Topic Covered

• Introduction
• Surface Area of a Cuboid and a Cube


Wherever we look, usually we see solids. So far, in all our study, we have been dealing with figures that can be easily drawn on our notebooks or blackboards.

These are called plane figures. We have understood what rectangles, squares and circles are, what we mean by their perimeters and areas, and how we can find them. We have learnt these in earlier classes.

It would be interesting to see what happens if we cut out many of these plane figures of the same shape and size from cardboard sheet and stack them up in a vertical pile. By this process, we shall obtain some solid figures (briefly called solids) such as a cuboid, a cylinder, etc.

In the earlier classes, you have also learnt to find the surface areas and volumes of cuboids, cubes and cylinders. We shall now learn to find the surface areas and volumes of cuboids and cylinders in details and extend this study to some other solids such as cones and spheres.

Surface Area of a Cuboid and a Cube

Have you looked at a bundle of many sheets of paper? How does it look? Does it look like what you see in Fig. 13.1?

That makes up a cuboid. How much of brown paper would you need, if you want to cover this cuboid? Let us see:
First we would need a rectangular piece to cover the bottom of the bundle. That would be as shown in Fig. 13.2 (a)

Then we would need two long rectangular pieces to cover the two side ends. Now, it would look like Fig. 13.2 (b).


Now to cover the front and back ends, we would need two more rectangular pieces of a different size. With them, we would now have a figure as shown in Fig. 13.2(c).

This figure, when opened out, would look like Fig. 13.2 (d).

Finally, to cover the top of the bundle, we would require another rectangular piece exactly like the one at the bottom, which if we attach on the right side, it would look like Fig. 13.2(e).

This shows us that the outer surface of a cuboid is made up of six rectangles (in fact, rectangular regions, called the faces of the cuboid), whose areas can be found by multiplying length by breadth for each of them separately and then adding the six areas together.

Now, if we take the length of the cuboid as l, breadth as b and the height as h, then the figure with these dimensions would be like the shape you see in Fig. 13.2(f).

So, the sum of the areas of the six rectangles is:

`color{green}("Area of rectangle" 1 (= l × h))`
`color{red}(text( +))`
`color{green}("Area of rectangle" 2 (= l × b))`
`color{red}(text( +))`
`color{green}("Area of rectangle" 3 (= l × h))`
`color{red}(text( +))`
`color{green}("Area of rectangle" 4 (= l × b))`
`color{red}(text( +))`
`color{green}("Area of rectangle" 5 (= b × h))`
`color{red}(text( +))`
`color{green}("Area of rectangle" 6 (= b × h))`
`color{blue}(= 2(l × b) + 2(b × h) + 2(l × h))`
`color{orange}(= 2(lb + bh + hl))`

This gives us:

`color{navy}("Surface Area of a Cuboid" = 2(lb + bh + hl))`

where l, b and h are respectively the three edges of the cuboid.

`ul"Note :"` The unit of area is taken as the square unit, because we measure the magnitude of a region by filling it with squares of side of unit length.

For example, if we have a cuboid whose length, breadth and height are 15 cm, 10 cm and 20 cm respectively, then its surface area would be:

`color{green}(2[(15 × 10) + (10 × 20) + (20 × 15)] cm^2)`
`color{orange}(= 2(150 + 200 + 300) cm^2)`
`color{blue}(= 2 × 650 cm^2)`
`color{red}(= 1300 cm^2)`

Recall that a cuboid, whose length, breadth and height are all equal, is called a cube. If each edge of the cube is a, then the surface area of this cube would be

`color{green}(2(a × a + a × a + a × a), i.e., 6a^2)` (see Fig. 13.3), giving us

`color{orange}("Surface Area of a Cube" = 6a^2)`

where a is the edge of the cube.

Suppose, out of the six faces of a cuboid, we only find the area of the four faces, leaving the bottom and top faces. In such a case, the area of these four faces is called the lateral surface area of the cuboid.

So, lateral surface area of a cuboid of length l, breadth b and height h is equal to `color{green}(2lh + 2bh or 2(l + b)h.)` Similarly, lateral surface area of a cube of side a is equal to `color{orange}(4a^2.)`

Keeping in view of the above, the surface area of a cuboid (or a cube) is sometimes also referred to as the total surface area. Let us now solve some examples.
Q 3230578412

Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santa Claus on it (see Fig. 13.4). She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively how many square sheets of paper of side 40 cm would she require?
Class 9 Chapter 13 Example 1

Since Mary wants to paste the paper on the outer surface of the box; the quantity of paper required would be equal to the surface area of the box which is of the shape of a cuboid. The dimensions of the box are:

Length =80 cm, Breadth `= 40 cm`, Height `= 20 cm.`
The surface area of the box `= 2(lb + bh + hl)`
`= 2[(80 × 40) + (40 × 20) + (20 × 80)] cm^2`
`= 2[3200 + 800 + 1600] cm^2`
`= 2 × 5600 cm^2 = 11200 cm^2`
The area of each sheet of the paper `= 40 × 40 cm^2`
`= 1600 cm^2`
`textTherefore, number of sheets required = ("surface area of box")/text(area of one sheet of paper)`

`= 11200/1600 = 7`
So, she would require 7 sheets.
Q 3250578414

Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm (see Fig. 13.5). Find how much he would spend for the tiles, if the cost of the tiles is Rs. 360 per dozen.
Class 9 Chapter 13 Example 2

Since Hameed is getting the five outer faces of the tank covered with tiles, he would need to know the surface area of the tank, to decide on the number of tiles required.

Edge of the cubical tank `= 1.5 m = 150 cm (= a)`
So, surface area of the tank `= 5 × 150 × 150 cm^2`
Area of each square tile `= side × side = 25 × 25 cm^2`

`"So, the number of tiles required" = text(surface area of the tank)/text(area of each tile)`

`= (5 ×150 ×150)/(25 × 25)= 180`

Cost of 1 dozen tiles, i.e., cost of 12 tiles `= Rs. 360`
Therefore, cost of one tile `= Rs. (360)/(12) = Rs. 30`

So, the cost of 180 tiles `= 180 ×Rs 30 = Rs 5400`