Distance around a square field = 200 rods
Length of this rod = 140 cm
Total distance around a squared field
= 200 \(\times\) 140= 28000 cm
So, perimeter of this squared field
= 28000 cm= 280 m
Sides of this field = \(\frac{280}{4}\) = 70 m

Given, length of the room = 4 m
and breadth of the room = 3 m 50 ern = 3 m + 50 cm
\(=1m+50\times \frac { 1 }{ 100 } m\quad \left[ \because \quad 1cm=\frac { 1 }{ 100 } m \right] \)
\(=3m+\frac { 50 }{ 100 } m\quad =3m+0.50m=3.50m\)
∵ Carpet needed to cover the floor of room = Area of the floor
= Length \(\times\) Breadth = 4 m \(\times\) 3.50 m = 14 sq m
Hence, the carpet needed to cover the floor of the room is 14 sq m.

Given, length of the table-top = 2 m
and breadth of the table-top = 1m 50 cm = 1m + 50 cm
\(=1m+50\times \frac { 1 }{ 100 } m\quad \left[ \because \quad 1cm=\frac { 1 }{ 100 } m \right] \)
\(=1m+\frac { 50 }{ 100 } m\quad =1m+0.50m=1.50m\)
∴  Area of the table-top = Length \(\times\)Breadth
= 2 m \(\times\)1.50 m =3sq m
Hence, the area of the table-top is 3 sq m

The area of a square = 1 sq cm
Area of figure = Number of (x) squares + Number of (✔️) squares + \(\frac{1}{2}\) Number of (0) squares
=65 +8+\(\frac{1}{2}\times\)12= 65 +8+6 = 79 cm²
[since, X and ✔️ squares are covered fully by the figures and 0 square are covered half]
Hence, the area of square is 79 cm²

1800 m

8

Given, the figure is a rectangle, whose length = 30 cm and breadth = 20 cm
∴ Perimeter of rectangle = 2x (Length + Breadth)
= 2 \(\times\) (30 + 20) cm
= 2 \(\times\) 50cm =100cm

Now, distance covered by in one round = Perimeter of the park
∴ Perimeter of square park = 4 \(\times\) Length of a side
= 4 \(\times\) 75 m = 300 m
and distance covered by Bulbul in one round
= Perimeter of rectangle = 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (60 + 45) m = 2 \(\times\) 105 m = 210 m
Since, 300 m > 210 m
Hence, Bulbul covers less distance.

Here, length of string will be the perimeter of square.
∴ Perimeter of square = 30 cm
We know that, a square has 4 equal sides.
∴ Perimeter of a square = 4 \(\times\) Length of a side
\(Now,length\quad of\quad one\quad side=\frac { Perimeter\quad of\quad a\quad square }{ 4 } \)
\(=\frac { 30 }{ 4 } cm=7.5cm\)
Hence, length of each side of a square is 7.5 cm.

Given, perimeter of a regular pentagon = 100 cm
We know that, a regular pentagon has 5 equal sides.
∴ Perimeter of a regular pentagon = 5 x Length of a side
So, we can divide given perimeter by 5 to get the one side of a regular pentagon.
\(\therefore \quad Length\quad of\quad one\quad side=\frac { Perimeter\quad of\quad a\quad regular\quad pentagon }{ 5 } \)
\(=\frac { 100 }{ 5 } cm=20cm\)
Hence, all the sides of a regular pentagon is of 20 cm

Given, length of the wooden strip = 32 cm
and breadth of the wooden strip = 21 cm
Now, wooden strip required = Perimeter of the photograph
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (32 cm + 21 cm)
=2\(\times\)53cm =106 cm
Hence, the required length of wooden strip is 106 cm.

Given, length of lid of a rectangular box = 40 cm
and breadth of the lid of a rectangular box = 10 cm
Length of the tape required = Perimeter of the lid of the rectangular box
= 2\(\times\) (Length + Breadth)
= 2 \(\times\)(40 cm + 10 cm) = 2 x 50 cm = 100 cmor 1 m
[∵ 1 cm = \(\frac{1}{100}\)m or 100 cm = 1 m]
Hence, the length of tape required is 100 cm or 1 m.

Let ABCD be a park whose lengths are BC, AD and widths are AB, CD respectively.

Here, AB = CD = 80 m and BC = DA = 150 m
Now, sum of the lengths of four sides
=AB+ BC + CD +DA
= 80 m + 150 m + 80 m + 150 m
= (80 + 150 + 80 + 150) m = 460 m
∴  Perimeter of the park = Sum of the lengths of four sides of the park =460m
Hence, the distanee eovered by Meera is 460 m.

Let ABCD be the page of a notebook.
On measuring, AB = CD = 15 cm and BC = DA = 20 cm

Then, the sum of the lengths of four sides
= AB + BC + CD + DA
= 15 cm + 20 cm + 15 cm + 20 cm
= (I5 + 20 + 15 + 20) cm = 70 cm
Hence, the sum of the four sides is 70 em.
Now, perimeter of the page = Sum of the lengths of four sides
=70 cm

Given, perimeter of a regular hexagon =30 cm
A regular hexagon has 6 sides.
∴ The perimeter of regular hexagon = 6 \(\times\) Length of a side
⇒30 = 6 \(\times\) Length of a side
∴ The length of a side=\(\frac{30}{6}\)=5cm
Hence, length of each side of a regular hexagon is 5 cm.