Question

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

Answer 1

Radius of top bucket ( R ) = 15 cm.

Radius of bottom  bucket ( r ) = 5 cm.

Height of the bucket :-

H  = 24 cm

Slant Height ( L ) = $\sqrt{( H )^{2} + ( R - r )^{2} }$

=> $\sqrt{(24)^{2} + (15 - 5)^{2}}$

=> $\sqrt{576 + (10)^{2} }$

=> $\sqrt{ 576 + 100}$

=> $\sqrt{676}$

=> 26 cm.

Curved Surface area of bucket = πL ( R + r ) cm².

=>  $\frac{22}{7}$× 26 ( 15 + 5 ) cm².

=> ( 22 × 26 ) × $\frac{22}{7}$ cm².

=> ( 22 × 26 × 20 ) / 7 cm².

=> 1634.28 cm².

Answer 2

Answer:

1634.3 cm² .

Step-by-step explanation:

Let radii be R and r .

It is given R = 15 cm , r = 5 cm and h = 24 cm

We know slant height :

l = [ √ h² + ( R - r )² ]

l = √ 24² + 10²

l = 26 cm .

Now ,

C.S.A. = π ( R + r ) l

= 3.14 × 20 × 26 cm²

= 1634.3 cm² .