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Chapter 12 · Question 2

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})

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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})

Answer Revealed
Direct Answer:
Diameter =3.5= 3.5 cm, so radius r=3.52=1.75r = \frac{3.5}{2} = 1.75 cm. The top consists of a hemisphere surmounted on a cone. Height of hemispherical part =r=1.75= r = 1.75 cm. Height of cone h=5−1.75=3.25h = 5 - 1.75 = 3.25 cm. Slant height of cone l=r2+h2=(1.75)2+(3.25)2=3.0625+10.5625=13.625=3.7l = \sqrt{r^2 + h^2} = \sqrt{(1.75)^2 + (3.25)^2} = \sqrt{3.0625 + 10.5625} = \sqrt{13.625} = 3.7 cm (approx.). Area to be coloured =CSA of hemisphere+CSA of cone=2πr2+πrl=πr(2r+l)=227×3.52(2×3.52+3.7)=112×(3.5+3.7)=112×7.2=39.6= \text{CSA of hemisphere} + \text{CSA of cone} = 2\pi r^2 + \pi r l = \pi r(2r + l) = \frac{22}{7} \times \frac{3.5}{2} \left(2 \times \frac{3.5}{2} + 3.7\right) = \frac{11}{2} \times (3.5 + 3.7) = \frac{11}{2} \times 7.2 = 39.6 cm2^2 (approx.).

Simple Explanation

The playing top has a hemisphere on top and cone below. The diameter is 3.53.5 cm, so the radius of both parts is 1.751.75 cm. The hemisphere alone is 1.751.75 cm tall, leaving 3.253.25 cm for the cone. The slant height of the cone (the sloping side) is 1.752+3.252≈3.7\sqrt{1.75^2 + 3.25^2} \approx 3.7 cm. The area to paint is just the outer curved surfaces: 2πr22\pi r^2 (hemisphere) +πrl+ \pi r l (cone) =112×7.2≈39.6= \frac{11}{2} \times 7.2 \approx 39.6 cm2^2. Rasheed needs to colour about 39.639.6 square centimetres.

Exam-Ready Structure

This is a direct surface area application for a cone-hemisphere combination, approached systematically: 1. Read the dimensions: Diameter of the top =3.5= 3.5 cm, so the common radius r=3.52=1.75r = \frac{3.5}{2} = 1.75 cm. Total height =5= 5 cm. 2. Since the hemisphere is on top of the cone, the height of the hemisphere equals its radius r=1.75r = 1.75 cm. Therefore, height of the conical part h=5−1.75=3.25h = 5 - 1.75 = 3.25 cm. 3. The slant height ll of the cone is required to compute its curved surface area: l=r2+h2=(1.75)2+(3.25)2=3.0625+10.5625=13.625≈3.7l = \sqrt{r^2 + h^2} = \sqrt{(1.75)^2 + (3.25)^2} = \sqrt{3.0625 + 10.5625} = \sqrt{13.625} \approx 3.7 cm. 4. The area to be coloured is the total exposed (visible) surface of the combined solid. Since the hemisphere sits on the flat circular top of the cone, that joint face is not visible or colour-able: Area to be coloured =CSA of hemisphere+CSA of cone=2πr2+πrl= \text{CSA of hemisphere} + \text{CSA of cone} = 2\pi r^2 + \pi r l. 5. Substitute values with π=227\pi = \frac{22}{7}: Area =2×227×(1.75)2+227×1.75×3.7= 2 \times \frac{22}{7} \times (1.75)^2 + \frac{22}{7} \times 1.75 \times 3.7. Factorise πr\pi r: Area =227×1.75×(2×1.75+3.7)=227×1.75×(3.5+3.7)=227×1.75×7.2= \frac{22}{7} \times 1.75 \times (2 \times 1.75 + 3.7) = \frac{22}{7} \times 1.75 \times (3.5 + 3.7) = \frac{22}{7} \times 1.75 \times 7.2. Since 1.75=741.75 = \frac{7}{4}, Area =227×74×7.2=224×7.2=5.5×7.2=39.6= \frac{22}{7} \times \frac{7}{4} \times 7.2 = \frac{22}{4} \times 7.2 = 5.5 \times 7.2 = 39.6 cm2^2 (approx.). 6. Important observation: The total surface area of the combined solid is NOT the sum of the TOTAL surface areas of the cone and hemisphere — that would incorrectly count the joint face twice. Only exposed surfaces matter.

Key Points

  • Radius r=1.75r = 1.75 cm from diameter 3.53.5 cm
  • Height of hemispherical part =r=1.75= r = 1.75 cm; height of cone h=3.25h = 3.25 cm
  • Slant height l=r2+h2=13.625≈3.7l = \sqrt{r^2 + h^2} = \sqrt{13.625} \approx 3.7 cm
  • Area to colour =2πr2+πrl=πr(2r+l)≈39.6= 2\pi r^2 + \pi r l = \pi r (2r + l) \approx 39.6 cm2^2
  • Only exposed (visible) surfaces count; the joint face is not coloured

Common Mistakes

  • Summing the total surface areas of the individual solids — this double-counts the joint face. Use only curved surface areas for joined parts
  • Using the total height of 55 cm as the cone height without subtracting the hemispherical part
  • Forgetting that the hemisphere sits ON TOP of the cone (surmounted by), so the hemisphere radius occupies part of the total height

Related Questions

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})

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)

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