Sure! Let’s solve each problem step by step. We’ll use the standard volume formulas for cylinders, cones, and spheres, and use
π ≈ 3.14 as instructed. We’ll round each answer to the
nearest hundredth.
---
🔹 Formula Review:
-
Cylinder:
\( V = \pi r^2 h \)
-
Cone:
\( V = \frac{1}{3} \pi r^2 h \)
-
Sphere:
\( V = \frac{4}{3} \pi r^3 \)
---
##
✔ Problem 1: Cylinder (radius = 3 in, height = 9 in)
\( V = \pi r^2 h = 3.14 × 3^2 × 9 \)
= \( 3.14 × 9 × 9 \)
= \( 3.14 × 81 \)
=
254.34 in³
✔ Answer: 254.34
---
##
✔ Problem 2: Cone (radius = 4 yd, height = 7 yd)
\( V = \frac{1}{3} \pi r^2 h = \frac{1}{3} × 3.14 × 4^2 × 7 \)
= \( \frac{1}{3} × 3.14 × 16 × 7 \)
= \( \frac{1}{3} × 3.14 × 112 \)
= \( \frac{351.68}{3} \)
≈
117.23 yd³
✔ Answer: 117.23
---
##
✔ Problem 3: Sphere (radius = 10 cm)
\( V = \frac{4}{3} \pi r^3 = \frac{4}{3} × 3.14 × 10^3 \)
= \( \frac{4}{3} × 3.14 × 1000 \)
= \( \frac{4}{3} × 3140 \)
= \( \frac{12560}{3} \)
≈
4186.67 cm³
✔ Answer: 4186.67
---
##
✔ Problem 4: Cylinder (diameter = 8 ft → radius = 4 ft, height = 6 ft)
\( V = \pi r^2 h = 3.14 × 4^2 × 6 \)
= \( 3.14 × 16 × 6 \)
= \( 3.14 × 96 \)
=
301.44 ft³
✔ Answer: 301.44
---
##
✔ Problem 5: Sphere (diameter = 24 mm → radius = 12 mm)
\( V = \frac{4}{3} \pi r^3 = \frac{4}{3} × 3.14 × 12^3 \)
= \( \frac{4}{3} × 3.14 × 1728 \)
= \( \frac{4}{3} × 5425.92 \)
= \( \frac{21703.68}{3} \)
≈
7234.56 mm³
✔ Answer: 7234.56
---
##
✔ Problem 6: Cone (diameter = 16 m → radius = 8 m, height = 8 m)
\( V = \frac{1}{3} \pi r^2 h = \frac{1}{3} × 3.14 × 8^2 × 8 \)
= \( \frac{1}{3} × 3.14 × 64 × 8 \)
= \( \frac{1}{3} × 3.14 × 512 \)
= \( \frac{1607.68}{3} \)
≈
535.89 m³
✔ Answer: 535.89
---
## 📝 Final Answers:
1.
254.34 in³
2.
117.23 yd³
3.
4186.67 cm³
4.
301.44 ft³
5.
7234.56 mm³
6.
535.89 m³
Let me know if you’d like to see the work written out differently or need diagrams explained!
Parent Tip: Review the logic above to help your child master the concept of volume of cylinder and cone worksheet.
