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Chapter 12 · Question 3

A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire rocket is 2626 cm, while the height of the conical part is 66 cm. The base of the conical portion has a diameter of 55 cm, while the base diameter of the cylindrical portion is 33 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take π=3.14\pi = 3.14)

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Q3

A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire rocket is 2626 cm, while the height of the conical part is 66 cm. The base of the conical portion has a diameter of 55 cm, while the base diameter of the cylindrical portion is 33 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take π=3.14\pi = 3.14)

Answer Revealed
Direct Answer:
For the cone: radius r=52=2.5r = \frac{5}{2} = 2.5 cm, height h=6h = 6 cm. Slant height l=r2+h2=2.52+62=6.25+36=42.25=6.5l = \sqrt{r^2 + h^2} = \sqrt{2.5^2 + 6^2} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5 cm. For the cylinder: radius r′=32=1.5r' = \frac{3}{2} = 1.5 cm, height h′=26−6=20h' = 26 - 6 = 20 cm. Orange area: The cone's base overhangs the cylinder, so a ring of area πr2−π(r′)2\pi r^2 - \pi (r')^2 also needs orange. Orange area =CSA of cone+πr2−π(r′)2=π[rl+r2−(r′)2]=3.14×[2.5×6.5+2.52−1.52]=3.14×[16.25+6.25−2.25]=3.14×20.25=63.585= \text{CSA of cone} + \pi r^2 - \pi (r')^2 = \pi [r l + r^2 - (r')^2] = 3.14 \times [2.5 \times 6.5 + 2.5^2 - 1.5^2] = 3.14 \times [16.25 + 6.25 - 2.25] = 3.14 \times 20.25 = 63.585 cm2^2. Yellow area: The cylinder is open at the top (mounted by the cone), so only the CSA and the bottom base are painted yellow. Yellow area =CSA of cylinder+π(r′)2=2πr′h′+π(r′)2=πr′(2h′+r′)=3.14×1.5×(2×20+1.5)=4.71×41.5=195.465= \text{CSA of cylinder} + \pi (r')^2 = 2\pi r' h' + \pi (r')^2 = \pi r'(2h' + r') = 3.14 \times 1.5 \times (2 \times 20 + 1.5) = 4.71 \times 41.5 = 195.465 cm2^2.

Simple Explanation

The rocket has a wide cone (r=2.5r = 2.5 cm) on top of a narrower cylinder (r′=1.5r' = 1.5 cm). The cone is 66 cm tall, so the cylinder below it is 2020 cm tall. The catch: the cone's base is wider than the cylinder, so a ring of the cone's base sticks out below the cone and needs orange paint too. Orange area: curved cone surface plus the extra ring =3.14×20.25=63.585= 3.14 \times 20.25 = 63.585 cm2^2. Yellow area: curved cylinder surface plus one circular base =4.71×41.5=195.465= 4.71 \times 41.5 = 195.465 cm2^2. So the orange portion is about 63.663.6 cm2^2 and the yellow portion is about 195.5195.5 cm2^2.

Exam-Ready Structure

This problem introduces the subtlety of non-matching base radii in combined solids: 1. Given dimensions: Total rocket height =26= 26 cm, height of cone =6= 6 cm, so height of cylinder h′=26−6=20h' = 26 - 6 = 20 cm. Cone base diameter =5= 5 cm, so cone radius r=2.5r = 2.5 cm. Cylinder base diameter =3= 3 cm, so cylinder radius r′=1.5r' = 1.5 cm. 2. Slant height of cone: l=r2+h2=2.52+62=42.25=6.5l = \sqrt{r^2 + h^2} = \sqrt{2.5^2 + 6^2} = \sqrt{42.25} = 6.5 cm. 3. Orange area (conical portion): The cone sits on the cylinder, but the cone's base (πr2\pi r^2) is larger than the cylinder's top face (π(r′)2\pi (r')^2). The part of the cone base that lies outside the cylinder (i.e., the overhang) is a circular ring and must also be painted orange because it is visible. Thus, orange area =CSA of cone+(base area of cone−base area of cylinder)=πrl+πr2−π(r′)2=π[rl+r2−(r′)2]= \text{CSA of cone} + (\text{base area of cone} - \text{base area of cylinder}) = \pi r l + \pi r^2 - \pi (r')^2 = \pi [rl + r^2 - (r')^2]. Substituting: =3.14×[2.5×6.5+6.25−2.25]=3.14×[16.25+4.0]=3.14×20.25=63.585= 3.14 \times [2.5 \times 6.5 + 6.25 - 2.25] = 3.14 \times [16.25 + 4.0] = 3.14 \times 20.25 = 63.585 cm2^2. 4. Yellow area (cylindrical portion): The top of the cylinder is covered by the cone and is not visible. So only the curved surface area and one exposed base (the bottom) count: Yellow area =CSA of cylinder+area of bottom base=2πr′h′+π(r′)2=πr′(2h′+r′)= \text{CSA of cylinder} + \text{area of bottom base} = 2\pi r' h' + \pi (r')^2 = \pi r'(2h' + r'). Substitute: =3.14×1.5×(40+1.5)=4.71×41.5=195.465= 3.14 \times 1.5 \times (40 + 1.5) = 4.71 \times 41.5 = 195.465 cm2^2. 5. Cross-check: The top base of the cylinder is hidden beneath the cone — so unlike the usual TSA formula, we do not include it. But the cone's overhanging base ring IS visible, so it is included in the orange area.

Key Points

  • Cone: r=2.5r = 2.5 cm, h=6h = 6 cm, slant l=6.5l = 6.5 cm; Cylinder: r′=1.5r' = 1.5 cm, h′=20h' = 20 cm
  • Orange area includes CSA of cone AND the overhanging ring of the cone base: πrl+πr2−π(r′)2\pi r l + \pi r^2 - \pi (r')^2
  • Yellow area includes CSA of cylinder AND one exposed base: 2πr′h′+π(r′)22\pi r' h' + \pi (r')^2
  • Orange =3.14×20.25=63.585= 3.14 \times 20.25 = 63.585 cm2^2; Yellow =4.71×41.5=195.465= 4.71 \times 41.5 = 195.465 cm2^2
  • When constituent radii differ, account for the visible overhang and hidden joint faces

Common Mistakes

  • Forgetting the overhanging ring of the cone base when computing orange area — the base of the wider cone that extends beyond the cylinder IS visible and must be painted
  • Counting both bases of the cylinder — the top base is hidden under the cone and should not be included in yellow area

Related Questions

Q2

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 55 cm in height and the diameter of the top is 3.53.5 cm. Find the area he has to colour. (Take π=227\pi = \frac{22}{7})

Q6

A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 55 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 1010 cm, find the apparent capacity of the glass and its actual capacity. (Use π=3.14\pi = 3.14)

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