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In Fig., from a cuboidal solid metallic block, of dimensions 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block.(Take π = \(\frac{22}7\))

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Here, 

dimension of cuboid = 15 cm × 10 cm × 5 cm 

Radius of cylindrical hole = \(\frac{7}2\) cm 

Surface area of the remaining block = surface area of cuboidal solid metallic block – 2 × base area of cylindrical hole 

= 2(15 × 10 + 10 × 5 + 5 × 15) - 2 × \(\frac{22}7\times(\frac{7}2)^2\)

= 583 cm2

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