How does surface area vary from volume in a cuboid?

Surface area is the outer covering, and volume is the internal capacity.

When is the total surface area applied in everyday use?

When calculating material for painting, wrapping, or coating all sides of a box.

In what scenarios is lateral surface area more applicable?

It is convenient while tiling walls or painting only the vertical faces of a container.

Are two cuboids able to have the same surface area but a different volume?

Yes, varying sizes can create equal surface areas with unequal internal capacities.

Surface Area of a Cuboid: Formula, Solved Examples

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Surface Area of a Cuboid

The surface area of a cuboid is a basic principle in geometry, especially applicable in both academic and practical uses such as packaging, construction, and architecture. Knowledge of how to calculate the total surface area and lateral surface area of a cuboid allows learners and professionals to solve various problems related to this important concept of cuboids. Here, we’ll break down everything about this formula of this simple yet crucial three-dimensional figure of geometry.

1.0What is a Cuboid?

A cuboid is a three-dimensional (3D) geometric shape bounded by six rectangular faces. It is also referred to as a rectangular prism. Every face of a cuboid is a rectangle, and all the angles are right angles (90°). The faces opposite to each other are congruent (similar in shape and size), and the adjacent faces are at right angles to each other. Some common examples of a cuboid are a brick, a shoebox, a book, a matchbox, etc.

2.0Surface Area: Introduction

The surface area of any 3D object is the sum of the area of all its faces. In a cuboid, there are six faces—three pairs of congruent rectangles. Surface area of cuboids or any other three-dimensional figures is classified into two major categories:

Lateral Surface Area (LSA): The lateral surface area of a 3D figure is the total area of all the side faces, excluding the top and bottom (or base and top) surfaces.
Total Surface Area (TSA): The total surface area of a 3D shape is the sum of the areas of all the faces (top, bottom, sides, and curved surfaces if there are any).

3.0Formula for Surface Area of a Cuboid

The formula for the surface area of a cuboid is an essential entity for understanding the properties of a cuboid. Derived from calculations of the areas of all the rectangular faces of the cuboid, and adding them together. As mentioned earlier, the surface area of a three-dimensional figure is classified into two categories; hence, the mathematical formula for both of these categories also differs. For example:

Lateral Surface Area (LSA) of a Cuboid:

The sum of the area of the four faces of a cuboid, meaning it doesn’t include the area of the top or base of the cuboid. The formula for the lateral surface area of a cuboid can be expressed as:

$L S A = 2 h (l + b)$

Total Surface Area (LSA) of a Cuboid

The sum of the area of all six faces, that is, it includes the area of the top and bottom of the cuboid. The total surface area of a cuboid is expressed as:

$TS A = 2 (l b + bh + h l)$

Here in both formulas:

l = length
b = breadth
h = height

4.0Surface Area vs Volume of Cuboid

Feature	Surface Area	Volume
Definition	It is defined as the sum of the areas of all the faces of a cuboid.	Volume is the capacity of a cuboid.
What it Measures	It measures the outer surface or boundary coverage of the cuboid.	It measures how much space it can enclose within its boundaries.
Formula	$L S A = 2 h (l + b)$ $TS A = 2 (l b + bh + h l)$	$Volume of Cuboid = l \times b \times h$
Units	The surface area is measured in square units, for example, m², cm², etc.	The volume of 3D figures is measured in cubic units, such as m³, cm³, etc.
Nature	It works only in two dimensions to measure the outer area.	Volumes are exclusive to three-dimensional figures.
Application	Used to compute how many litres of water a tank can hold	Used to calculate how many litres of water a tank can hold

5.0Solved Examples

Problem 1: A rectangular water tank has dimensions: length = 5 m, breadth = 3 m, and height = 2.5 m. The top of the tank is open. Calculate the total area that needs to be painted on the inside walls and base of the tank. Also, find the cost of painting at ₹12 per square meter.

Solution: Given that, in a water tank

length = 5 m,

breadth = 3 m, and

height = 2.5 m

Required Surface Area = Lateral surface area of the tank + Area of the Base of the tank

Required Surface Area =2h(l+b)+lb

Required Surface Area = $2 \times 2.5 (5 + 3) + 5 \times 3 = 5 \times 8 + 15$

Required Surface Area =40+15= $55 m^{2}$

Now, the cost of painting the tank = 1255=Rs 660

Problem 2: A gift box is in the shape of a cuboid with dimensions 40 cm × 30 cm × 20 cm. If decorative paper is used to wrap the entire surface of the box, how much paper is needed? Also, if the paper comes in sheets of size 60 cm × 40 cm, how many sheets are required?

Solution: Given that the dimensions of the cuboid are 40 cm × 30 cm × 20 cm, and the gift wrapper is 60 cm × 40 cm.

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

= $2 (40 \times 30 + 30 \times 20 + 20 \times 40)$

=2(1200+600+800)=2(2600)

$= 5200 c m^{2}$

Area of one Sheet = 60 × 40 = 2400 cm²

Now,

$the number of sheets required = \frac{T o t a l s u r f a ce a re a o f t h e g i f t b o x}{A re a o f o n e S h ee t} = \frac{5200}{2400} = 2.17$

The number of sheets required is approximately 3.

Problem 3: A company needs to design a cuboidal box to hold 12 cylindrical cans, each of radius 3.5 cm and height 10 cm. The cans are placed upright in 2 rows of 6 cans each (side by side). Determine:

(a) Dimensions of the box

(b) Total surface area of the box

(c) Cost of making the box if the surface cost is ₹5 per 100 cm².

Solution: Given that each row has 6 cans side by side, so:

(a) Length = 6 × 2r = 6 × 7 = 42 cm

Breadth = 2r = 7 cm (1 can diameter)

Height = height of can = 10 cm

(b) Total surface area of the box (TSA)=2(lb+bh+hl)

$TS A = 2 (42 \times 7 + 7 \times 10 + 10 \times 42)$

$TS A = 2 \times 784 = 1568 c m^{2}$

(c) Cost of making a box per 100 cm² = 5 Rs

Cost for 1568 cm² = $\frac{1568}{100} \times 5 = R s 78.4$

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