Surface Area of a Cuboid - Examples, Exercises and Solutions

Every cuboid is made up of rectangles. These rectangles are the faces of the cuboid.

As we know that in a rectangle the parallel faces are equal to each other, we can conclude that for each face found there will be two rectangles.

Let's first look at the face painted orange,

It has width and height, 5 and 3, so we already know that they are two rectangles of size 5x6

Now let's look at the side faces, they also have a height of 3, but their width is 6,

And then we understand that there are two more rectangles of 3x6

Now let's look at the top and bottom faces, we see that their dimensions are 5 and 6,

Therefore, there are two more rectangles that are size 5x6

That is, there are
2 rectangles 5X6

2 rectangles 3X5

2 rectangles 6X3

Remember that the formula for the surface area of a cuboid is:

(length X width + length X height + width X height) 2

We input the known data into the formula:

2*(3*2+2*5+3*5)

2*(6+10+15)

2*31 = 62

Let's go through the options:

A - In this option, we can see that there are two flaps on the same side.

If we try to turn this net into a box, we'll get a box where on one side there are two faces one on top of the other while the other side is "open",
meaning this net cannot be turned into a complete and full box.

B - This net looks valid at first glance, but we need to verify that it matches the box we want to draw.

In the original box, we see that we have four flaps of size 9*4, and only two flaps of size 4*4,
if we look at the net we can see that the situation is reversed, there are four flaps of size 4*4 and two flaps of size 9*4,
therefore we can conclude that this net is not suitable.

C - This net at first glance looks valid, it has flaps on both sides so it will close into a box.

Additionally, it matches our drawing - it has four flaps of size 9*4 and two flaps of size 4*4.

Therefore, we can conclude that this net is indeed the correct net.

D - In this net we can see that there are two flaps on the same side, therefore this net will not succeed in becoming a box if we try to create it.

Let's see what rectangles we have:

8*5

8*12

5*12

Let's review the formula for the surface area of a rectangular prism:

(length X width + length X height + width X height) * 2

Now let's substitute all this into the exercise:

(8*5+12*8+12*5)*2=
(40+96+60)*2=
196*2= 392

This is the solution!
 

We identified that the faces are

3*3, 3*11, 11*3
As the opposite faces of an cuboid are equal, we know that for each face we find there is another face, therefore:

3*3, 3*11, 11*3

or

(3*3, 3*11, 11*3 ) *2

To find the surface area, we will have to add up all these areas, therefore:

(3*3+3*11+11*3 )*2

And this is actually the formula for the surface area!

We calculate:

(9+33+33)*2

(75)*2

150

First, we recall the formula for the surface area of a cuboid:

(width*length + height*width + height*length) *2

As in the cuboid the opposite faces are equal to each other, the given data is sufficient to arrive at a solution.

We replace the data in the formula:

(8*5+3*5+8*3) *2 =

(40+15+24) *2 =

79*2 = 

158

To solve the problem, let's recall the formula for calculating the surface area of a box:

(width*length + height*width + height*length) *2

Let's substitute the known values into the formula, and we'll denote the missing side as X:

2*(3*7+7*X+3*X) = 122

2*(21+7x+3x) = 122

2(21+10x) = 122

Let's expand the parentheses:

42+20x=122

Let's move terms:

20x=122-42

20x=80

Let's simplify:

x=4

And that's the solution!

To calculate the surface area of the rectangular prism, we will need to identify its three faces (each face appears twice):

1*3

1*8

3*8

The formula for the surface area of a rectangular prism is the sum of all the areas of the faces, that is:

We replace the data in the formula:

2*(1*3+1*8+3*8)=
2*(3+8+24) = 
2*35 = 

70

And this is the solution!

The first step is to calculate the relevant data for all the components of the box.

The length of the box = 6

Given that the height of a cuboid is equal to half its length we are able to deduce the height of the box as follows : 6/2= 3

Hence the height = 3

In order to determine the width, we insert the known data into the formula for the volume of the box:

height*length*width = volume of the cuboid.

3*6*width = 72

18*width=72

We divide by 18:

Hence the width = 4

We are now able to return to the initial question regarding the surface of the cuboid.

Remember that the formula for the surface area is:

(height*length+height*width+length*width)*2

We insert the known data leaving us with the following result:

(3*6+4*3+4*6)*2=

(12+24+18)*2=

(54)*2=

108

There is no limitation or rule regarding the dimensions that a cuboid can have.

Therefore the correct answer is D.