write the substance of poem 'truth'
Answers
Answer:
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Explanation:
The title, 'The Truth', suggests that truth is the greatest virtue. Being truthful is very important. But truth generally is bitter. Therefore, if one has to speak the bitter truth it should not be spoken in harsh and unpalatable words, for harsh and bitter words can deeply hurt the listener.
Answer:
Answer:
i) The curved surface area of cylinderical petrol tank is 59.4 m².
ii) The steel actually used to make the tank is 95.04 m².
Step-by-step-explanation:
We have given that,
For a cylinderical petrol tank,
Diameter ( d ) = 4.2 m
∴ Radius ( r ) = d ÷ 2 = 4.2 ÷ 2 = 2.1 m
Height ( h ) = 4.5 m
We have to find,
i) Curved surface area of tank
ii) Total steel used to make tank
i)
Now, we know that,
\displaystyle{\pink{\sf\:Curved\:surface\:area\:of\:cylinder\:=\:2\:\pi\:r\:h}}Curvedsurfaceareaofcylinder=2πrh
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:2\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\cancel{2.1}\:\times\:4.5}⟹CSA
cylinder
=2×
7
22
×
2.1
×4.5
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:2\:\times\:22\:\times\:0.3\:\times\:4.5}⟹CSA
cylinder
=2×22×0.3×4.5
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:22\:\times\:0.3\:\times\:9}⟹CSA
cylinder
=22×0.3×9
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:198\:\times\:0.3}⟹CSA
cylinder
=198×0.3
\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:59.4\:m^2}⟹CSA
cylinder
=59.4m
2
\therefore\:\underline{\boxed{\red{\sf\:Curved\:surface\:area\:of\:cylinder\:=\:59.4\:m^2\:}}}∴
Curvedsurfaceareaofcylinder=59.4m
2
────────────────────────
ii)
Now,
The tank is closed with steel.
We have to find how much steel was used to make the tank.
Now,
\displaystyle{\pink{\sf\:Total\:surface\:area\:of\:cylinder\:=\:2\:\pi\:r\:(\:r\:+\:h\:)}}Totalsurfaceareaofcylinder=2πr(r+h)
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:2\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\cancel{2.1}\:(\:2.1\:+\:4.5\:)}⟹TSA
cylinder
=2×
7
22
×
2.1
(2.1+4.5)
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:2\:\times\:22\:\times\:0.3\:\times\:6.6}⟹TSA
cylinder
=2×22×0.3×6.6
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:22\:\times\:0.3\:\times\:13.2}⟹TSA
cylinder
=22×0.3×13.2
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:22\:\times\:3.96}⟹TSA
cylinder
=22×3.96
\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:87.12\:m^2}⟹TSA
cylinder
=87.12m
2
\therefore\:\underline{\boxed{\green{\sf\:Total\:surface\:area\:of\:cylinder\:=\:87.12\:m^2\:}}}∴
Totalsurfaceareaofcylinder=87.12m
2
Now, from the given condition,
\displaystyle{\pink{\sf\:Steel\:used\:to\:make\:tank\:-\:Steel\:wasted\:=\:Total\:surface\:area\:of\:cylinder}}Steelusedtomaketank−Steelwasted=Totalsurfaceareaofcylinder
\displaystyle{\implies\sf\:Steel\:used\:-\:\dfrac{1}{12}\:Steel\:used\:=\:TSA_{cylinder}}⟹Steelused−
12
1
Steelused=TSA
cylinder
\displaystyle{\implies\sf\:Steel\:used\:\left(\:1\:-\:\dfrac{1}{12}\:\right)\:=\:87.12}⟹Steelused(1−
12
1
)=87.12
\displaystyle{\implies\sf\:Steel\:used\:\left(\:\dfrac{12\:-\:1}{12}\:\right)\:=\:87.12}⟹Steelused(
12
12−1
)=87.12
\displaystyle{\implies\sf\:Steel\:used\:\times\:\dfrac{11}{12}\:=\:87.12}⟹Steelused×
12
11
=87.12
\displaystyle{\implies\sf\:Steel\:used\:=\:\dfrac{\cancel{87.12}\:\times\:12}{\cancel{11}}}⟹Steelused=
11
87.12
×12
\displaystyle{\implies\sf\:Steel\:used\:=\:7.92\:\times\:12}⟹Steelused=7.92×12
\displaystyle{\therefore\:\underline{\boxed{\blue{\sf\:Steel\:used\:to\:make\:tank\:=\:95.04\:m^2\:}}}}∴
Steelusedtomaketank=95.04m
2