Question

a metal piece is 77 cm long . the inner diametr of a cross section is 4 cm the outer diameter being 4.4 cm .. find its

i. inner curved surface area
ii. other curved surface area​

Answer 1

$\huge { \bold{answer}}$

$\bold {length \: of \: inner \: pipes \: = 77cm}$

$\bold{inner \: diameter = 4cm}$

$\therefore \bold{inner \: radius = \frac{4}{2} = 2cm}$

outer diameter =4.4 cm

outer radius ,R =2.2 cm

$\therefore \bold{inner \: curved \: surface \: area = 2 \pi \: rh}$

$\: \: \: \: \: \: \: \: \: \implies \: 2 \times \frac{22}{7} \times 2 \times 77$

$\: \: \: \: \: \implies2 \times 22 \times 2 \times 11$

$\: \: \: \: \: \implies \: 968 {cm}^{2}$

$\bold{now}$

$\bold{outer \: curved \: surface \: area = 2 \pi \: rh}$

$\: \: \: \: \: \: \implies2 \times 22 \times 2.2 \times 11$

$\: \: \: \: \implies 1064.8 {cm}^{2}$

Answer 2

Answer:

answer

\bold {length \: of \: inner \: pipes \: = 77cm}lengthofinnerpipes=77cm

\bold{inner \: diameter = 4cm}innerdiameter=4cm

\therefore \bold{inner \: radius = \frac{4}{2} = 2cm}∴innerradius=

2

4

=2cm

outer diameter =4.4 cm

outer radius ,R =2.2 cm

\therefore \bold{inner \: curved \: surface \: area = 2 \pi \: rh}∴innercurvedsurfacearea=2πrh

\: \: \: \: \: \: \: \: \: \implies \: 2 \times \frac{22}{7} \times 2 \times 77⟹2×

7

22

×2×77

\: \: \: \: \: \implies2 \times 22 \times 2 \times 11⟹2×22×2×11

\: \: \: \: \: \implies \: 968 {cm}^{2} ⟹968cm

2

\bold{now}now

\bold{outer \: curved \: surface \: area = 2 \pi \: rh}outercurvedsurfacearea=2πrh

\: \: \: \: \: \: \implies2 \times 22 \times 2.2 \times 11⟹2×22×2.2×11

\: \: \: \: \implies 1064.8 {cm}^{2} ⟹1064.8cm

2