Question

From a cuboidal solid metallic block of dimensions 15 cm ×10 cm × 5 cm a cylindrical hole of diameter 7cm is drilled out. Find the surface area of the remaining block. (Use π= 227 ).

Answer 1

FIGURE IS IN THE ATTACHMENT
SOLUTION:

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

Diameter of a cylindrical hole(d)= 7 cm
Radius of a cylindrical hole = d/2 = 7/2

Surface area of a cuboid = 2(lb + bh + hl)
Surface area of a cuboid = 2(15×10 + 10×5 + 5×15)
= 2 (150 + 50 + 75) = 2 × 375 = 550 cm²

Curved surface Area of a cylinder = 2πrh
= 2 × (22/7) × (7/2) × 5 = 22 × 5 = 110 cm²

Area of  2 cylindrical holes = 2(πr²)
= 2 × (22/7) × (7/2)²
= 2 × (22/7) × (7/2) × (7/2)
= 11 × 7
= 77 cm²

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
= 583 cm².

Hence, the Surface area of the remaining block = 583 cm².

HOPE THIS WILL HELP YOU....

Answer 2

Surface Area of remaining block = TSA of cuboid - 2 (Base Area of Cylinder) + CSA of Cylinder
= 2 (5*10 + 10*15 + 5*15) - 2 (22/7*7/2*7/2) + 2*22/7*7/2*5
= 583 cm2