(i)

(iii)

12 cm

16 cm²

Let side of squared garden be x m.
Given that, perimeter of a squared garden = 48 m
∴ 4 x Side of a square = 48
⇒ 4x=48⇒x=\(\frac{48}{4}=12m\)
Now, area of the squared garden = (X)²
=(12)²=144m²
Also given, area of small flower bed cover inside the garden = 18 m²
∴ Area of the 9B.rden not covered by flower bed
= Area of squared garden - Area of flower bed
=144 m² -18 m² = 126 m²
The fractional part of the garden covered by flower bed=\(\frac { Area\quad covered\quad by\quad the\quad flower }{ Remaming\quad Area\quad of\quad the\quad squared\quad garden } \)
\(=\frac { 18 }{ 126 } =\frac { 2 }{ 14 } =\frac { 1 }{ 7 } \)
Hence, ratio of the area covered by the flower bed and the remaining area is 1 : 7.

Given, length of floor of the room (l)= 9.68 m
and width of floor ofthe room (b)= 6.2 m
Area of the room = 9.68\(\times\) 6.2 sq m . ... (i)
Also, given that length of each tile = 22 cm
and width of each tile = 10 cm
Now, area of each tile = 22 \(\times\) 10 sq cm ... (ii)
Number of tiles required to cover the floor of the room
\(=\frac { 9.68\times 6.2\times 100\times 100 }{ 22\times 10 } \left[ \because 1m=100cm \right] \)
\(=\frac { 968\times 62\times 10 }{ 22\times 10 } =\frac { 968\times 62 }{ 22 } =2728\)
Total cost = Rs. 2728 \(\times\) 25 = Rs. 68200

Given, length of field (1)= 34 m
and width (b)= 18 m
Perimeter of rectangular field = 2 (34 + 18) m
= 104 m ... (i)
Area of rectangular field = 34 \(\times\) 18 sq m
= 612 sq m ... (ii)
Cost of fencing of this rectangular field at Rs. 2.25 per m
=Rs.104\(\times\)2.25=Rs.234
Now, cost of cultivating the field at Rs.4.50 per sq m
= 612 \(\times\) 4.50=Rs. 2754

Given, length of a tile = 12 cm and breadth of a tile = 5 cm
∴ Area of one tile = Length x Breadth = 12 cm \(\times\) 5 cm =60 sq cm
(a) Here, length of the rectangular region = 100 cm
and breadth of the rectangular region = 144 cm
:. Area of the rectangular region = Length \(\times\) Breadth
=100 cm \(\times\) 144 cm
= 14400 sq cm
Now, number of required tiles
\(=\frac { Area\quad of\quad the\quad rectangular\quad region }{ Area\quad of\quad the\quad one\quad tile } \)
\(=\frac { 14400 }{ 60 } =240\)
Hence, the number of required tiles is 240.
(b) Given, length of the rectangular region = 70 cm
and breadth of the rectangular region = 36 cm
∴  Area of the rectangular region = Length \(\times\) Breadth
= 70 cm \(\times\) 36 cm
= 2520 sq cm
Now, number of the required tiles
\(=\frac { Area\quad of\quad the\quad rectangular\quad region }{ Area\quad of\quad the\quad one\quad tile } \)
\(=\frac { 2520 }{ 60 } =42\)
Hence, the number of the required tiles is 42.

Given, length of the piece of land = 5 m
and breadth of the piece of land = 4 m
Area of the piece of land = Length \(\times\) Breadth
= 5 m \(\times\) 4 m= 20 sq m
Given, side of one square flower bed = 1m
∴  Area of one square flower bed = Side \(\times\) Side
= (1m \(\times\) 1rn) = 1sq m
Then, area of 5 such flower beds = 5 \(\times\) Area of one square flower bed
= 5 \(\times\) 1sq m = 5 sq m
Now, area of the remaining part of land
= Area of the piece of land - Area of 5 square flower beds
= (20 - 5)sq m =15 sq m
Hence, the area of the remaining part of the land is 15 sq m.

(a) Here, length of the rectangle = 9 m
and breadth of the rectangle = 6 m
∴ Area of the rectangle = Length \(\times\) Breadth = 54 sq m
Hence, the area of the rectangle is 54 sq m.
(b) Here, length of the rectangle = 17 m
and breadth of the rectangle = 3 m
∴ Area of the rectangle = Length \(\times\) Breadth
= 17m \(\times\) 3m = 51 sq m
Hence, the area of the rectangle is 51 sq m.
(c) Here, length of the rectangle = 14 m
and breadth of the rectangle = 4 m
∴ Area of the rectangle = Length \(\times\) Breadth
= 14 m \(\times\) 4 m = 56 sq m
Hence, the area of the rectangle is 56 sq m.
Now, we have 56> 54> 51
Hence, the rectangle having sides 4 m and 14 m has the largest area and the rectangle having sides 17m and 3 m has the smallest area.

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.