Cuboids - 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

The formula to calculate the volume of a cuboid is:

height*length*width

We replace the data in the formula:  

3*5*7

7*5 = 35

35*3 = 105

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 calculate the volume of a cuboid, as we mentioned, we need the length, width, and height.

It is important to note that in the exercise we are given the height and the base area of the cuboid.

The base area is actually the area multiplied by the length. That is, it is the data that contains the two pieces of information we are missing.

Therefore, we can calculate the area by height * base area

15*3 = 45

This is the solution!

Volume formula for a rectangular prism:

Volume = length X width X height

Therefore, first we will place the data we are given into the formula:

45 = 2.5*4*X

We divide both sides of the equation by 2.5:

18=4*X

And now we divide both sides of the equation by 4:

4.5 = X

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!

The formula for the volume of a box is height*length*width

In the specific question, we are given the volume and the height,

and we are looking for the area of the base,

As you will remember, the area is length * width

If we replace all the data in the formula, we see that:

4 * the area of the base = 80

Therefore, if we divide by 4 we see that

Area of the base = 20

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!

Since we are given the area of rectangle CAEG and length AE, we can find GE:

Let's denote GE as X and substitute the data in the rectangle area formula:

$3\times x=15$

Let's divide both sides by 3:

$x=5$

Therefore GE equals 5

Since we are given the area of rectangle ABFE and length AE, we can find EF:

Let's denote EF as Y and substitute the data in the rectangle area formula:

$3\times y=24$

Let's divide both sides by 3:

$y=8$

Therefore EF equals 8

Now we can calculate the volume of the box:

$3\times5\times8=15\times8=120$

Based on the given data, we can argue that:

$BD=HF=2$

We know the area of ABCD and also the length of DB

We'll substitute in the formula to find CD, let's call the side CD as X:

$2\times x=12$

We'll divide both sides by 2:

$x=6$

Therefore, CD equals 6

Now we can calculate the volume of the box:

$6\times2\times3=12\times3=36$

We know the area of DBHF and also the length of HF

We will substitute into the formula in order to find BF, let's call the side BF as X:

$4\times x=20$

We'll divide both sides by 4:

$x=5$

Therefore, BF equals 5

Now we can calculate the volume of the box:

$8\times4\times5=32\times5=160$