Question

An iron pipe is 20 cm long, and has exterior diameter equal to 25 cm. if the thickness of the pipe is 1 cm. find the total surface area of the pipe.

Answer 1

Answer:

• The total surface area of the pipe is 3168 cm² .

Given :

• Length of pipe = 20 cm

• Outer Diameter = 25 cm

• The thickness of Pippi = 1 cm

To find :

• Total Surface Area of the pipe =?

Step-by-step explanation:

Length of pipe = 20 cm

Inner diameter = 25 - (1 + 1)

25 - (1 + 1)= 23 cm

Inner radius = 23/2 = 11.5 cm

Now,

Let the external radius be R and the internal radius be r.

Total surface area of the pipe = 2πh(R + r) + 2π(R² - r²)

Substituting the values in the above formula, we get,

⇒ 2*22/7*20(12.5 + 11.5) + 2*22/7*{(12.5)² - (11.5²)

⇒ (44*480)/7 + (44*24)/7

⇒ 21120/7 + 1056/7

⇒ 22176/7

= 3168 cm²

Thus, the total surface area of the pipe is 3168 square cm.

Answer 2

AnswEr :

3168cm².

$\bf{\pink{\underline{\underline{\bf{Given\::}}}}}$

An iron pipe is 20 cm long, and has exterior diameter equal to 25 cm. If the thickness of the pipe is 1 cm.

$\setlength{\unitlength}{1.05 cm}}\begin{picture}(12,4)\thicklines\put(3,6){ . }\put(6,6){\circle{4}}\put(6,6){\circle{0.8}}\put(5.34,6){\line(0,1){2.5}}\put(6.66,6){\line(0,1){2.5}}\put(6,8.5){\circle{4}}\put(6,8.5){\circle{0.8}}\put(5.6,6){\line(0,1){2.5}}\put(6.4,6){\line(0,1){2.5}}\put(7,7){\vector(0,1){1.5}}\put(7,7){\vector(0,-1){1}}\put(7.3,7.2){ 20\:cm }\put(5.6,4.6){ 25\:cm }\put(6,5){\vector(1,0){0.5}}\put(6,5){\vector(-1,0){0.5}}\end{picture}}$

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The Total surface area of the pipe.

$\bf{\blue{\underline{\underline{\bf{Explanation\::}}}}}$

$\bf{We\:have}\begin{cases}\sf{Height\:of\:Iron\:pipe\:(h)=20\:cm}\\ \sf{External\:diameter\:of\:pipe\:(d)=25\:cm}\\ \sf{Thickness\:of\:pipe=1\:cm}\end{cases}}$

&

$\bf{We\:get}\begin{cases}\sf{External\:radius\:of\:pipe\:(R)=\cancel{\dfrac{25}{2} }cm=12.5cm}\\\\ \sf{Internal\:diameter\:of\:pipe\:(d)=(25-2)cm=23\:cm}\\ \\\sf{Internal\:radius\:of\:pipe\:(r)=\cancel{\dfrac{23}{2}} cm=11.5\:cm}\end{cases}}$

Formula use :

$\bf{\large{\boxed{\bf{Curved\:surface\:area\:of\:cylinder\:=2\pi rh\:\:\:(sq.unit)}}}}}$

A/q

$\mapsto\tt{T.S.A.=E.C.S.A+I.C.S.A.+2(Area\:of\:ring)}}\\\\\\\\\mapsto\tt{T.S.A.=2\pi Rh+2\pi rh+2(\pi R^{2} -\pi r^{2} )}\\\\\\\\\mapsto\tt{T.S.A.=2\pi h(R+r)+2\pi \big[(R+r)(R-r)\big]}\\\\\\\\\mapsto\tt{T.S.A.=2\pi (R+r)(h+R-r)}\\\\\\\\\mapsto\tt{T.S.A.=2*\dfrac{22}{7} (12.5+11.5)(20+12.5-11.5)}\\\\\\\\\mapsto\tt{T.S.A.=\dfrac{44}{7} *24(20+1)}\\\\\\\\\mapsto\tt{T.S.A.=\bigg(\dfrac{44}{\cancel{7}} *24*\cancel{21}\bigg)cm^{2} }\\\\\\\\\mapsto\tt{T.S.A.=(44*24*3)cm^{2} }$

$\mapsto\tt{\orange{T.S.A.=3168\:cm^{2} }}$

Thus,

$\bf{\pink{\large{\underline{\sf{The\:total\:surface\:area\:of\:the\:pipe\:is\:3168\:cm^{2} }}}}}$