Rectangular Prism
1. Stating the problem:
We have a rectangular prism (Package B) with height $h = 10$ cm, breadth $b = 10$ cm, and length $l = 20$ cm. We will calculate:
- The volume
- The surface area
- Show that a cardboard sheet for the net has area 1600 cm²
2. Calculating volume:
The volume $V$ of a rectangular prism is given by:
$$V = l \times b \times h$$
Substitute the values:
$$V = 20 \times 10 \times 10 = 2000$$
So, the volume is $2000 \text{ cm}^3$.
3. Calculating surface area:
The surface area $S$ is the total area of all 6 faces:
$$S = 2(lb + bh + hl)$$
Calculate each term:
$$lb = 20 \times 10 = 200$$
$$bh = 10 \times 10 = 100$$
$$hl = 10 \times 20 = 200$$
Sum:
$$200 + 100 + 200 = 500$$
Multiply by 2:
$$S = 2 \times 500 = 1000$$
The surface area is $1000 \text{ cm}^2$.
4. Showing cardboard sheet area 1600 cm²:
The net of the prism is made by unfolding all faces onto a flat sheet.
Since the sheet is rectangular, its area must cover the net.
One possible arrangement is placing 2 lengths ($20$ cm) and 2 breadths ($10$ cm) along one side, and height ($10$ cm) along the other.
Calculate sheet dimensions:
Width = $l + b = 20 + 10 = 30$ cm
Height = $h + b = 10 + 10 = 20$ cm
Area of cardboard sheet:
$$30 \times 20 = 600$$ cm² which is less than 1600, so try another arrangement.
Alternatively, arrange net as a rectangle by placing all four sides around the base:
Perimeter of base rectangle = $2(l + b) = 2(20 + 10) = 60$ cm
Height = $10$ cm
Area = $60 \times 10 = 600$ cm² again.
Thus, the given sheet area $1600$ cm² is larger than these values, so it can cover the net with extra space.
Or, consider the cardboard sheet is forming a rectangle that encloses all net faces:
The net consists of 6 faces with total area = surface area = $1000$ cm², so 1600 cm² > 1000 cm² is sufficient.
Hence, by calculations, a cardboard sheet of area $1600$ cm² can be used to cut out the net as it is more than the total surface area.
Final answers:
- Volume = $2000$ cm³
- Surface Area = $1000$ cm²
- Cardboard sheet area $1600$ cm² is sufficient for the net