Question

The inner and the outer radii of a cylindrical piper are 2.5cm and 3.5 cm respectively fond the area of the cross sections of the pipes take pi 3.14​

Answer 1

r2 (outer radius)= 3.5 cm

r1 (Inner radius)= 2.5 cm

Area of cross-section =

$\pi( {r2}^{2} - {r1}^{2} )$

$= 3.14 ( {3.5}^{2} - {2.5}^{2} )$

$= 3.14 (12.25 - 6.25)$

$= 3.14 \times 6 \\ =18.84 \: {cm}^{2}$

Answer 2

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r2 (outer radius)= 3.5 cm

r1 (Inner radius)= 2.5 cm

Area of cross-section =

\pi( {r2}^{2} - {r1}^{2} )π(r22−r12)

= 3.14 ( {3.5}^{2} - {2.5}^{2} )=3.14(3.52−2.52)

= 3.14 (12.25 - 6.25)=3.14(12.25−6.25)

= 3.14 \times 6 \\ =18.84 \: {cm}^{2=3.14×6=18.84cm2