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Chapter 12 · Question 4

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 22 cm and the diameter of the base is 44 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π=3.14\pi = 3.14)

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Q4

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 22 cm and the diameter of the base is 44 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π=3.14\pi = 3.14)

Answer Revealed
Direct Answer:
The base diameter is 44 cm, so the common radius r=2r = 2 cm. Height of cone h=2h = 2 cm. Volume of the toy =Volume of hemisphere+Volume of cone=23πr3+13πr2h=13πr2(2r+h)=13×3.14×22×(2×2+2)=13×3.14×4×6=13×75.36=25.12= \text{Volume of hemisphere} + \text{Volume of cone} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r + h) = \frac{1}{3} \times 3.14 \times 2^2 \times (2 \times 2 + 2) = \frac{1}{3} \times 3.14 \times 4 \times 6 = \frac{1}{3} \times 75.36 = 25.12 cm3^3. For the circumscribing cylinder: radius =r=2= r = 2 cm, height =r+h=2+2=4= r + h = 2 + 2 = 4 cm. Volume of cylinder =πr2H=3.14×4×4=50.24= \pi r^2 H = 3.14 \times 4 \times 4 = 50.24 cm3^3. Difference in volumes =50.24−25.12=25.12= 50.24 - 25.12 = 25.12 cm3^3.

Simple Explanation

The toy is a sphere-half (hemisphere) with a cone on top, both sharing a radius of 22 cm. The hemisphere contributes (2/3)πr3=16.75(2/3)\pi r^3 = 16.75 cm3^3 and the cone contributes (1/3)πr2h=8.37(1/3)\pi r^2 h = 8.37 cm3^3, totalling 25.1225.12 cm3^3. The cylinder that just fits around the toy has the same 22 cm radius and a height of 44 cm (the hemisphere radius plus the cone height). Its volume is π×4×4=50.24\pi \times 4 \times 4 = 50.24 cm3^3. The empty space between the cylinder and the toy is 50.24−25.12=25.1250.24 - 25.12 = 25.12 cm3^3 — exactly equal to the toy's own volume.

Exam-Ready Structure

This question combines volume of a combination solid with comparison against a circumscribing shape: 1. Determine common dimensions: The diameter of the base that is common to both hemisphere and cone is 44 cm, so r=2r = 2 cm. The height of the cone h=2h = 2 cm. 2. Volume of the toy =Vhemisphere+Vcone= V_{\text{hemisphere}} + V_{\text{cone}}. Use the direct-sum principle for volume (unlike surface area, volumes of constituents simply add): Vtoy=23πr3+13πr2h=13πr2(2r+h)V_{\text{toy}} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2(2r + h). Substitute: Vtoy=13×3.14×(2)2×(2×2+2)=13×3.14×4×6=75.363=25.12V_{\text{toy}} = \frac{1}{3} \times 3.14 \times (2)^2 \times (2 \times 2 + 2) = \frac{1}{3} \times 3.14 \times 4 \times 6 = \frac{75.36}{3} = 25.12 cm3^3. 3. The cylinder 'circumscribing' the toy means the cylinder just touches the toy's boundaries. Its radius equals the toy's radius r=2r = 2 cm. Its height extends from the bottom of the hemisphere to the tip of the cone: H=r+h=2+2=4H = r + h = 2 + 2 = 4 cm. 4. Volume of the cylinder =πr2H=3.14×(2)2×4=3.14×4×4=50.24= \pi r^2 H = 3.14 \times (2)^2 \times 4 = 3.14 \times 4 \times 4 = 50.24 cm3^3. 5. Difference =Vcylinder−Vtoy=50.24−25.12=25.12= V_{\text{cylinder}} - V_{\text{toy}} = 50.24 - 25.12 = 25.12 cm3^3. 6. Observation: The difference equals the volume of the toy. This is a specific property when h=rh = r in this configuration — each of the three volumes (hemisphere, cone, and difference) is one-third of the cylinder volume.

Key Points

  • Common radius r=2r = 2 cm (from diameter 44 cm); cone height h=2h = 2 cm
  • Volume of toy =23πr3+13πr2h=13πr2(2r+h)=25.12= \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2(2r + h) = 25.12 cm3^3
  • Circumscribing cylinder: same radius, height =r+h=4= r + h = 4 cm
  • Cylinder volume =πr2H=50.24= \pi r^2 H = 50.24 cm3^3; difference =25.12= 25.12 cm3^3
  • Difference equals toy volume — each constituent occupies one-third of the cylinder

Common Mistakes

  • Forgetting to add the cone height AND the hemisphere radius to get the cylinder's height — the cylinder must span the entire toy from bottom of hemisphere to tip of cone
  • Using the formula for total surface area of the combined solid (subtracting joint areas) in a volume problem — volumes always add directly

Related Questions

Q5

A vessel is in the form of an inverted cone. Its height is 88 cm and the radius of its top, which is open, is 55 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.50.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Q1

A toy is in the form of a cone of radius 3.53.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.515.5 cm. Find the total surface area of the toy. (Take π=227\pi = \frac{22}{7})

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