Verified answer

Answer:

[tex]\boxed{ \bf{ \:Lateral\:surface\:area\:of\:Cylinder = \: 4658.50 \: {cm}^{2} \: }} \\ \\ [/tex]

Step-by-step explanation:

Let assume that radius and height of cylinder be r and h respectively.

Given that, the perimeter of whose base is 121 cm

[tex]\qquad\sf \: 2\pi \: r \: = \: 121 \\ \\ [/tex]

[tex]\qquad\sf \: 2 \times \dfrac{22}{7} \times r \: = \: 121 \\ \\ [/tex]

[tex]\qquad\sf \: r = \dfrac{121 \times 7}{2 \times 22} \\ \\ [/tex]

[tex]\qquad\sf \: r = \dfrac{11 \times 7}{2 \times 2} \\ \\ [/tex]

[tex]\sf\implies \sf \: r = \dfrac{77}{4} \: cm \\ \\ [/tex]

Further given that, height is equal to the diameter.

[tex]\sf \: h = 2r \\ \\ [/tex]

[tex]\sf \: h = 2 \times \dfrac{77}{4} \\ \\ [/tex]

[tex]\sf\implies \sf \: h = \dfrac{77}{2} \: cm \\ \\ [/tex]

Now,

[tex]\sf \: Lateral\:surface\:area\:of\:Cylinder \\ \\ [/tex]

[tex]\sf \: = \: 2 \: \pi \: r \: h \\ \\ [/tex]

[tex]\sf \: = \: 2 \times \dfrac{22}{7} \times \dfrac{77}{4} \times \dfrac{77}{2} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{22}{7} \times \dfrac{77}{2} \times \dfrac{77}{2} \\ \\ [/tex]

[tex]\sf \: = \: 11 \times 11 \times \dfrac{77}{2} \\ \\ [/tex]

[tex]\sf \: = \: 4658.50 \: {cm}^{2} \\ \\ [/tex]

Hence,

[tex]\sf\implies Lateral\:surface\:area\:of\:Cylinder = \: 4658.50 \: {cm}^{2} \\ \\ [/tex]

Given that,

The perimeter of whose base is 121 cm and its height is equal to the diameter

We know that,

[tex] \qquad \sf \purple{Perimeter = 2\pi r}[/tex]

Let,

[tex] \qquad \sf \: 2\pi r \: = \: 121[/tex]

[tex] \displaystyle \qquad \sf \: 2 \times \frac{22}{7} \times r \: = \: 121 [/tex]

[tex]\displaystyle \qquad \sf \:\frac{44}{7} \times r \: = \: 121[/tex]

[tex]\displaystyle \qquad \sf \: r \: = \: 121 \times \frac{7}{44} [/tex]

[tex]\displaystyle \qquad \sf \: r \: = \: \cancel{\frac{847}{44} }[/tex]

[tex]\displaystyle \qquad \bf \: r \: = \: \frac{77}{4} [/tex]

Also given that,

Height(h) is equals to Diameter

We know that,

[tex] \qquad \sf \purple{Diameter = 2× Radius}[/tex]

Let,

[tex] \qquad \sf \: h = 2 \times radius[/tex]

[tex]\displaystyle \qquad \sf \: h\: = \: 2 \times \frac{77}{4} [/tex]

[tex] \displaystyle\qquad \sf \: h \: = \: \frac{77}{2} [/tex]

Now,

We know the formula lateral surface area of a circular cylinder is 2πrh

According to the question by using formula we get,

[tex] \qquad \sf \: \purple{L.S.A _{(Circular \: Cylinder)} \: = \: 2\pi rh}[/tex]

[tex] \displaystyle \qquad \sf \: \implies\: 2 \times \frac{22}{7} \times \frac{77}{4} \times \frac{77}{2} [/tex]

[tex]\displaystyle \qquad \sf \: \implies\: \frac{44}{7} \times \frac{77}{4} \times \frac{77}{2} [/tex]

[tex]\displaystyle \qquad \sf \: \implies\: \frac{121}{1} \times \frac{77}{2}[/tex]

[tex]\displaystyle \qquad \sf \: \implies\: \frac{9317}{2}[/tex]

[tex]\displaystyle \qquad \bf \: \implies\: \red{ 4658.5}[/tex]

Hence,

[tex] \displaystyle \bf \longmapsto \underline{\boxed{ \bf The \: L.S.A \: of \: Circular \: cylinder \: is \: 4658.5 \: cm²}} \: \bigstar[/tex]

[tex] \: [/tex]