Question

The radius of the base of a cylindrical waterdrum open ar the top is 35cm and the height is 1.3m find the inner surface area of the waterdrum

Answer 1

Inner Surface area = Curved Surface Area of cylinder + area of base

= 2πrh+πr^2

=πr(2h+r)

=35π(2×130+35)

=32,436.94 sq. cm

Answer 2

$\Huge{\red{\underline{\underline{\rm{Solution}}}}}$

$\blacksquare$ $\large\bold{\underline{\textsf{Given-}}}$

RADIUS = 35 cm = 0.35 m

HEIGHT = 1.3 m

$\blacksquare$ $\large\bold{\underline{\textsf{To\:Find-}}}$

INNER SURFACE AREA OF THE WATER DRUM

HERE WE GO

$\huge\bold{\mathtt{\red{Formula\:Used-}}}$

2πrh + π{r}^{2}

On substituting the values,we get -

$\blacksquare$ 2π(0.35)(1.3) + π {0.35}^{2}

$\blacksquare$ 2π (0.455) + π(0.1225)

$\blacksquare$ 2(3.14)(0.455) + 3.14(0.1225)

$\blacksquare$ 2.8574 + 0.38465

$\blacksquare$ 3.24205 {m}^{2}

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