Question

the radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm respectively the curved surface area of the bucket is​

Answer 1

ANS ]➡ r1 (smaller radius)=7

➡ r2 (bigger radius) = 28

➡ slant height.(L) =45

curved surface Area of Frustum

= π(r1+r2)l

=3.14(7+28)45

=4945.5⚫

Answer 2

Answer:-

Known Terms:-

l = Length

r₁ = Radius of top

r₂ = Radius of bottom

C.S.A = Curved surface area

Given:-

Slant height, l = 45 cm

Radius of top, r₁ = 28 cm

Radius of bottom, r₂ = 7 cm

To Find:-

Curved surface area

Formula to be used:-

Curved surface area of frustum of a cone that is πl×(r₁ + r₂)

Solution:-

Putting all the values, we get

= πl × (r₁ + r₂)

= π (45) (28 + 7)

= $\frac{22}{7}$ × 45 × (28 + 7)

= $\frac{22}{7}$ × 45 × 35

= 4950 cm²

Hence, the curved surface area of bucket is 4950 cm².