Q. 725.0( 1 Vote )

# If the radii of t

Given: The radii of the end of a bucket are 5 cm and 15 cm and it is 24 cm high

Bucket is in the shape of frustum.

TSA of a frustum of a cone = πl(r1 + r2) + πr12 + πr22 (here l , r1, r2 are the slant height, radii of the frustum)

Let S be the TSA of the bucket

S = πl(r1 + r2) + π(r2)2 (here , top of the bucket is not closed but bottom is closed, π(r2)2 = 0 )

l = √(h2 + (R-r)2)

S = π × √(h2 + (R-r)2) × (r1 + r2) + π(r2)2

S = π × √(242 + (15-5)2) × (5 + 15) + π × (5)2

S = π × √(576 + 100) × (20) + π × 25

S = π × √(676) × (20) + π × 25

S = π × 26 × (20) + π × 25

S = π × 520 + π × 25

S = π × (520 + 25)

S= 3.14 × 545 = 1711.3 cm2

The surface area of the bucket is 1711.3 cm2

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