question_answer

In Fig. 5, from a cuboidal solid metallic block, of dimensions \[15\text{ }cm\times 10\text{ }cm\times 5\text{ }cm\], a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. \[\left[ \text{Use}\,\pi =\frac{22}{7} \right]\]

Answer:

We have cuboidal solid metallic block having dimensions \[15\text{ }cm\times 10\text{ }cm\times 5\text{ }cm\]. and diameter of cylinder is 7 cm.

Now, Total surface area of cuboidal block

\[=2(lb+bh+hl)\]

\[=2(15\times 10+10\times 5+5\times 15)\]

\[=2(150+50+75)\]

\[=2\times 275=550\,\,c{{m}^{2}}\].

2 (Area of circular base) \[=2\times \pi {{r}^{2}}\]

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

\[=77\,\,c{{m}^{2}}\].

And, curved surface area of cylinder \[=2\pi rh\]

\[=2\times \frac{22}{7}\times \frac{7}{5}\times 5\]

\[=110\,\,c{{m}^{2}}\]

Hence, required surface area = T.S.A. of block\[\]Area of base + C.S.A. of cylinder

\[=550-77+110\]

\[=583\text{ }c{{m}^{2}}\]