Question

9. A cylindrical container with internal radius of
its base 10 cm, contains water up to a height
of 7 cm. Find the area of the wet surface of
the cylinder.​

Answer 1

Step-by-step explanation:

if it is a closed cylindrical container then

area of the wet surface of the cylinder = tsa of cylinder =

$2\pi \: r(h + r)$

2*22/7*10(7+10)

440/7(17)

now solve this

Answer 2

$\huge\underline\red{solution} : - \\ \\ \underline\red{given} : - \: A \: cylinder \: container \: with \: \\ internal \: radius \: of \: its \: base \: 10 \: cm, \: \\ contains \: water \: up \: to \: height \: of \: 7cm. \\ \red{ 1)} \purple{ \: radius = 10 \: cm \: \: and \: \: height = 7 \: cm }\\ \\ \underline \red{find}: - \: area \: of \: the \: of \: the \: cylinder. \\ \\ \red{formula} : - \\ \\ \: Area \: of \: cylinder \: = 2 \: \pi \: r \: h \: +2 \pi \: {r}^{2} \\ \\ \: Area \: of \: cylinder \: = 2 \times \frac{22}{7} \times 10 \times 7 \: + 2\times\frac{22}{7} \times {(10)}^{2} \\ \\ \: Area \: of \: cylinder \: = 2 \times 22 \times 10 \: + \frac{44}{7} \times 100 \\ \\ \: Area \: of \: cylinder \: = 440 + \frac{4400}{7} \\ \\ \: Area \: of \: cylinder \: = \frac{440 \times 7}{7} + \frac{4400}{7} \\ \\ \: Area \: of \: cylinder \: = \frac{3080 }{7} + \frac{4400}{7} \\ \\ Area \: of \: cylinder \: = \frac{3080 + 4400 }{7} \\ \\ Area \: of \: cylinder \: = \frac{7480 }{7} \\ \\ \: Area \: of \: cylinder \: = 1068.571429 \: {cm}^{2} \\ \\ \underline{ hence \: the \: are \: of \: cylindricl \: container \: is} \\ \underline \red{1068.571429 \: {cm}^{2} .} \: \\ \\ \fcolorbox{red}{white}{i \: hope \: it \: helps \: you.}$