We have learnt in the earlier classes about 3-Dimension structures. The 3D shapes are those which do not lie completely in a plane. Any 3D shape has dimensions namely length, breadth and height.

**Surface Area of Cuboid and Cube**

We have learnt
in the earlier classes about 3-Dimension structures. The 3D shapes are those which
do not lie completely in a plane. Any 3D shape has dimensions namely length, breadth
and height.

** **

Cuboid: A cuboid is a closed solid figure bounded by six rectangular plane regions. For example,
match box, Brick, Book.

A cuboid
has 6 faces, 12 edges and 8 vertices. Ultimately, a cuboid has the shape of a rectangular
box.

**Total Surface Area **(TSA) of a cuboid is the sum of the areas of all the faces that enclose the cuboid. If we leave out the areas of the top and bottom of the cuboid we get what is known as its Lateral Surface Area (LSA).

In the Fig
7.10, *l*, *b* and *h* represents length, breadth and height respectively.

(i) Total
Surface Area (TSA) of a cuboid

= 2 (*lb* + *bh* + *lh* ) *sq. units.*

(ii) Lateral
Surface Area (LSA) of a cuboid

= 2 (*l*+*b*)*h* *sq. units*.

We are using
the concept of Lateral Surface Area (LSA) and Total Surface Area (TSA) in real life
situations. For instance a room can be cuboidal in shape that has different length,
breadth and height. If we require to find areas of only the walls of a room, avoiding
floor and ceiling then we can use LSA. However if we want to find the surface area
of the whole room then we have to calculate the TSA.

If the length, breadth and height of a cuboid are *l*, *b*
and *h* respectively. Then

(i) Total Surface Area = 2 (*lb* + *bh* + *lh* ) *sq.units.*

(ii) Lateral Surface Area = 2 (*l*+*b*)*h* *sq.units*.

**Note**

â€¢ The top and bottom area in a cuboid is independent of height. The
total area of top and bottom is 2*lb*. Hence LSA is obtained by removing 2*lb*
from 2(*lb*+*bh*+*lh*).

â€¢ The units of length, breadth and height should be same while calculating
surface area of the cuboid.

**Example 7.4**

Find the
TSA and LSA of a cuboid whose length, breadth and height are 7.5 *m*, 3 *m*
and 5 *m* respectively.

*Solution*

Given the
dimensions of the cuboid;

that is length
(*l)* = 7.5 *m*, breadth (*b)* = 3 *m* and height (*h)*
= 5 *m*.

TSA =
2(*lb* + *bh* + *lh*)

= 2[(7.5 Ã— 3) + (3 Ã— 5) + (7.5 Ã— 5)]

= 2(22.5 +15 + 37.5)

= 2Ã—75

= 150 *m*^{2}

LSA =
2(*l* + *b*) Ã— *h*

=2(7.5+3)Ã—5

=2Ã—10.5Ã—5

= 105 *m*^{2}

**Example 7.5**

The length,
breadth and height of a hall are 25 *m*, 15 *m* and 5 *m *respectively.
Find the cost of renovating its floor and four walls at the rate of â‚¹80 per *m*^{2}.

*Solution*

Here, length
(*l* ) = 25 *m*, breadth (*b*) =15 *m*, height (*h*) =
5 *m*.

Area of four
walls = LSA of cuboid

= 2(*l*
+
*b*) Ã— *h*

= 2(25
+15)
Ã—
5

= 80
Ã—
5 =
400 *m*^{2}

Area of the
floor =
*l* Ã— *b*

= 25 Ã—15

= 375 *m*^{2}

Total renovating
area of the hall

= (Area of
four walls + Area of the floor)

= (400 + 375) *m*^{2}

= 775 *m*^{2}

Therefore,
cost of renovating at the rate of â‚¹80 per *m*^{2} =
80 Ã—
775

= â‚¹ 62,000

** **

**Cube: **A cuboid whose length, breadth and height are all equal is called as
a** **cube.

That is a
cube is a solid having six square faces. Here are some real-life examples.

A cube being
a cuboid has 6 faces, 12 edges and 8 vertices.

Consider
a cube whose sides are â€˜*a*â€™ units as shown in the Fig 7.14. Now,

(i)
Total Surface Area of the cube

= sum
of area of the faces (*ABCD*+*EFGH*+*AEHD*+*BFGC*+*ABFE*+*CDHG*)

= (*a*
^{2} + *a* ^{2} + *a* ^{2} +
*a* ^{2} + *a* ^{2} +
*a*^{2} )

= 6*a*^{2}
*sq. units*

(ii)
Lateral Surface Area of the cube

= sum
of area of the faces (*AEHD+BFGC+ABFE+CDHG*)

= (*a*
^{2} + *a* ^{2} + *a* ^{2} +
*a*^{2} )

= 4*a*^{2}
*sq. units*

If the side of a cube is *a* units, then,

(i) The Total Surface Area = 6*a*^{2} sq.units

(ii) The Lateral Surface Area = 4*a*^{2} sq.units

**Thinking Corner:** Can you get these
formulae from the corresponding formula of Cuboid?

**Example 7.6**

Find the
Total Surface Area and Lateral Surface Area of the cube, whose side is 5 *cm*.

*Solution*

The side
of the cube (*a*) = 5 *cm*

Total Surface
Area = 6*a*^{2} = 6(5^{2} ) = 150 *sq.
cm*

Lateral Surface
Area = 4*a*^{2} = 4(5^{2} ) = 100 *sq.
cm*

A cube has
the Total Surface Area of 486 *cm*^{2}. Find its lateral surface area.

Here, Total
Surface Area of the cube = 486 *cm*^{2}

6*a*^{2}
=
486
â‡’ *a*^{2} = 486/6 and so, *a*^{2} = 81 . This gives *a* = 9.

The side
of the cube = 9 *cm*

Lateral Surface
Area =
4*a*^{2} = 4 Ã— 9^{2} = 4 Ã— 81 = 324 *cm*^{2}

**Example 7.8**

Two identical
cubes of side 7 *cm* are joined end to end. Find the Total and Lateral surface
area of the new resulting cuboid.

Side of a
cube = 7 *cm*

Now length
of the resulting cuboid (*l*) = 7+7 =14 *cm*

Breadth (*b*)
= 7 *cm*, Height (*h*) = 7 *cm*

So, Total
Surface Area = 2(*lb* + *bh* +
*lh*)

= 2 [(14Ã—7)+(7Ã—7)+(14Ã—7)]

= 2(98 + 49 + 98)

=2Ã—245

= 490 *cm*^{2}

Lateral Surface
Area =
2(*l* + *b*) Ã— *h*

= 2(14+7)Ã—7
=2Ã—21Ã—7

= 294 *cm*^{2}

Tags : Formula, Example Solved Problems | Mensuration | Maths , 9th Maths : UNIT 7 : Mensuration

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

9th Maths : UNIT 7 : Mensuration : Surface Area of Cuboid and Cube | Formula, Example Solved Problems | Mensuration | Maths

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright Â© 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.