
A solid cylinder has a Total Surface Area $462c{m^2}$. Its Curved Surface Area is One third of its Total Surface Area. Find radius of the solid cylinder .
Answer
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Hint- In this particular type of question use the formula of total surface area and substitute the value of curved surface area as one third of 462 $c{m^2}$ . Then we have to solve the given equation to get to the desired answer.
Complete step-by-step solution -
Total surface area = curved surface area + Area of circular bases
Total surface area= $462c{m^2}$
Curved surface area = $\dfrac{1}{3} \times 462 = 154cm$ ( given )
$ \Rightarrow {\text{Total surface area = 2}}\pi {\text{rh}} + 2\pi {r^2}$
$ \Rightarrow 462 = 154 + 2\pi {r^2}{\text{ ( from above )}}$
$
\Rightarrow 2\pi {r^2} = 462 - 154 \\
\Rightarrow 2\pi {r^2} = 308 \\
\Rightarrow \pi {r^2} = 154 \\
\Rightarrow \dfrac{22}{7} \times {{r}^{2}} =154 \\
\Rightarrow {{r}^{2}} = \dfrac{7}{22} \times 154 \\
\Rightarrow {r^2} = 49 \\
\Rightarrow r = 7cm \\
$
Note- It is important to note that the total surface area of a cylinder is made up of the curved surface area and the two circular bases . Remember to recall the formula of both total surface area and curved surface area to be used in this type of questions .
Complete step-by-step solution -
Total surface area = curved surface area + Area of circular bases
Total surface area= $462c{m^2}$
Curved surface area = $\dfrac{1}{3} \times 462 = 154cm$ ( given )
$ \Rightarrow {\text{Total surface area = 2}}\pi {\text{rh}} + 2\pi {r^2}$
$ \Rightarrow 462 = 154 + 2\pi {r^2}{\text{ ( from above )}}$
$
\Rightarrow 2\pi {r^2} = 462 - 154 \\
\Rightarrow 2\pi {r^2} = 308 \\
\Rightarrow \pi {r^2} = 154 \\
\Rightarrow \dfrac{22}{7} \times {{r}^{2}} =154 \\
\Rightarrow {{r}^{2}} = \dfrac{7}{22} \times 154 \\
\Rightarrow {r^2} = 49 \\
\Rightarrow r = 7cm \\
$
Note- It is important to note that the total surface area of a cylinder is made up of the curved surface area and the two circular bases . Remember to recall the formula of both total surface area and curved surface area to be used in this type of questions .
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