Surface Areas and Volumes

**Surface Areas of a Cube and a Cuboid**

We give gifts to our friends and relatives at one time or another. We usually wrap our gifts in nice and colourful wrapping papers. Look, for example, at the nicely wrapped and tied gift shown below.

Clearly, the gift is packed in box that is cubical or shaped like a [[mn:glossary]]cube[[/mn:glossary]]. Suppose you have a gift packed in a similar box. How would you determine the amount of wrapping paper needed to wrap the gift? You could do so by making an estimate of the surface area of the box. In this case, the total area of all the faces of the box will tell us the area of the wrapping paper needed to cover the box.

Knowledge of surface areas of the different solid figures proves useful in many real-life situations where we have to deal with them. In this lesson, we will learn the formulae for the surface areas of a cube and a [[mn:glossary]]cuboid[[/mn:glossary]]. We will also solve some examples using these formulae.

[[mn:know]]

- The word ‘cuboid’ is made up of ‘cube’ and ‘-oid’ (which means ‘similar to’). So, a cuboid indicates something that is similar to a cube.
- A cuboid is also called a ‘rectangular prism’ or a ‘rectangular parallelepiped’.

[[/mn:know]]

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**Formulae for the Surface Area of a Cuboid**

Consider a cuboid of length *l*, breadth *b* and height *h*.

The formulae for the surface area of this cuboid are given as follows:

**Lateral surface area of the cuboid = 2****h**** (****l + b****)**

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

Here, lateral surface area refers to the area of the solid excluding the areas of its top and bottom surfaces, i.e., the areas of only its four standing faces are included. Total surface area refers to the sum of the areas of all the faces.

[[mn:know]]

Two mathematicians named Henri Lebesgue and Hermann Minkowski sought the definition of surface area at around the twentieth century.

[[/mn:know]]

[[mn:scientist]]

Henri Lebesgue (1875−1941) was a French mathematician who is famous for his theory of integration. His contribution is one of the major achievements of modern analysis which greatly expands the scope of Fourier analysis. He also made important contributions to topology, the potential theory, the *Dirichlet* problem, the calculus of variations, the set theory, the theory of surface area and the dimension theory.

Hermann Minkowski (1864−1909) was a Polish mathematician who developed the geometry of numbers and made important contributions to the number theory, mathematical physics and the theory of relativity. His idea of combining time with the three dimensions of space, laid the mathematical foundations for Albert Einstein’s theory of relativity.

[[/mn:scientist]]

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**Example Based on the Surface Area of a Cuboid**

[[mn:know]]

The concept of surface area is widely used in chemical kinetics, regulation of digestion, regulation of body temperature, etc.

[[/mn:know]]

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**Formulae for the Surface Area of a Cube**

Consider a cube with edge *a*.

The formulae for the surface area of this cube are given as follows:

**Lateral surface area of the cube = 4****a**^{2}

**Total surface area of the cube =** **6****a**^{2}

Here, lateral surface area refers to the area of the solid excluding the areas of its top and bottom surfaces, i.e., the areas of only its four standing faces are included. Total surface area refers to the sum of the areas of all the faces.

[[mn:know]]

- A cube can have 11 different nets.
- Cubes and cuboids are [[mn:glossary]]convex polygons[[/mn:glossary]] that satisfy
**Euler’s formula**, i.e.,**F + V − E = 2**.

[[/mn:know]]

[[mn:more]]**Length of the diagonal in a cube and in a cuboid**

A cuboid has four diagonals (say AE, BF, CG and DH). The four diagonals are equal in length.

Let us consider the diagonal AE.

In rectangle ABCD, length of diagonal AC =

Now, ACEG is a rectangle with length AC and breadth CE or *h*.

So, length of diagonal AE =

∴ **Length of the diagonal of a cuboid =**

A cube is a particular case of cuboid in which the length, breadth and height are equal to *a*.

∴ **Length of the diagonal of a cube** =

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[[mn:solvedExample]]

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**Example 1**:

**There are twenty-five cuboid-shaped pillars in a building, each of dimensions 1 m ×1 m × 10 m. Find the cost of plastering the surface of all the pillars at the rate of Rs 16 per m**^{2}**.**

**Solution**:

Length (*l*) of one pillar = 1 m

Breadth (*b*) of one pillar = 1 m

Height (*h*) of one pillar = 10 m

∴ Lateral surface area of one pillar= 2*h* (*l* + *b*)

= 2 × 10 × (1 + 1) m^{2}

= 40 m^{2}

⇒ Lateral surface area of twenty-five pillars = (25 × 40) m^{2} = 1000 m^{2}

Cost of plastering 1 m^{2} of surface = Rs 16

⇒ Cost of plastering 1000 m^{2} of surface = Rs (16 × 1000) = Rs 16000

Thus, the cost of plastering the twenty-five pillars of the building is Rs 16000.

**Example 2**:

**Find the length of the diagonal of a cube whose surface area is 294 m**^{2}**.**

**Solution**:

Let the edge of the given cube be *a*.

∴ Surface area of the cube = 6*a*^{2}

It is given that the surface area of the cube is 294 m^{2}.

So, 6*a*^{2} = 294

Now, length of the diagonal of the cube =

[[/mn:solvedExampleLow]]

[[mn:solvedExampleMedium]]

**Example 1: **

**A metallic container (open at the top) is a cuboid of dimensions 7 cm ****×**** 5 cm ****×**** 8 cm. What amount of metal sheet went into making the container? Also, find the cost required for painting the outside of the container, excluding the base, at the rate of Rs 17 per 3 cm ^{2}.**

**Solution: **

Length (*l*) of the container = 7 cm

Breadth (*b*) of the container = 5 cm

Height (*h*) of the container = 8 cm

The container is open at the top. Therefore, while calculating the amount of metal sheet used, we will exclude the top part.

∴ Amount of metal sheet used = Total surface area − Area of the top part

= 2 (*lb* + *bh + lh*) − *lb*

= [2 × (7 × 5 + 5 × 8 + 7 × 8) − 7 × 5] cm^{2}

= [2 × (35 + 40 + 56) − 35] cm^{2}

= (2 × 131 − 35) cm^{2 }

= 227 cm^{2}

Thus, 227 cm^{2} of metal went into making the given container.

Now, area to be painted = Lateral surface area of the cuboid

= 2*h* (*l + b*)

= [2 × 8 × (7 + 5)] cm^{2}

= (16 × 12) cm^{2}

= 192 cm^{2}

Cost of painting 3 cm^{2} of surface = Rs 17

⇒ Cost of painting 1 cm^{2} of surface = Rs

⇒ Cost of painting 192 cm^{2} of surface = = Rs 1088

Therefore, the cost of painting the outside of the container is Rs 1088.

**Example 2: **

**If the total surface area of a cube is 24 x^{2}, then find the surface area of the cuboid formed by joining **

**i)two such cubes. **

**ii)three su...**

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