In figure below, from a cuboidal solid metalic block, of dimensions \( 15 \mathrm{~cm} \times 10 \mathrm{~cm} \) \( \times 5 \mathrm{~cm} \), a cylindrical hole of diameter \( 7 \mathrm{~cm} \) is drilled out. Find the surface area of the remaining block. (Take \( \pi=22 / 7) \).

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Given:

From a cuboidal solid metalic block, of dimensions \( 15 \mathrm{~cm} \times 10 \mathrm{~cm} \) \( \times 5 \mathrm{~cm} \), a cylindrical hole of diameter \( 7 \mathrm{~cm} \) is drilled out.

To do:

We have to find the surface area of the remaining block.

Solution:

Dimensions of the cuboidal solid metallic block are $15\ cm \times\ 10\ cm \times 5\ cm$

Radius of the hole $= \frac{7}{2}\ cm$

Height of the cylinder $= 5\ cm$

Length of the block $l = 15\ cm$

Breadth of the block $b = 10\ cm$

Height of the block $h= 5\ cm$

Therefore,

Surface area of the block $= 2(lb + bh + hl)$

$= 2(15 \times 10 + 10 \times 5 + 5 \times 15)$

$= 2(150 + 50 + 75)$

$= 2 \times 275$

$= 550\ cm^2$

Area of circular holes on both sides of the cylinder $= 2 \times \pi r^2$

$=2 \times \frac{22}{7} \times (\frac{7}{2})^2$

$=77 \mathrm{~cm}^{2}$

Surface area of the cylindrical hole $=2 \pi r h$

$=2 \times \frac{22}{7} \times \frac{7}{2} \times 5$

$=110 \mathrm{~cm}^{2}$

Therefore,

Surface area of the remaining block $=550-77+110$

$=660-77$

$=583 \mathrm{~cm}^{2}$

The surface area of the remaining block is $583\ cm^2$.

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Updated on: 10-Oct-2022

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