Question

In Fig. 14.58,from a cuboidal solid metallic block,of dimensions $15cm\times 10cm\times 5cm$,a cylindrical hole of diameter 7 cm is drilled out.Find the surface area of the remaining block.

Surface areas and volumes

Answer 1

Length of the cuboidal solid metallic block(l) $=15cm$.
Breadth of the cuboidal solid metallic block(b) $=10cm$.
Height of the cuboidal solid metallic block = height of the cylinder (h)$=5cm$.

Diameter of a cylindrical hole(d)= 7 cm
Radius of a cylindrical hole = $\frac{d}{2}=\frac{7}{2}$

Surface area of a cuboid = $2(lb+bh+hl)$
Surface area of a cuboid = $2(15\times 10+10\times 5+5\times 15)=2(150+50+75)=2\times 375=550c{m}^2$

Curved surface Area of a cylinder = $2\pi rh=2\times \frac{22}{7}\times \frac{7}{2}\times 5=22\times 5=110c{m}^2$

Area of 2 cylindrical holes = $2\pi r^2=2\times \frac{22}{7}\times (\frac{7}{2}{)}^2=11\times 7=77c{m}^2$

Surface area of the remaining block = surface area of the cuboidal block +CSA of cylinder - Area of 2 cylindrical holes

Surface area of the remaining block $=550+110-77$
$=660-77$
$=583c{m}^2$.

Hence, the Surface area of the remaining block $=583c{m}^2$