Given, length of rectangular park = 175 m
and breadth of rectangular park = 125 m
Then, the perimeter of rectangular park
= 2 (Length + Breadth)
=2 (175 +125) m = 2 \(\times\) 300 cm = 600 m
Cost of fencing per metre = Rs. 12
∴ Cost offencing 600 m =600 \(\times\) 12 = Rs. 7200
Hence, the cost of fencing a rectangular park is Rs. 7200.

Given, side of square park = 250 m
Then, the perimeter of square park
= 4 \(\times\) Length of side of park
= 4 \(\times\)250 m = 1000 m
Cost of fencing =Rs. 20 per metre
∴Cost of fencing = 1000 \(\times\) 20 = Rs.20000
Hence, the cost of fencing a square park is Rs.20000

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.

The perimeter of the given figure is 28 cm

Given, length of rectangular field (I) = 30 m
and breadth of rectangular field (b) = 20 m
Total distance covered by the athlete in one round = Perimeter of field (rectangle)
= 2 (I + b) = 2 (30 + 20) = 100 m
Since, the distance covered in one round = 100 m
So, the distance covered in 10 rounds = 10 \(\times\) 100 = 1000 m.

24 cm² (approx.)

39 cm² (approx.)

Given, figure is covered by 9 full squares.
∴ Area = (9 \(\times\) 1)sq units = 9 sq units

10

Perimeter of an equilateral triangle = 3 X Length of side
[∵  in equilateral triangle, all sides are equal, so length of each side will be 8 cm]
=3 \(\times\) 8 = 24 cm
Hence, the perimeter of rectangle is 24 cm.

Let the given figure is divided into rectangles A, Band C and their length and breadth are written on the figure.
For rectangle A,
Length = 2 cm and breadth = 1 cm
∴ Area of the rectangle A = Length \(\times\) Breadth
=2 \(\times\) 1=2 sq cm
For rectangle B,
Length = 5 ern and breadth = 1 cm

∴  Area of the rectangle B
= Length \(\times\) Breadth = 5 \(\times\)1 = 5 sq cm
For rectangle C,
Length = 2 cm and breadth = 1cm
∴ Area of the rectangle C = Length \(\times\) Breadth
=2 \(\times\) 1=2 sqcm
Now, total area of the given figure = Area of rectangle A + Area of rectangle B + Area of rectangle C
= (2 + 5 + 2) sq cm = 9 sq cm
Hence, the area of the given figure is 9 sq cm

Given, length of table-top = 2 m 25 cm
\(=2m+25\times \frac { 1 }{ 100 } m\quad \left[ \because \quad 1cm=\frac { 1 }{ 100 } m \right] \)
\(=2m+\frac { 25 }{ 100 } m =2m+0.25m\)
= (2 + 0.25) m = 2.25 m
Breadth of table-top = 1m 50 cm = 1 m + 50 cm
\(=1m+50\times \frac { 1 }{ 100 } m\quad \left[ \because 1cm=\frac { 1 }{ 100 } m \right] \)
\(=1m+\frac { 50 }{ 100 } m\quad =1m+0.50m\)
= (1+ 050) m = 1.50 m
∴ Perimeter of table top = 2\(\times\)(Length + Breadth)
= 2\(\times\) (2.25 m + 1.50 rn)
= 2 \(\times\) 3.75 m = 7.50 m
Hence, the perimeter of the table top is 7.50 m.

Number of tiles =\(\frac { Area\quad of\quad the\quad floor }{ Area\quad of\quad one\quad tile } =\frac { 40\times 30 }{ 2\times 2 } =\frac { 1200 }{ 4 } \)
=300

Area of a square = Side \(\times\) Side⇒64= Side \(\times\) Side
⇒8\(\times\) 8= Side\(\times\)Side ⇒ Side = 8m
Perimeter = 4\(\times\)ide= 4\(\times\)8 =32 m

Area of rectangular park = I x b = 30 x 20 = 600 sq m.

Area = 5 x 3 = 15 sq cm

The region enclosed by a plane closed figure is called its area.