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 \[=550-77+110\] \[=583\text{ }c{{m}^{2}}\]
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