Question

volume of cones and pyramids
find the volume of each figure. round your answers to the nearest hundredth, if necessary.
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)

Explanation:

Step1: Recall the formula for the volume of a cone

The volume \( V \) of a cone is given by the formula \( V=\frac{1}{3}\pi r^{2}h \), where \( r \) is the radius of the base and \( h \) is the height of the cone.

Step2: Solve for problem 1 (Cone with \( r = 2\) ft, \( h=5\) ft)

Substitute \( r = 2\) and \( h = 5\) into the formula:
\( V=\frac{1}{3}\pi(2)^{2}(5)=\frac{1}{3}\pi\times4\times5=\frac{20\pi}{3}\approx\frac{20\times3.1416}{3}\approx20.94 \) cubic feet.

Step3: Solve for problem 2 (Cone with \( r = 22\) m, \( h = 22\) m)

Substitute \( r=22\) and \( h = 22\) into the formula:
\( V=\frac{1}{3}\pi(22)^{2}(22)=\frac{1}{3}\pi\times484\times22=\frac{10648\pi}{3}\approx\frac{10648\times3.1416}{3}\approx11162.73 \) cubic meters.

Step4: Solve for problem 3 (Cone with \( r = 8\) cm, \( h = 8\) cm)

Substitute \( r = 8\) and \( h=8\) into the formula:
\( V=\frac{1}{3}\pi(8)^{2}(8)=\frac{1}{3}\pi\times64\times8=\frac{512\pi}{3}\approx\frac{512\times3.1416}{3}\approx536.17 \) cubic centimeters.

Step5: Solve for problem 4 (Cone with \( r = 1\) ft, \( h = 8\) ft)

Substitute \( r = 1\) and \( h = 8\) into the formula:
\( V=\frac{1}{3}\pi(1)^{2}(8)=\frac{8\pi}{3}\approx\frac{8\times3.1416}{3}\approx8.38 \) cubic feet.

Step6: Solve for problem 5 (Cone with \( r = 8\) m, \( h = 11\) m)

Substitute \( r = 8\) and \( h = 11\) into the formula:
\( V=\frac{1}{3}\pi(8)^{2}(11)=\frac{1}{3}\pi\times64\times11=\frac{704\pi}{3}\approx\frac{704\times3.1416}{3}\approx737.26 \) cubic meters.

Step7: Solve for problem 6 (Cone with \( r = 12\) m, \( h = 24\) m)

Substitute \( r = 12\) and \( h = 24\) into the formula:
\( V=\frac{1}{3}\pi(12)^{2}(24)=\frac{1}{3}\pi\times144\times24 = 1152\pi\approx1152\times3.1416\approx3644.25 \) cubic meters.

Step8: Solve for problem 7 (Cone with \( r = 4\) yd, \( h = 10\) yd)

Substitute \( r = 4\) and \( h = 10\) into the formula:
\( V=\frac{1}{3}\pi(4)^{2}(10)=\frac{1}{3}\pi\times16\times10=\frac{160\pi}{3}\approx\frac{160\times3.1416}{3}\approx167.55 \) cubic yards.

Step9: Solve for problem 8 (Cone with \( r = 4\) in, \( h = 7\) in)

Substitute \( r = 4\) and \( h = 7\) into the formula:
\( V=\frac{1}{3}\pi(4)^{2}(7)=\frac{1}{3}\pi\times16\times7=\frac{112\pi}{3}\approx\frac{112\times3.1416}{3}\approx117.29 \) cubic inches.

Step10: Solve for problem 9 (Cone with \( r=\frac{6}{2}=3\) km, \( h = 9\) km)

First, find the radius: since the diameter is 6 km, \( r=\frac{6}{2}=3\) km. Substitute \( r = 3\) and \( h = 9\) into the formula:
\( V=\frac{1}{3}\pi(3)^{2}(9)=\frac{1}{3}\pi\times9\times9 = 27\pi\approx27\times3.1416\approx84.82 \) cubic kilometers.

Step11: Solve for problem 10 (Cone with \( r = 9\) ft, \( h = 18\) ft)

Substitute \( r = 9\) and \( h = 18\) into the formula:
\( V=\frac{1}{3}\pi(9)^{2}(18)=\frac{1}{3}\pi\times81\times18=486\pi\approx486\times3.1416\approx1526.04 \) cubic feet.

Answer:

1. \(\approx 20.94\) ft³
2. \(\approx 11162.73\) m³
3. \(\approx 536.17\) cm³
4. \(\approx 8.38\) ft³
5. \(\approx 737.26\) m³
6. \(\approx 3644.25\) m³
7. \(\approx 167.55\) yd³
8. \(\approx 117.29\) in³
9. \(\approx 84.82\) km³
10. \(\approx 1526.04\) ft³