of cross section | h \(\times\) p | p\(\times\) h + 2A | V = A\(\times\) h |

- Lateral Surface Area (LSA) = height \(\times\) perimeter of base
- Total Surface Area of the prism (TSA) = Lateral surface area + 2 \(\times\) area of cross section
- Volume of cuboid (V) = l \(\times\) b \(\times\) h

V = A \(\times\) H

Where, A = l \(\times\) b

- Volume of cube(V) = l \(\times\) l \(\times\) l

(V) = l^{2} \(\times\) l

(V) = A \(\times\) l

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Find the total surface area and the volume of the given cube.

Solution:

Here, length (l) = 8cm

Total surface area (TSA) = 6l^{2}

= 6× (8cm)^{2}

= 6× 64cm^{2}

= 384cm^{2}

Again,

Volume (V) = l^{3}

= (8cm)^{3}

= 512cm^{3}

Hence, TSA = 384cm^{2} and volume = 512cm^{3}

Find the lateral surface area of the given prism. Also find it's volume.

Solution:

Here,

Lateral Surface Area (LSA) = perimeter× height

= (1cm +cm + 4cm + 1cm + 8cm + 6cm + 8cm) × 6cm

= 34cm× 6cm

= 204cm^{2}

Also,

Volume (V) = Area of cross section× height

= (8cm× 6cm - 4cm× 3cm)× 6cm

= (48cm^{2} - 12cm^{2})× 6cm

= 36cm^{2}× 6cm

= 216cm^{3}

According to the dimension in the given prism, calculate

- the area of cross section
- lateral surface area
- total surface area
- volume

Solution:

Here, taking only front section only

Area of cross section (A) = (5× 4 + 9× 3 + 13× 3)cm^{2}

= (20 + 27 + 39)cm^{2}

= 86cm^{2}

Lateral surface area (LSA) = p× l

= (5 + 4 + 4 + 3 + 4 + 3 + 13 + 10)cm× 3cm

= 46cm + 3cm

= 138cm^{2}

Total surface area (TSA) = LAS + 2A

= 138cm^{2} + 172cm^{2}

= 310cm^{2}

Volume of the prism (V) = A× h

= 86cm^{2}× 3cm

= 258cm^{3}

Find the area of cross section, lateral surface area, total surface area and volume of the given prism.

Solution:

Here, taking front section of the figure

Area of cross section (A) = 9cm× 15cm + 18cm × 15cm

= 135cm^{2} + 270cm^{2}

= 405cm^{2}

Lateral Surface Area (LSA) = p× l

= (30 + 18 + 9 + 15 + 9 + 15)cm × 10cm

= 96cm× 10cm

= 960cm^{2}

Total Surface Area (TSA) = LSA + 2A

= 960cm^{2} + 2 × 405cm^{2}

= 960cm2 + 810cm^{2}

=1770cm^{2}

Volume of the prism (V) = A × h

= 405cm^{2}× 10cm

= 4050cm^{3}

Find the LSA,TSA and volume of the given cubiod.

Solution:

Here,

Length (l) = 10cm

Breadth (b) = 5cm

Heigth (h) = 8cm

Now,

LSA = 2h(l + b)

= 2× 8cm (10cm + 5cm)

= 16cm× 15cm

= 240cm^{2}

Again,

TSA = 2(lb + bh + lh)

= 2 (10cm× 5cm + 5cm× 8cm +10cm× 8cm)

= 2 (50cm^{2} + 40cm^{2} + 80cm^{2})

= 2× 170cm^{2}

= 340cm^{2}

Also,

Volume (V) = l× b× h

= 10cm× 5cm× 8cm

= 400cm^{3}

A box is of cuboid shaped whose length, breadth and height are 30cm, 20cm and 10cm respectively. How many soaps of size 3cm long, 2cm wide and 2cm thick can be fitted on the box?

Solution:

Here,

Length of box (l) = 30cm

Breadth of box (b) = 20cm

Height of box (h) = 10cm

Length of soap (l_{1}) = 3cm

Breadth of soap (b_{1}) = 2cm

Height of soap (h_{1}) = 2cm

Now,

Volume of box (V) = l× b× h

= 30cm× 20cm× 10cm

= 6000cm^{3}

Volume of soap (V_{1}) =l_{1}× b_{1}× h_{1}

= 3cm × 2cm ×2cm

= 12cm^{3}

∴ Number of soap that can be fitted on the box (N) = \(\frac{V}{V_1}\)

= \(\frac{600cm^3}{12cm^3}\)

= 50

Hence, 50 soap can be fitted on the box.

A room is 16m long, 15m wide and 13m high. Find the area of 4 walls of the room. Find the cost of painting the walls at Rs 120/m^{2}.

Solution:

Here,

Length (l) = 16m

Breadth (b) = 15m

Heigth (h) = 13m

Area of 4 walls (A) = 2h(l + b)

= 2× 13m (16m + 15m)

= 26m× 31m

= 806m^{2}

Cost of painting wall (C) = Rs 120m^{2}

Total Cost (T) = ?

∴ Total Cost (T) = C× A

= Rs 120 ×806m^{2}

= Rs 96,720

How many bricks of size 15cm × 10cm × 5cm will be required to build a wall of size 20,000cm × 500cm × 30cm? What is the cost of bricks if 1,000 bricks cost Rs 17,000?

Solution:

Here,

Volume of each brick (V_{1}) = l_{1} × b_{1} × h_{1}

=15cm× 10cm× 5cm

= 750cm^{3}

Volume of wall (V) = l × b × h

=20,000cm× 500cm× 30cm

= 300,000,000cm^{3}

Now,

Number of Bricks (N) = \(\frac{V}{V_1}\)

= \(\frac{300,000,000cm^3}{750cm^3}\)

= 400,000

Again,

The cost of 1,000 bricks = Rs 17,000

Cost of 1 bricks = \(\frac{Rs 17,000}{1,000}\)

Cost of 400,000 bricks = \(\frac{Rs 17,000}{1,000}\)× 400,000 = Rs 6,800,000

∴ Rs 6,800,000 is required to make the wall.

What is the difference between LSA and TSA. Illustrate with the examples.

LSA | TSA |

The lateral surface area of a three-dimensional object is the surface area of the object minus the area of its bases. | The total surface area of a three- dimensional object is the surface area of the object adding all of its areas. |

For example, dice. We found that the surface area of a six-sided dice.Since the dice has two bases, we subtractthe area of the two bases. | For example,dice. We found that the surface area of a six-sided dice. We add all the areas of a dice. |

After the earthquake of 25 April 2015, the northern wall of Singh Durbar collapsed down. 300m long, 1m wide and 4.5m high wall is to be constructed. In order to construct this wall, brick of size 20cm × 10cm × 5cm was chosen. If the cost of brick, transport and load/unload for a unit brick are Rs 13, Rs 1.20 and Rs 0.20 respectively, find the total cost of constructing the new wall.

Solution:

Here,

Length of wall (l) = 300m = 30000cm

Breadth of wall (b) = 1m = 100cm

Height of wall (h) = 4.5m = 450cm

Length of brick (l_{1}) = 300m = 30000cm

Breadth of brick (b_{1})= 1m = 100cm

Height of brick (h_{1}) = 4.5m = 450cm

Cost of brick = Rs 13

Cost of Transport = Rs 1.20

Cost of load/unload = Rs 0.20

Total cost of a brick = Rs (13 + 1.20 + 0.20) = Rs 14.4

Now,

Volume of a brick (V_{1}) =20cm× 10cm× 5cm

= 1000cm^{3}

Volume of wall (V) = 30000cm × 100cm× 450cm

= 1,350,000,000cm^{3}

∴ Number of bricks (N) = \(\frac{V}{V_1}\)

= \(\frac{1,350,000,000cm^3}{1000cm^3}\)

= 1,350,000

Cost of 1 brick = Rs 14.4

Cost of1,350,000 = Rs 14.4× 1,350,000 = Rs 19,440,000

∴ Total cost of constructing a wall is Rs 19,440,000.

A soap factory has a store room of 30m long, 8m wide and 6m high. How many packets of soap having dimension of 80cm × 60cm × 50cm fulfill the store room.

Solution:

Here,

Length of store (l) = 30m = 3000cm

Breadth of store (b) = 8m = 800cm

Heigth of store (h) = 6m = 600cm

Length of soap (l_{1}) = 80cm

Breadth of soap (b_{1}) = 60cm

Heigthof soap (h_{1}) = 50cm

Now,

Volume of a store room (V) = l× b× h

= 3000cm× 800cm× 600cm

= 1,440,000,000cm^{3}

Volume of packets of soap (v_{1}) = l_{1}× b_{1}× h_{1}

=80cm× 60cm× 50cm

= 2,40,000cm^{3}

Then,

Number of soap of packets that can be fulfill the store room (N) = \(\frac{V}{V_1}\)

=\(\frac{1,440,000,000cm^3}{2,40,000cm^3}\)

= 6000

∴ 6000 soap of packets that can be fulfill the store room.

What do you mean by a prism? Give any two examples.

A solid object having opposites faces congruent and parallel is known as a prism. Cube and Square prism are the examples of a prism.

Volume of a cubical tank is 125m^{3}. Find its total surface area.

Solution:

Volume of a cubical tank (V) = 125m^{3}

Total Surface Area (TSA) = ?

Now,

V = l^{3}

Or, 125m^{3} = l^{3}

or, l = (125m^{3})^{\(\frac{1}{3}\)}

∴ l = 5m

Now, TSA = 6l^{2}

= 6× (5m)^{2}

= 150m^{2}

A cubical tank is 6m long, 4m wide and 5m high. How much water does this tank hold?

Solution:

Length of tank (l) = 6m

Breadth of tank (b) = 4m

Height of tank (h) = 5m

Now,

Volume of a tank (V) = l× b× h

= 6m× 4m × 5m

= 120m^{3}

Hence, the tank hold 120m^{3} = 120× 1000 = 120,000 litre of water.

A prism having the perimeter of base 24cm has a height of 5cm. Find the lateral surface area.

Solution:

Perimeter (P) = 24cm

Height (h) = 5cm

Now,

Lateral surface area (LSA) = P× h

= 24cm × 5cm

= 120cm^{2}

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