What is the volume of a rectangular prism that has a height of 10 1/2

in., a length of 6 2/3
in., and a width of 2 1/4
in.? Is it large enough to hold an item that has a height of 8 in., a length of 6 in., and a width of 2 in.?(1 point)
Responses

3,780/24 in.3
and no, the item will not fit inside the prism.
Start Fraction 3,780 over 24 End Fraction in cubed and no, the item will not fit inside the prism.

157 1/2 in.3
and yes, the item should fit inside the prism.
157 Start Fraction 1 over 2 End Fraction in cubed and yes, the item should fit inside the prism.

15 in.3
and no, the item will not fit inside the prism.
15 in cubed and no, the item will not fit inside the prism.

160 in.3
and yes, the item should fit inside the prism.

1 answer

To find the volume of a rectangular prism, you can use the formula:

\[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Height} \]

First, let's convert the mixed numbers into improper fractions to facilitate the calculation:

  1. Height: \[ 10 \frac{1}{2} = \frac{21}{2} \text{ inches} \]

  2. Length: \[ 6 \frac{2}{3} = \frac{20}{3} \text{ inches} \]

  3. Width: \[ 2 \frac{1}{4} = \frac{9}{4} \text{ inches} \]

Now substituting these values into the volume formula:

\[ \text{Volume} = \left(\frac{20}{3}\right) \times \left(\frac{9}{4}\right) \times \left(\frac{21}{2}\right) \]

Now, multiply the numerators and the denominators:

\[ \text{Numerator: } 20 \times 9 \times 21 = 3780 \] \[ \text{Denominator: } 3 \times 4 \times 2 = 24 \]

Thus, the volume is:

\[ \text{Volume} = \frac{3780}{24} \text{ in}^3 \]

Now, we simplify the fraction:

Dividing both the numerator and the denominator by 12:

\[ \frac{3780 \div 12}{24 \div 12} = \frac{315}{2} = 157.5 \text{ in}^3 \]

Now we can compare this volume to the item that needs to fit inside. The dimensions of the item are:

  • Height: 8 in.
  • Length: 6 in.
  • Width: 2 in.

The volume of the item can be calculated as:

\[ \text{Volume of the item} = 8 \times 6 \times 2 = 96 \text{ in}^3 \]

Since the volume of the prism \(157.5 , \text{in}^3\) is greater than the volume of the item \(96 , \text{in}^3\), the item will fit inside the prism.

Thus, the answer is:

\[ 157 \frac{1}{2} , \text{in}^3 \text{ and yes, the item should fit inside the prism.} \]