Question

The radius and height of a cylinder are in the ratio 5: 7 and it volume is 4400 cm. Find the radius
and surface area of the cylinder.​

Answer 1

Given: Ratio of radius and height of a cylinder is 5:7. & Volume of park is 4400 cm.

Need to find: Radius and surface area of cylinder?

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❍ Let's consider radius and height of cylinder be 5x and 7x.

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As we know that,

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$\begin{gathered}\star\:{\underline{\boxed{\frak{Volume_{\:(cylinder)} = \pi {r}^{2}h}}}}\\\\\\ \bf{\dag}\:{\underline{\frak{Putting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf 4400 = \frac{22}{7} \times ({5x})^{2} \times 7x\\\\\\ :\implies\sf 4400 = \frac{22}{7} \times 25 {x}^{2} \times 7x \\ \\ \\ :\implies\sf4400 = 22\times 25 {x}^{2} \times x \\ \\ \\ :\implies\sf4400 = 22 \times 25 {x}^{3} \\\\\\ :\implies\sf x^3 = \frac{4400}{22 \times 25} \\\\\\ :\implies\sf x^3 = \frac{4400}{550} \\\\\\ :\implies\sf x^3 = 8\\\\\\ :\implies{\underline{\boxed{\frak{\purple{x = 2}}}}}\:\bigstar\\\\\end{gathered}$

Thus, Radius will be 5 × 2 = 10cm & height will be 7 × 2 = 14cm.

Now, Surface area,

$\begin{gathered}\star\:{\underline{\boxed{\frak{Surface \: Area_{\:(cylinder)} = 2 \pi {r}h + 2\pi {r}^{2}}}}}\\\\\\ \bf{\dag}\:{\underline{\frak{Putting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf 2\times \dfrac{22}{7} \times 10 \times 14+ 2 \times \frac{22}{7} \times {10}^{2}\\\\\\ :\implies\sf 2\times \dfrac{22}{7} \times 140 + 2 \times \frac{22}{7} \times 100 \\ \\ \\ :\implies\sf2\pi \times 140 + 2\pi \times 100\\ \\ \\ :\implies\sf280\pi + 200\pi\\\\\\ :\implies\sf 480\pi \\\\\\ :\implies{\underline{\boxed{\frak{\purple{1508.57 \ cm}}}}}\:\bigstar\\\\\end{gathered}$

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Therefore,

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Radius of the cylinder, 5x = 10cm

Surface area of the cylinder, 1508.57cm (approx)

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$\therefore\:{\underline{\sf{Hence,\:Radius\:and\:surface \: area \: of \: cylinder \:is\:\bf{10\:cm}\: \sf{and}\: \bf{1508.57\:cm}\: \sf{respectively}.}}}$

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$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$

Answer 2

the above ans is correct

Step-by-step explanation: