Question

finding the unknown side of a rectangular prism with the given volume

1. multiply to find the area of the base
2. divide the volume by the area of the base.

find the unknown side of each rectangular prism.
6 volume: 13.2 m³
length: 2 m
width: 1.8 m
height:

1. volume: 118.69 cm³

length:
width: 3 cm
height: 11 cm

1. volume: 13.2 m³

length: 1.24 m
width:
height: 4.5 m
find the length of the unknown side of each container. then answer the question.
9
10
how many cubes with sides of 3 cm are needed to fill up each container?

Explanation:

Step1: Find the area of the base

For problem 6:
The base of the rectangular prism is a rectangle with length \(l = 2m\) and width \(w=1.8m\).
The area of the base \(A = l\times w\).
\(A=2\times1.8 = 3.6m^{2}\)

Step2: Divide the volume by the area of the base

The volume of the prism \(V = 13.2m^{3}\).
Let the height be \(h\).
We know that \(V=A\times h\), so \(h=\frac{V}{A}\).
\(h=\frac{13.2}{3.6}=\frac{132}{36}=\frac{11}{3}\approx3.67m\)

For problem 7:
The base of the rectangular prism has width \(w = 3cm\) and let the length be \(l\).
The area of the base \(A=l\times w\).
The height \(h = 11cm\) and volume \(V=118.69cm^{3}\).
Since \(V = A\times h=(l\times w)\times h\), then \(l=\frac{V}{w\times h}\).
\(l=\frac{118.69}{3\times11}=\frac{118.69}{33}\approx3.60cm\)

For problem 8:
The base of the rectangular prism has length \(l = 1.24m\) and let the width be \(w\).
The area of the base \(A=l\times w\).
The height \(h = 4.5m\) and volume \(V = 13.2m^{3}\).
Since \(V=A\times h=(l\times w)\times h\), then \(w=\frac{V}{l\times h}\).
\(w=\frac{13.2}{1.24\times4.5}=\frac{13.2}{5.58}\approx2.37m\)

For container A:
Let the unknown side be \(x\).
The volume \(V = 4320cm^{3}\), with two known sides \(20cm\) and \(8cm\).
Since \(V=l\times w\times h\), then \(x=\frac{V}{20\times8}=\frac{4320}{160} = 27cm\)

For container B:
Let the unknown side be \(y\).
The volume \(V = 0.012m^{3}\), with two known sides \(0.3m\) and \(0.2m\).
Since \(V=l\times w\times h\), then \(y=\frac{V}{0.3\times0.2}=\frac{0.012}{0.06}=0.2m\)

For container C:
Let the unknown side be \(z\).
The volume \(V = 6375cm^{3}\), with two known sides \(13cm\) and \(30cm\).
Since \(V=l\times w\times h\), then \(z=\frac{V}{13\times30}=\frac{6375}{390}=16.35cm\)

For container D:
Let the unknown side be \(k\).
The volume \(V = 0.5m^{3}\), with two known sides \(2m\) and \(0.5m\).
Since \(V=l\times w\times h\), then \(k=\frac{V}{2\times0.5}=\frac{0.5}{1}=0.5m\)

For problem 10:
The volume of a small cube with side \(a = 3cm\) is \(V_{cube}=a^{3}=3^{3}=27cm^{3}\)
For container A: \(n_{A}=\frac{4320}{27}=160\)
For container B: First convert volume to \(cm^{3}\), \(V_{B}=0.012m^{3}=0.012\times100^{3}cm^{3}=12000cm^{3}\), \(n_{B}=\frac{12000}{27}\approx444.44\), but since we need whole - number of cubes, we consider the integer part (assuming no partial - cube filling in a practical sense, if we consider strict volume division \(\frac{12000}{27}=\frac{4000}{9}\approx444\))
For container C: \(n_{C}=\frac{6375}{27}=\frac{2125}{9}\approx236.11\approx236\) (using the same whole - cube logic)
For container D: Convert volume to \(cm^{3}\), \(V_{D}=0.5m^{3}=0.5\times100^{3}cm^{3}=50000cm^{3}\), \(n_{D}=\frac{50000}{27}\approx1851.85\approx1851\)

Answer:

1. \(h\approx3.67m\)
2. \(l\approx3.60cm\)
3. \(w\approx2.37m\)

Container A: \(27cm\)
Container B: \(0.2m\)
Container C: \(16.35cm\)
Container D: \(0.5m\)
Problem 10:
Container A: \(160\) cubes
Container B: \(444\) cubes
Container C: \(236\) cubes
Container D: \(1851\) cubes