Find the volume, in cubic inches, of the composite solid below, which consists of a 4

Question

-inch square solid rectangular bar that is 16
inches in length. The bar has a 2
-inch diameter cylinder hole cut out of the center of the bar from the top of the bar through the entire length of the bar. Use pi=3.14
to find the volume. Enter only the number.

An image shows a solid rectangular metal bar that has a 4-inch square top and a length of 16 inches. There is a 2-inch diameter cylinder hole cut out of the center of the bar from the top to the bottom of the bar.

Answer 1

To find the volume of the composite solid, we first calculate the volume of the rectangular bar and then subtract the volume of the cylindrical hole.

1. Volume of the rectangular bar: The volume \( V \) of a rectangular prism is given by: \[ V = \text{Length} \times \text{Width} \times \text{Height} \] For the bar:

   • Length = 16 inches
   • Width = 4 inches
   • Height = 4 inches (since it's a square cross-section)

   Therefore, the volume of the rectangular bar is: \[ V_{\text{bar}} = 16 , \text{inches} \times 4 , \text{inches} \times 4 , \text{inches} = 256 , \text{cubic inches} \]

2. Volume of the cylinder hole: The volume \( V \) of a cylinder is given by: \[ V = \pi r^2 h \] where:

   • \(\pi \approx 3.14\)
   • \( r \) is the radius of the cylinder (diameter = 2 inches, so radius = 1 inch)
   • \( h \) is the height/length of the cylinder (which is the same as the length of the bar, 16 inches)

   Therefore, the volume of the cylindrical hole is: \[ V_{\text{cylinder}} = 3.14 \times (1 , \text{inch})^2 \times 16 , \text{inches} \] \[ = 3.14 \times 1 \times 16 = 50.24 , \text{cubic inches} \]

3. Volume of the composite solid: To find the volume of the composite solid, subtract the volume of the hole from the volume of the rectangular bar: \[ V_{\text{composite}} = V_{\text{bar}} - V_{\text{cylinder}} \] \[ = 256 , \text{cubic inches} - 50.24 , \text{cubic inches} = 205.76 , \text{cubic inches} \]

Thus, the volume of the composite solid is: \[ \boxed{205.76} \]