Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
Given that 
The cost of painting is Rs10 per $c{m}^2$.
The length of the rectangular wall=400 cm
The breadth of the rectangular wall=900 cm
Area of the rectangle =  length $\times$ breadth
On substituting the values we get

Area of the rectangle = (400) $\times$ (900)

  = 360000 $c{m}^2$

Money required to paint the rectangular wall = (10) $\times$ (360000)

= Rs.3600000
The diameter of the circular wall = 126 cm.
Radius of the wall, $r$ = $1262$ = 63 cm
The area a circle is given by, Area = $\pi \times r^2$
On substituting the values we get,
Area of the circular wall
= $\frac{22}{7}\times 63\times 63$ = 12474 $c{m}^2$
Money required to paint the circular wall
 = 12474$\times$10 = Rs 124740
The total cost of painting
= 3600000 + 124740 = Rs 3724740
Hence, the total cost of painting the walls is Rs 37,24,740.

Steps: 2 Marks
Application: 1 Mark
Answer: 1 Mark
The given area will be inscribed between two rectangles of length and breadth.
We can imagine as a rectangle whose length has been increased on both the sides. 
So, the sides  of the outer triangle are,
Length = 35 + 5 = 40 m
And breadth = 40 + 5 = 45 m

40m and 45m, 40m and 35m respectively.
The area of the footpath = Area of outer rectangle - Area of inner rectangle
Area of the rectangle = length$\times$breadth
$\therefore$ On substituting the values we get;

Area required = (45) $\times$ (40) - (40) $\times$ (35)

= 1800 - 1400

= 400$m^2$
The area of the footpath is 400$m^2$.

Formula: 1 Mark
Steps: 2 Marks
Result: 1 Mark
Given that  :
A man wants to cover the floor with carpet which costs Rs 100.
His floor is in the form of a square.
Perimeter of the floor = 24 meters
 
Let the length of each side of the square be 'a'.
Perimeter of a square = 4a = 24 meters
Length of one side, a = $\frac{24}{4}$ = 6 meters
Area of a sqaure = a $\times$ a = 6 $\times$ 6 = 36 square meters
Total Cost of carpeting the floor = 36 $\times$ 100 = Rs 3600. 
The total cost of carpeting the floor is  Rs 3600. 

To find out the minimum length of the wire to be bought by Puja so that the entire farm is covered, we have to find out the perimeter of the entire farm.

From the above figure, the length of the side AF = (Length of side BC) - (Length of side ED) = (20-5) m = 15m.
Length of side AB = (Length of side DC + Length of side FE) = (5+10) m = 15m

As we know, the perimeter of a plane figure is the length of its boundary.

Perimeter of the farm = (15+20+5+5+10+15) m = 70 m.

So, the minimum length of the wire to be bought by Puja is 70 m.

The area grazed by both the goats will be circular in shape as they are moving at a constant distance from a fixed point.
Hence, the area grazed when a goat is tied at a point with a rope would be a circle with radius equal to the length of the rope.
Area of a circle = $\pi r^2$.
Hence, the required ratio = $\frac{AreaatA}{AreaatB}$
= $(\frac{\pi r_A}{\pi r_B}{)}^2$ 
=$(\frac{r_A}{r_B}{)}^2$
= $(\frac{5}{6}{)}^2$ = 25:36 .

Circumference of tyre = 2 πr = Distance travelled in each rotation = $2\times \frac{22}{7}\times 7$ = 44 cm.

Total distance covered = Number of rotations $\times$ Circumference of tyre

Number of rotations = $\frac{0.121km}{44cm}=\frac{0.121\times 1000m}{44\times 0.01m}$ = 275

 Area of a parallelogram = Base $\times$ corresponding height ​.
Therefore,  Base = $\frac{\text{Area of a parallelogram}}{\text{Corresponding height}}$
$\Rightarrow$ Base = $\frac{24}{8}$ = 3 m.

Area of parallelogram = Base $\times$ corresponding height 
Area of parallelogram = 8 $\times$ (1.75 $\times$ 8) 
Area of parallelogram = 112 $m^2$ ​.
Total cost of carpeting the hall = Cost per $m^2$ $\times$​ Area
= ₹ 2/$m^2$ $\times$ 112 $m^2$ = ₹ 224.