In the given figure, 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. [Use π=227]

A
550 cm2
B
583 cm2
C
500 cm2
D
400 cm2

Solution:

Given that a cuboidal solid metallic block has dimensions

15 cm × 10 cm × 5 cm

and a cylindrical hole of diameter 7 cm is drilled out.

Let l, b and h be the length, breadth and height of the cuboid respectively and r and h be the radius and height of the cylinder respectively.
Here,

l = 15 cm, b = 10 cm, h = 5 cm, r = \frac{7}{2} cm .

The surface area of the cuboid is given by,

S A = 2 l b + b h + h l

⇒ S A = 2 15 × 10 + 10 × 5 + 15 × 5 ⇒ S A = 2 (150 + 50 + 75) ⇒ S A = 2 (275) ⇒ S A = 550 c m 2

The curved surface area of the cylinder is given by,

C S A =

2 πrh

⇒ C S A = 2 × 227 × 72 × 5 ⇒ C S A = 22 × 5 ⇒ C S A = 110 c m 2

The area of the two ends of the cylindrical holes is,

A = 2 π r 2

⇒ A = 2 227 × 72 × 72 ⇒ A = 22 × 7 2 ⇒ A = 11 × 7 ⇒ A = 77 c m 2

Here,
Surface area of remaining block = Surface area of cuboid + curved surface area of cylinder – area of ends of two cylindrical holes
⇒ Surface area =

(550 + 110 - 77) {cm}^{2}

⇒ Surface area = 583

{cm}^{2}

Hence the surface area of remaining solid is 583

{cm}^{2}

.
Therefore, the correct answer is option (2).

In the given figure, 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. [ $Use π = \frac{22}{7}]$

Solution:

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