Question
A solid iron cylinder has total surface area of 1848 sq.m. Its curved surface area is five – sixth of its total surface area. Find the radius and height of the iron cylinder.
✓
Answer
T.S.A of the cylinder $=1848 sq \cdot cm$
$
2 \pi r(h+r)=1848
$
Curved surface area $=\frac{5}{6} \times 1848 sq \cdot cm$
$
\begin{aligned}
& 2 \pi r h=5 \times 308 \\
& 2 \pi r h=1540 sq \cdot m
\end{aligned}
$
Substitute the value of $2 \pi r h$ in (1)
$
\begin{aligned}
& 2 \pi r(h+r)=1848 \\
& 2 \pi r h+2 \pi r^2=1848 \\
& 1540+2 \pi r^2=1848 \\
& 2 \pi r^2=1848-1540 \\
& 2 \times \frac{22}{7} \times r^2=308
\end{aligned}
$
$
\begin{aligned}
& r^2=\frac{308 \times 7}{2 \times 22}=49 \\
& r=7
\end{aligned}
$
Radius of the cylinder $=7 m$
$
\begin{aligned}
& 2 \pi r h=1540 \\
& 2 \times \frac{22}{7} \times 7 \times h=1540 \\
& h=\frac{1540}{2 \times 22}=35 m
\end{aligned}
$
Radius of the cylinder $=7 m$
Height of the cylinder $=35 m$
$
2 \pi r(h+r)=1848
$
Curved surface area $=\frac{5}{6} \times 1848 sq \cdot cm$
$
\begin{aligned}
& 2 \pi r h=5 \times 308 \\
& 2 \pi r h=1540 sq \cdot m
\end{aligned}
$
Substitute the value of $2 \pi r h$ in (1)
$
\begin{aligned}
& 2 \pi r(h+r)=1848 \\
& 2 \pi r h+2 \pi r^2=1848 \\
& 1540+2 \pi r^2=1848 \\
& 2 \pi r^2=1848-1540 \\
& 2 \times \frac{22}{7} \times r^2=308
\end{aligned}
$
$
\begin{aligned}
& r^2=\frac{308 \times 7}{2 \times 22}=49 \\
& r=7
\end{aligned}
$
Radius of the cylinder $=7 m$
$
\begin{aligned}
& 2 \pi r h=1540 \\
& 2 \times \frac{22}{7} \times 7 \times h=1540 \\
& h=\frac{1540}{2 \times 22}=35 m
\end{aligned}
$
Radius of the cylinder $=7 m$
Height of the cylinder $=35 m$
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