Volume of a Orthohedron: Subtraction or addition to a larger shape

To solve this problem, we will calculate the volume of each rectangular prism separately and then sum these volumes:

• Rectangular Prism 1: The prism with a square base has dimensions as follows:
• Side length of the square base = 5 units
• Height = 9 units
• The volume of a rectangular prism with a square base is given by $V = \text{side}^2 \times \text{height}$.
• Substituting the given values: $V_1 = 5^2 \times 9 = 25 \times 9 = 225 \text{ cubic units}$.
• Rectangular Prism 2: The second prism's dimensions:
• Length = 3 units
• Width = 2 units
• Height = 4 units
• The volume of this rectangular prism is calculated as $V = \text{length} \times \text{width} \times \text{height}$.
• Substituting the given values: $V_2 = 3 \times 2 \times 4 = 24 \text{ cubic units}$.

Finally, adding the volumes of the two prisms gives us the total volume:

$V_{\text{total}} = V_1 + V_2 = 225 + 24 = 249 \text{ cubic units}$.

Therefore, the volume of the new shape is $249$ cubic units.

To solve this problem, we'll follow these steps:

• Step 1: Calculate the volume of one smaller cuboid.
• Step 2: Multiply the volume of one smaller cuboid by the number of cuboids (5) to find the total volume of the large cuboid.

Now, let's work through each step:
Step 1: The dimensions for each smaller cuboid are given as:

• Length: $4 \, \text{cm}$
• Width: $2 \, \text{cm}$
• Height: $5 \, \text{cm}$

Using the volume formula for a cuboid, we find the volume of one smaller cuboid:
$V_{\text{small}} = 4 \times 2 \times 5 = 40 \, \text{cm}^3$
Step 2: Since there are 5 such smaller cuboids, the total volume for the large cuboid is:
$V_{\text{large}} = 5 \times 40 = 200 \, \text{cm}^3$

Therefore, the volume of the large cuboid is $200 \, \text{cm}^3$.

To solve the problem, we'll start by calculating the volume of the small cuboid.

Step 1: Volume of the small cuboid
The small cuboid dimensions are $AB = 5$, $BC = 4$, and $BK = 3$. Hence, its volume is:

$V = 5 \times 4 \times 3 = 60 \, \text{cm}^3$

Step 2: Volume of the large cuboid
Since the small cuboid fits four times into the large cuboid, the volume of the large cuboid is:

$4 \times 60 = 240 \, \text{cm}^3$

Step 3: Volume of the large cuboid minus the small cuboid
Subtracting the volume of the small cuboid from the large cuboid, we have:

$240 - 60 = 180 \, \text{cm}^3$

Thus, the volume of the large cuboid minus the small cuboid is $\boxed{180 \, \text{cm}^3}$.

To solve this problem, we will derive the dimensions of the cuboid using the given volume and side BK, and then find the area of triangle ABC.

We start with the given information:
- Volume of the cuboid: $120 \, \text{cm}^3$
- Side $BK = 4 \, \text{cm}$

The volume formula of the cuboid is:
$V = l \times w \times h = 120 \, \text{cm}^3$

We assume $BK = h = 4 \, \text{cm}$, leading to:
$l \times w \times 4 = 120$
$l \times w = \frac{120}{4} = 30$

Now, to get the area of triangle ABC, which forms a right triangle with sides $l$ and $w$ being the base and height, respectively, we use:
$A = \frac{1}{2} \times l \times w = \frac{1}{2} \times 30 = 15 \, \text{cm}^2$

Thus, the area of triangle ABC is $15 \, \text{cm}^2$.

To solve this problem, we'll follow these steps:

• Step 1: Calculate the base area of the rectangular prism.
• Step 2: Use the base area to find the height of the water, $Y$.

Now, let's work through each step:

Step 1: The volume of the rectangular prism is given as 150 cm$^3$. Let's denote the height of the prism without any liquid as $H$, and the gap from the water line to the top of the prism is 3 cm. Thus, the total height is $H - 3$. The water volume given is 50 cm$^3$, which means the volume occupied by the air is:

$150 - 50 = 100$ cm$^3$.

Since the gap is 3 cm (the height of air column), the base area $A$ can be calculated as:

$A \times 3 = 100$.

This implies $A = \frac{100}{3} \approx 33.\overline{3}$ cm$^2$.

Step 2: Now, to find $Y$ (the height of water):

We use the formula for the volume of water in the prism:

$A \times Y = 50$,

where $A = 33.\overline{3}$ cm$^2$. Therefore,

$Y = \frac{50}{33.\overline{3}} = 1.5$ cm.

Therefore, the solution to the problem is $Y = 1.5 \, \text{cm}$.