Question

In fig, from a cuboidal solid metallic block of dimensions 15 cm x 10 cm x 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remainings block.[ Use π =22/7 ]

Answer 1

Dimensions of the cuboidal solid metallic block are 15 cm ×  10 cm × 5 cm

Radius of the hole = 72 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 × 10 + 10 × 5 + 5 × 15)

= 2(150 + 50 + 75)

= 2 × 275

= 550 cm²

Area of circular holes on both sides of the cylinder = 2 × πr²

= 2 × \(\frac{22}7\) × (\(\frac 72\))²

= 77 cm²

Surface area of the cylindrical hole = 2πrh

= 2 × \(\frac{22}7\) × \(\frac 72\) × 5

= 110 cm²

Therefore,

Surface area of the remaining block = 550 − 77 + 110

= 660 − 77

= 583 cm²

The surface area of the remaining block is 583 cm².

Answer 2

Total surface area = 2( lb + bh + hl ) + 2πrh