Question

in a cuboidal solid metallic block of dimensions 15cm×10cm×5cm a cylinderical hole of diameter 7cm is drilled out.Find the surface area of remaining block​

Answer 1

QUESTION :-

in a cuboidal solid metallic block of dimensions 15cm×10cm×5cm a cylinderical hole of diameter 7cm is drilled out.Find the surface area of remaining block.

GIVEN :-

• L = 15cm

• B = 10cm

• h = 5cm

• r = d/2 .°. r = 7/2 = 3.5cm.

Find :-

the surface area of remaining block.

SOLUTION :-

_____________________

=> TSA of cuboidal block = 2(lb+bh+lh)

=> TSA of cuboidal block =2(15×10+10×5+15×5)

=>TSA of cuboidal block = 2(150 + 50+75)

=> TSA of cuboidal block =2(275)

=>.°. TSA of cuboidal block = 550cm².

______________________

=> TSA of cylinder = 2πrh

=> TSA of cylinder = 2 × 22/7 × 3.5× 5

=> TSA of cylinder = 44 × 0.5 × 5

=>.°. TSA of cylinder = 110cm²

___________________

=> Area of two circular bases = 2πr²

=> Area of two circular bases = 2× 22/7 × 3.5 × 3.5

=> Area of two circular bases = 44 × 0.5 × 3.5

=>.°. Area of two circular bases = 77cm².

___________________

=> Required area = TSA of cuboidal block + TSA of cylinder -(Area of two circular bases )

=> Required area = 550 + 110 - (77)

=> Required area = 550 + 33

=>.°. Required area = 583cm².

___________________

Hence, the surface area of remaining block is 583cm².

Answer 2

${\boxed{\mathtt{\green{GIVEN}}}}$

Dimensions of Cuboidal block = 15cm×10cm× 5cm

Diameter of cylindrical hole = 7cm

${\boxed{\mathtt{\green{To \: Find}}}}$

Area of remaining block.

${\boxed{\mathtt{\green{Solution}}}}$

➾ Area of remaining block = T.S.A of cuboid + C.S.A of cylindrical hole - 2 ( Area of circular hole)

$\mathtt{T.S.A\: of \: Cuboid}$

${\boxed{\mathtt{Formula\: = \:2\:(\:lb \:+\: bh\: +\:hl)}}}$

➾ 2 ( 15×10 + 10×5 + 5×15 )

➾ 2 ( 150 + 50 + 75 )

➾ 2 ( 275 )

$\boxed{ \: 550 {cm}^{2} }$

$\mathtt{C.S.A \: of \: conical\: hole}$

${\boxed{ \mathtt{formula \: 2.\pi.r.h}}}$

➾ 2× π × (7/2) × 5

➾ 35 π

➾( 7 × 5 × 22 )/7

${ \boxed {\mathtt{⇢ \: 110 {cm}^{2} }}}$

$\mathtt{Area \: of \: two\: holes\:}$

${\boxed{ \mathtt {formula \: = \: \pi. {r}^{2} }}}$

➾[ π (7/2)² ]×2

➾[( 22/7) (49/4) ]×2

➾ (77/2) × 2

${ \boxed{ \mathtt{ \: 77 {cm}^{2} }}}$

Remaining area = 550 + 110 - 77

➾ 660 - 77

➾ 583 cm²

${\boxed {\mathtt{ \red{ \: 583 {cm}^{2} \: is \: ans}}}}$