Let ABCD be the square park and the region shaded in green be the path.
Here, side length of the outer square, AB = 11 m (given)
and the path is 2 m wide.
$\Rightarrow$ Side length of the inner square, HG
= 11 - 2 - 2 = 7 m
We know that area of a square = $sid{e}^2$
$\Rightarrow$ Area of the path = Area of the outer square - Area of the inner square
$={11}^2-7^2=121-49=72m^2$

$EF$ divides the rectangle ABCD into two trapeziums.

Area of trapezium ABFE
$=\frac{1}{2}\times \left(Sumofparallelsides\right)\times (Distancebetweenthem$
$=\frac{1}{2}\times (AE+BF)\times AB$

$=\frac{1}{2}\times (1+2)\times 4$

$=\frac{1}{2}\times 3\times 4=6c{m}^2$

Area of trapezium CDEF $=\frac{1}{2}\times (FC+ED)\times CD$

$=\frac{1}{2}\times (1+2)\times 4$

$=\frac{1}{2}\times 3\times 4=6c{m}^2$

$\text{Area of both the trapeziums are equal.}$

$\therefore$  $EF$ divides the rectangle $ABCD$ in two equal parts and the area of both the parts are $6c{m}^2$.

Area of the two semi-circles = $\frac{1}{2}(\pi \times r^2)+\frac{1}{2}(\pi \times r^2)=\pi \times r^2=\frac{22}{7}\times 7\times 7=154c{m}^2$

Area of rectangular part = length $\times$ breadth = 10 $\times$14 = 140 $c{m}^2$

Total area = Area of the two semi-circles + Area of the rectangle  = 154+140 = 294 $c{m}^2$​.

Given: The metal sheet in the form of a square of side $5m$.
The area of a square of side '$a$' is given by $a^2$.
So, area of the sheet  $=5^2=25m^2$

Total cost of laying bricks in the park  = Area of the park  x  rate

Area of a the park =  (7 x 7) m²

        = 49 m²

Cost = 49 m² x ( ₹ 10) / m²

        =  ₹ 490

Area of parallelogram = b x h = 48 sq m.

The circumference of the tyre = 2πr (Distance travelled in each rotation)
Total Distance = Distance covered in 1 rotation x Number of rotations
$\Rightarrow 0.396=2\times \frac{22}{7}\times r\times 300$
$\Rightarrow r=\frac{0.396\times 7}{2\times 22\times 300}$
Hence r = 0.00021 km or  21 cm.

When a circular disc is cut into four equal parts, the perimeter of each part is more than $\frac{2\pi r}{4}$. This is because the length of two radii are also added to the perimeter in each part.

Perimeter of each part = $\frac{2\pi r}{4}$+ r + r

 = $\frac{\pi r}{2}$ + 2r
 = $\frac{22\times 7}{7\times 2}$ + 2r

 = $11+2\times$7
                                    
 = 11 + 14

 = 25 cm

Here we have two figures: a square of side 7 cm and a circle of diameter 7 cm, or radius $\frac{7}{2}$ cm
Area of the shaded portion = Area of the square – Area of the circle

=  $a^2–\pi r^2$    (where a is the side of the square and r is the radius of the circle)

= $7^2–\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}$

= 49 – 38.5

= 10.5 $c{m}^2$

Result : 1 Mark
Let the square given be of side A cm.

Perimeter of the square whose side is A cm is 4 $\times$ A

Perimeter of the given square

= 4 $\times$ 10

= 40 m

= 4000 cm