Answer:

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: Thickness\:of\:sheet = 0.00005 \: m\qquad \: \\ \\& \qquad \:\sf \:Volune\:of\:cistern = 389.76 \: {m}^{3} \\ \\& \qquad \:\sf \:Area\:of\:sheet\:required = 341.12 \: {m}^{2} \end{aligned}} \qquad \: \\ \\ [/tex]

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-1}}[/tex]

Given that, the volume of a block of gold is 0.5 [tex] m^3[/tex]. It is hammered into a sheet to cover an area of 1 hectare.

Let assume that thickness of the sheet be h m

Thus, we have

[tex]\sf \: Volume\:of\:block = 0.5 \: {m}^{3} \\ \\ [/tex]

[tex]\sf \: Area\:of\:sheet = 1 \: hectare \: = \: 10000 \: {m}^{2} \\ \\ [/tex]

Now,

[tex]\sf \: Volume\:of\:block = Area\:of\:sheet \times thickness \\ \\ [/tex]

So, on substituting the values, we get

[tex]\sf \: 0.5 = 10000 \times h \\ \\ [/tex]

[tex]\sf \: h = \dfrac{0.5}{10000} \\ \\ [/tex]

[tex]\sf\implies \sf \: h = 0.00005 \: m \\ \\ [/tex]

Hence,

[tex]\sf\implies \sf \: Thickness\:of\:sheet = 0.00005 \: m \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

[tex]\large\underline{\sf{Solution-2}}[/tex]

Given that, the dimensions of of a rectangular cistern are 11.2m × 6m × 5.8m.

So,

[tex]\sf \: Length\:of\:cistern, \: l = 11.2 \: m \\ \\ [/tex]

[tex]\sf \: Breadth\:of\:cistern, \: b = 6 \: m \\ \\ [/tex]

[tex]\sf \: Height\:of\:cistern, \: h = 5.8 \: m \\ \\ [/tex]

So,

[tex]\sf \: Volune\:of\:cistern \: \\ \\ [/tex]

[tex]\qquad\sf \: = \: l \times b \times h \\ \\ [/tex]

[tex]\qquad\sf \: = \: 11.2 \times 6 \times 5.8 \\ \\ [/tex]

[tex]\qquad\sf \: = \: 389.76 \: {m}^{3} \\ \\ [/tex]

Now,

[tex]\sf \: Area\:of\:sheet\:required \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2(lb + bh + hl) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2(11.2 \times 6 + 6 \times 5.8 + 5.8 \times 11.2) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2(67.2 +34.8 + 64.96) \\ \\ [/tex]

[tex]\qquad\sf \: = \: 2 \times 170.56 \\ \\ [/tex]

[tex]\qquad\sf \: = \: 341.12 \: {m}^{2} \\ \\ [/tex]

Hence,

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: Thickness\:of\:sheet = 0.00005 \: m\qquad \: \\ \\& \qquad \:\sf \:Volune\:of\:cistern = 389.76 \: {m}^{3} \\ \\& \qquad \:\sf \:Area\:of\:sheet\:required = 341.12 \: {m}^{2} \end{aligned}} \qquad \: \\ \\ [/tex]