Question

The tsa of a cylinder is 231 cm if the csa of the cylinder is 2/3 of tsa find radius and height

Answer 1

Answer : Radius = 3.5 cm and

Height = 7 cm

Solution :
_________

Given that : Total Surface Area of cylinder = 231 cm² and curved surface area of cylinder is 2/3 of TSA

Then Curved surface area of cylinder

$231 \times \frac{2}{3} = 154 \: {cm}^{2}$

As we know that :

Total Surface Area of Cylinder = Curved surface area of cylinder+2πr²

$154 + 2 \times \frac{22}{7} \times {r}^{2} = 231 \\ \\ = > 2 \times \frac{22}{7} \times {r}^{2} = 77 \\ \\ = > {r}^{2} = \frac{77 \times 7}{22 \times 2} \\ \\ = > {r}^{2} = \frac{49}{4} \\ \\ = > r = \frac{7}{2} = 3.5 \: cm$

Now, As we know that : Curved surface area of cylinder = 2πrh

According to the question :

$= > 2 \times \frac{22}{7} \times 3.5 \times h = 154 \\ \\ = > h = \frac{154}{22} = 7 \: cm$

So, the radius of the cylinder will be 3.5 cm and height will be 7 cm.

Answer 2

$\large\underline\pink{Given:-}$

Cylinder of Height = 4 cm

Cylinder of Radius = 3.5 cm

$\large\underline\pink{To find:-}$

Fine the ratio of the TSA and CSA of a cylinder ....?

$\large\underline\pink{Solutions:-}$

$\: \: \: \: \: \therefore \: \: Total \: \: surface \: \: area \: \: cylinder \: \: = \: \: {2} \: \pi \: r \: {({r} \: + \: {h})}$

$\: \: \: \: \: \leadsto \: \: {2} \: \times \: \frac{22}{7} \: \times \: {3.5} \: {({3.5} \: + \: {4})}$

$\: \: \: \: \: \leadsto \: \: {2} \: \times \: \frac{22}{7} \: \times \: {3.5} \: \times \: {7.5}$

$\: \: \: \: \: \leadsto \: \: {2} \: \times \: {22} \: \times \: {0.5} \: \times \: {7.5}$

$\: \: \: \: \: \leadsto \: \: {44} \: \times \: {3.75}$

$\: \: \: \: \: \leadsto \: \: {165} \: {cm}^{2}$

$\: \: \: \: \: \therefore \: \: Curved \: \: surface \: \: area \: \: of Cylinder \: \: = \: \: {2} \: \pi \: r \: h$

$\: \: \: \: \: \leadsto \: \: {2} \: \times \: \frac{22}{7} \: \times \: {3.5} \: \times \: {4}$

$\: \: \: \: \: \leadsto \: \: {2} \: \times \: {22} \: \times \: {0.5} \: \times \: {4}$

$\: \: \: \: \: \leadsto \: \: {44} \: \times \: {2}$

$\: \: \: \: \: \leadsto \: \: {88} \: {cm}^{2}$

$\: \: \: \: \: Ratio \: \: = \: \: \frac{TSA \: \: of \: \: Cylinder}{CSA \: \: of \: \: Cylinder}$

$\: \: \: \: \: \leadsto \: \: \frac{165}{88}$

$\: \: \: \: \: \: \: Hence, \\ \: \:\therefore \: \: The \: \: ratio \: \: of \: \: the \: \: TSA \: \: and \: \: CSA \: \: of \: \: a \: \: cylinder \: \: {165} \: : \: {88}$