In the given figure, ∆AOB is right-angled triangle right angled at O.
So, 

$\Rightarrow$ $A{O}^2$ + $O{B}^2$ = $A{B}^2$.

 $\Rightarrow$ $5^2$ + $O{B}^2$ = ${13}^2$.

$\Rightarrow$ $O{B}^2$ = 169 - 25 = 144

$\Rightarrow$ OB = 12cm

So now the total area of the hexagon = 4$\times$ Area of ∆AOB+ 2$\times$ Area of rectangle BCPO

$\Rightarrow$ Area = 4 $\times$ ($\frac{1}{2}$ $\times$ OB $\times$ AO) + 2 $\times$ (BC $\times$ OB)

$\Rightarrow$ Area = 4 $\times$ ($\frac{1}{2}$ $\times$ 12 $\times$ 5) + 2 $\times$ (13 $\times$ 12)

$\Rightarrow$ Area = 4 $\times$ (30) + 2 $\times$ (156)

$\Rightarrow$ Area = 432$c{m}^2$.

First, we need to calculate the area of one carpet:

Area of one carpet = 8 $\times$ 6 = 48 $m^2$.

Since total area of the floor of the hall = 1200 $m^2$.

Therefore, number of carpets required = $\frac{areaoffloor}{areaofonecarpet}$ = $\frac{1200}{48}$ = 25.

Volume of Cylinder is given by: ${\text{V}}_{\text{cylinder}}=\pi \times {\text{radius}}^2\times \text{height}$
Volume of Cube is given by: ${\text{V}}_{\text{cube}}={\text{edge}}^3$
Volume of Cube is given by ${\text{V}}_{\text{cuboid}}=\text{length}\times \text{breadth}\times \text{height}$
Therefore,
$\Rightarrow {\text{V}}_{\text{cylinder}}=\frac{22}{7}\times r^2\left(2r\right)=\frac{44}{7}{r}^3$
$\Rightarrow {\text{V}}_{\text{cube}}=(\frac{3}{4}r{)}^3=\frac{27}{64}{r}^3$
$\Rightarrow {\text{V}}_{\text{cuboid}}=r\times \frac{5}{4}r\times \frac{7}{3}r=\frac{35}{12}{r}^3$
Comparing the above volumes we get that  ${\text{V}}_{\text{cube}}$ < ${\text{V}}_{\text{cuboid}}$ < ${\text{V}}_{\text{cylinder}}$

Let us assume the height and radius of the original cylinder is l and r respectively.

Let V₁ be the original volume and V₂ be the volume after the change in height and radius.

Therefore volume of the original cylinder is given by: $V=\pi r^{2}l$

After, height of a cylinder becomes $\frac{1}{8}$of the original height and the radius is doubled

We have, $r_2=2rand{l}_2=\frac{l}{8}$

So the new volume is given by:

$V_2=\pi r_2^2{l}_2=\pi \left(2r{)}^2\right(\frac{l}{8})$

$\Rightarrow V_2=\frac{4\pi r^{2}l}{8}$

$\Rightarrow V_2=\frac{\pi r^{2}l}{2}=\frac{V_1}{2}$

$\Rightarrow V_2=\frac{V_1}{2}$
Hence the volume of the cylinder will be halved.

Let the dimension of a cuboid be l, b, and h.

Since the six surface areas are given:

$\Rightarrow$  l $\times$ b = 12.......................................(1)

$\Rightarrow$  b $\times$ h = 36.......................................(2)

$\Rightarrow$  l $\times$ h = 48.......................................(3)

Now multiplying equation (1),(2) and (3), we get

$\Rightarrow (l\times b)\times (b\times h)\times \left)\right(l\times h)=12\times 36\times 48$

$\Rightarrow$ $(l\times b\times h{)}^2=20736$

$\Rightarrow (l\times b\times h)=\sqrt{20736}=144c{m}^3$

Since volume of a cuboid is calculated as ${}^{\prime }l\times b\times h^{\prime }$, the required volume is $144c{m}^3$.

Volume of one small cuboid = l $\times$ b $\times$ h = 20cm $\times$ 25cm $\times$ 40cm

Since the edge length of the cubical box = 2 m = 200 cm

Now, Volume of the cubical box = $200cm\times 200cm\times 200cm$

So the number of cuboids that can be just accommodated in the box = $\frac{Volumeofthecubicalbox}{volumeofthecuboid}$.

$\Rightarrow$ Number of cuboids = $\frac{200\times 200\times 200}{20\times 25\times 40}=400$

Let h be the height of a trapezium, 'a' be the length of a larger parallel side and 'b' be the length of a smaller parallel side.

Area of trapezium is given by:

$\text{A}=\frac{1}{2}\times (a+b)\times h$     

Since larger parallel side is twice in length as the smaller parallel side
 $\Rightarrow a=2b$

 $\Rightarrow \text{A}=\frac{1}{2}\times (2b+b)\times h$ 

  $\Rightarrow \text{A}=\frac{3b\times h}{2}$ 

  $\Rightarrow 570=\frac{3b\times 5}{2}$  

    $\Rightarrow b=76cm$

$\Rightarrow a=2b=2\times 76=152cm$

Let 6x and 5x be the sides of a rectangle.

$\Rightarrow 6x\times 5x=750$

$\Rightarrow 30x^2=750$

$\Rightarrow x^2=\frac{750}{30}=25$

$\Rightarrow x=5$

Length = 6x = 30m

Breadth = 5x = 25m

Perimeter = 2(l + b) = 2(30 + 25) = 110m 

Let n be the number of mats required 
base of mat = 3 cm
height of mat = 4 cm

Area of the parallelogram  = base $\times$ height

Area of hall = Area of mat $\times$ number of mats
$\Rightarrow$ 180 = 4 x 3 x n 

$\Rightarrow n=\frac{180}{12}=15$

Given that each mat costs Rs. 7

Then the cost of 15 mats costs = 7 x 15 = Rs.105