question_answer

Find the area to be painted in the following block with a cylindrical hole. Given that length is 15 cm, weight 12 cm, and radius of the hole 2.8 cm.

Answer:

Here,            l = 15 cm, b = 12 cm, h = 20 cm

and            r = 2.8 cm

Surface area of cuboid = 2(lb + bh + hl)

\[=2\left( 15\times 12+12\times 20+20\times 15 \right)\]

\[=\text{ }2\left( 180+240+300 \right)\]

\[=\text{ }2\times 720\]

= 1440 cm2

Area of two holes \[=\text{ }27\pi {{r}^{2}}\]

\[=2\times \frac{22}{7}\times 2.8\times 2.8\]

= 49.28 cm2

So,

Required area for paint = S.A. of cuboid - area of holes

\[=\text{ }1440\text{ }-\text{ }49.28\]

= 1390.72 cm2