Question

6 hint: the cylinder is a hole in the figure. find the volume of the triangular prism with
2 ft 6 ft 4 ft
5 ft
7 hint: the cone is a hole in the figure. find the volume of the triangular pyramid with the hole cut out.
9 m 9 m 6 m
8 m
8 a hollow sphere sits snugly in a foam cube so that the sphere touches each side of the cube. find the volume of the foam.
8 in 8 in 8 in

Explanation:

Step1: Recall volume formulas

Volume of triangular prism $V_{prism}=Bh$, where $B$ is the base - area and $h$ is the height. Volume of cylinder $V_{cylinder}=\pi r^{2}h$. For a triangular base with base length $b = 2$ ft and height $h_{base}=5$ ft, $B=\frac{1}{2}\times2\times5 = 5$ square - ft, and the length of the prism $h = 6$ ft. The radius of the cylinder $r = 1$ ft and its length is also $h = 6$ ft.

Step2: Calculate volume of triangular prism

$V_{prism}=Bh=\frac{1}{2}\times2\times5\times6= 30$ cubic - ft.

Step3: Calculate volume of cylinder

$V_{cylinder}=\pi r^{2}h=\pi\times1^{2}\times6 = 6\pi$ cubic - ft.

Step4: Calculate net volume

$V = V_{prism}-V_{cylinder}=30 - 6\pi\approx30-6\times3.14=30 - 18.84 = 11.16$ cubic - ft.

Step1: Recall volume formulas

Volume of triangular pyramid $V_{pyramid}=\frac{1}{3}Bh$, where $B$ is the base - area and $h$ is the height. The base of the triangular pyramid is a right - triangle with legs $a = 9$ m and $b = 6$ m, so $B=\frac{1}{2}\times9\times6 = 27$ square - m and $h = 8$ m. Volume of cone $V_{cone}=\frac{1}{3}\pi r^{2}h$. The radius of the cone $r = 2$ m and its height $h = 8$ m.

Step2: Calculate volume of triangular pyramid

$V_{pyramid}=\frac{1}{3}\times27\times8=72$ cubic - m.

Step3: Calculate volume of cone

$V_{cone}=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi\times2^{2}\times8=\frac{32\pi}{3}$ cubic - m.

Step4: Calculate net volume

$V = V_{pyramid}-V_{cone}=72-\frac{32\pi}{3}\approx72-\frac{32\times3.14}{3}=72 - \frac{100.48}{3}\approx72 - 33.49 = 38.51$ cubic - m.

Step1: Recall volume formulas

Volume of cube $V_{cube}=s^{3}$, where $s = 8$ in. Volume of sphere $V_{sphere}=\frac{4}{3}\pi r^{3}$, and since the sphere fits snugly in the cube, the diameter of the sphere is equal to the side - length of the cube, so $r=\frac{s}{2}=4$ in.

Step2: Calculate volume of cube

$V_{cube}=8^{3}=512$ cubic - in.

Step3: Calculate volume of sphere

$V_{sphere}=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi\times4^{3}=\frac{4}{3}\pi\times64=\frac{256\pi}{3}$ cubic - in.

Step4: Calculate net volume

$V = V_{cube}-V_{sphere}=512-\frac{256\pi}{3}\approx512-\frac{256\times3.14}{3}=512 - \frac{803.84}{3}\approx512 - 267.95 = 244.05$ cubic - in.

Answer:

$30 - 6\pi\approx11.16$ cubic - ft