(28) Find the surface area of a cylinder whose perimeter of the base 85 m and height 12 m.
Answers
Answer:
Answer:
\green{\tt{\therefore{T.S.A\:of\:cylinder=2170.475\:cm^{2}}}}∴T.S.Aofcylinder=2170.475cm
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\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}
Step−by−stepexplanation:
\begin{gathered} \green{\underline \bold{Fiven :}} \\ \tt: \implies Perimeter \: of \: base = 85 \: m \\ \\ \tt: \implies Height \: of \: cylinder = 12 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies T.S.A \: of \: cylinder = ?\end{gathered}
Fiven:
:⟹Perimeterofbase=85m
:⟹Heightofcylinder=12m
ToFind:
:⟹T.S.Aofcylinder=?
• According to given question :
\begin{gathered} \bold{As \: we \: know \: that} \\ \tt: \implies Perimeter \: of \: base = 2\pi r \\ \\ \tt: \implies 85 = 2\pi r \\ \\ \tt: \implies \frac{85}{2\pi} = r \\ \\ \tt: \implies r = \frac{85}{2\pi} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies T.S.A \: of \: cylinder = 2\pi r(h + r) \\ \\ \tt: \implies T.S.A \: of \: cylinder = 2\pi \times \frac{85}{2\pi} (12 + \frac{85}{2\pi)} \\ \\ \tt: \implies T.S.A \: of \: cylinder = 85 \times (12 + \frac{85 }{2 \times 3.14 }) \\ \\ \tt: \implies T.S.A \: of \: cylinder = 85 \times (12 + 13.535) \\ \\ \tt: \implies T.S.A \: of \: cylinder = 85 \times 25.535 \\ \\ \green{\tt: \implies T.S.A \: of \: cylinder = 2170.475 \: {cm}^{2} }\end{gathered}
Asweknowthat
:⟹Perimeterofbase=2πr
:⟹85=2πr
:⟹
2π
85
=r
:⟹r=
2π
85
Asweknowthat
:⟹T.S.Aofcylinder=2πr(h+r)
:⟹T.S.Aofcylinder=2π×
2π
85
(12+
2π)
85
:⟹T.S.Aofcylinder=85×(12+
2×3.14
85
)
:⟹T.S.Aofcylinder=85×(12+13.535)
:⟹T.S.Aofcylinder=85×25.535
:⟹T.S.Aofcylinder=2170.475cm
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