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Chapter 12 · Question 7

A metallic bucket, open at the top, is in the shape of a frustum of a cone. The height of the bucket is 2424 cm, the radius of the upper circular end is 1414 cm, and the radius of the lower circular end is 77 cm. Find: (i) the area of the metallic sheet used to make the bucket (ignoring the thickness of the metal), and (ii) the volume of water the bucket can hold. (Take π=227\pi = \frac{22}{7})

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Q7

A metallic bucket, open at the top, is in the shape of a frustum of a cone. The height of the bucket is 2424 cm, the radius of the upper circular end is 1414 cm, and the radius of the lower circular end is 77 cm. Find: (i) the area of the metallic sheet used to make the bucket (ignoring the thickness of the metal), and (ii) the volume of water the bucket can hold. (Take π=227\pi = \frac{22}{7})

Answer Revealed
Direct Answer:
(i) The bucket is a frustum of a cone, open at the top. Let upper radius r1=14r_1 = 14 cm, lower radius r2=7r_2 = 7 cm, and height h=24h = 24 cm. Slant height of the frustum: l=h2+(r1−r2)2=242+(14−7)2=576+49=625=25l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{24^2 + (14 - 7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25 cm. The metallic sheet area consists of the curved surface area of the frustum plus the area of the circular base (the lower end): Area of sheet =CSA of frustum+Area of circular base=π(r1+r2)l+πr22=π×(14+7)×25+π×72=π×21×25+π×49=525π+49π=574π= \text{CSA of frustum} + \text{Area of circular base} = \pi (r_1 + r_2) l + \pi r_2^2 = \pi \times (14 + 7) \times 25 + \pi \times 7^2 = \pi \times 21 \times 25 + \pi \times 49 = 525\pi + 49\pi = 574\pi cm2^2. Using π=227\pi = \frac{22}{7}: Area =574×227=82×22=1804= 574 \times \frac{22}{7} = 82 \times 22 = 1804 cm2^2. (ii) Volume of the bucket (frustum) =πh3(r12+r22+r1r2)=π×243(142+72+14×7)=8π×(196+49+98)=8π×343=2744π= \frac{\pi h}{3}(r_1^2 + r_2^2 + r_1 r_2) = \frac{\pi \times 24}{3}(14^2 + 7^2 + 14 \times 7) = 8\pi \times (196 + 49 + 98) = 8\pi \times 343 = 2744\pi cm3^3. Using π=227\pi = \frac{22}{7}: Volume =2744×227=392×22=8624= 2744 \times \frac{22}{7} = 392 \times 22 = 8624 cm3^3. Since 11 litre =1000= 1000 cm3^3, the bucket can hold 8.6248.624 litres.

Simple Explanation

Think of the bucket as a cone with its tip cut off. The top is wide (1414 cm radius) and the bottom is narrower (77 cm radius), and it is 2424 cm tall. The slant side is 242+(14−7)2=625=25\sqrt{24^2 + (14 - 7)^2} = \sqrt{625} = 25 cm. The sheet metal needed: curved side π(14+7)×25=525π\pi(14 + 7) \times 25 = 525\pi plus the circular bottom π×49=49π\pi \times 49 = 49\pi, totalling 574π=574×227=1804574\pi = 574 \times \frac{22}{7} = 1804 cm2^2. The water it holds: π×243×(142+72+14×7)=8π×343=2744π=2744×227=8624\frac{\pi \times 24}{3} \times (14^2 + 7^2 + 14 \times 7) = 8\pi \times 343 = 2744\pi = 2744 \times \frac{22}{7} = 8624 cm3^3, which is about 8.68.6 litres.

Exam-Ready Structure

A frustum of a cone is the portion of a right circular cone between the base and a plane parallel to the base. This problem demonstrates the computation of its surface area and volume: 1. Dimensions of the frustum: Upper radius r1=14r_1 = 14 cm, lower radius r2=7r_2 = 7 cm, height h=24h = 24 cm. 2. (i) Area of metallic sheet: The bucket is open at the top, so the sheet area is the curved surface area plus the bottom circular base. First, compute the slant height: l=h2+(r1−r2)2=242+72=576+49=625=25l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 cm. Curved surface area of frustum =π(r1+r2)l=π×21×25=525π= \pi (r_1 + r_2) l = \pi \times 21 \times 25 = 525\pi cm2^2. Area of circular bottom (radius r2r_2) =πr22=π×49=49π= \pi r_2^2 = \pi \times 49 = 49\pi cm2^2. Total sheet area =525π+49π=574π=574×227=1804= 525\pi + 49\pi = 574\pi = 574 \times \frac{22}{7} = 1804 cm2^2. 3. (ii) Volume of water the bucket can hold: Volume of frustum =πh3(r12+r22+r1r2)= \frac{\pi h}{3}(r_1^2 + r_2^2 + r_1 r_2). Substitute: V=π×243[(14)2+(7)2+14×7]=8π(196+49+98)=8π×343=2744πV = \frac{\pi \times 24}{3}[(14)^2 + (7)^2 + 14 \times 7] = 8\pi(196 + 49 + 98) = 8\pi \times 343 = 2744\pi cm3^3. Using π=227\pi = \frac{22}{7}: V=2744×227=8624V = 2744 \times \frac{22}{7} = 8624 cm3^3. In litres: 8624÷1000=8.6248624 \div 1000 = 8.624 litres. 4. Key frustum formulas: Slant height l=h2+(r1−r2)2l = \sqrt{h^2 + (r_1 - r_2)^2}, CSA =π(r1+r2)l= \pi (r_1 + r_2) l, Volume =πh3(r12+r22+r1r2)= \frac{\pi h}{3}(r_1^2 + r_2^2 + r_1 r_2). The total surface area of a closed frustum would also include πr12\pi r_1^2, but here the top is open.

Key Points

  • Frustum: r1=14r_1 = 14 cm, r2=7r_2 = 7 cm, h=24h = 24 cm, slant l=242+72=25l = \sqrt{24^2 + 7^2} = 25 cm
  • CSA of frustum =π(r1+r2)l=525π= \pi (r_1 + r_2) l = 525\pi cm2^2; base area =πr22=49π= \pi r_2^2 = 49\pi cm2^2
  • Sheet metal area =574π=1804= 574\pi = 1804 cm2^2 (open at top, so only CSA + bottom base)
  • Volume of frustum =πh3(r12+r22+r1r2)=2744π=8624= \frac{\pi h}{3}(r_1^2 + r_2^2 + r_1 r_2) = 2744\pi = 8624 cm3=8.624^3 = 8.624 litres
  • Frustum is the portion of a cone cut by a plane parallel to its base; key formulas above

Common Mistakes

  • Adding πr12\pi r_1^2 (top area) to the metallic sheet — the bucket is open at the top, so the top circle is not part of the surface
  • Using πh2\frac{\pi h}{2} instead of πh3\frac{\pi h}{3} in the volume formula — keep the 13\frac{1}{3} from the original cone formula
  • Forgetting that the slant height ll is the hypotenuse of a right triangle with sides hh and (r1−r2)(r_1 - r_2)

Related Questions

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)

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.

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