The radii of circular ends of a bucket of height 24 are 15 5 . Find its curved surface area.

Given that,The radii of circular ends of a bucket ₁ = 15 and ₂ = 5 Height of the bucket = 24 cm Slant height l=h2−r1−r22=242+(15−5)2=242+102 =576+100 =676 =26cm Slant height =26cm Curved surface area =πlr1+r2cm2=227×26×20=114407 =1634.29cm2 Hence the surface area is 1634. 29 ².

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The radii of circular ends of a bucket of height 24 𝑐𝑚 are 15 𝑐𝑚 𝑎𝑛𝑑 5 𝑐𝑚. Find its curved surface area.

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Detailed Solution

Given that,

The radii of circular ends of a bucket

𝑟₁ = 15 𝑐𝑚 and 𝑟₂ = 5 𝑐𝑚

Height of the bucket = 24 cm

$\begin{aligned} Slant height l = \sqrt{h^{2} - {(r_{1} - r_{2})}^{2}} \\ = \sqrt{24^{2} + (15 - 5)^{2}} \\ = \sqrt{24^{2} + 10^{2}} \\ = \sqrt{576 + 100} \\ = \sqrt{676} \\ = 26 cm \\ Slant height = 26 cm \end{aligned}$

$\begin{aligned} Curved surface area = π l (r_{1} + r_{2}) {cm}^{2} \\ = \frac{22}{7} \times 26 \times 20 \\ = \frac{11440}{7} \\ = 1634.29 {cm}^{2} \end{aligned}$