4

18 cm²

10 cm

Given, each side of square is of unit length. Figure contains length of 10 squares and width of 6 squares.
Now, length of rectangle, AD = (BC)
= Sum of length of a side of 10 squares
=1+1+1+1+1+1+1+1+1+1
=10\(\times\)1 =10 units
and breadth of rectangle, AB = (DC)
= Width of 6 squares = 6 \(\times\)1 = 6 units
(a) The perimeter of the rectangle ABCD
= AB+ BC+ CD+DA
= 6+10+ 6+10
= 32 units
(b) The area of the rectangle ABCD = Length \(\times\)Breadth
= AD\(\times\)AB=10\(\times\)6
= 60 units
(c) The total area of rectangle = 60 units

Now, we have to divide the rectangle into 10 equal parts i..e \(\frac{60}{10}=6\) square uni.t.s I.e. we have to take a group of 6-6 square blocks, which is shown in the figure
(d) Now, we find the perimeter of part l.

We know that perimeter of a figure is the total length of its boundary.
∴ Perimeter of part I
= 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1= 12 units
Similarly, we can find the perimeters of remaining 9 parts, all the parts have same perimeter i.e. 12 units.
Yes, all the parts have same perimeter.

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, size of lawn in front of Molly's house
=12m\(\times\)8m
Perimeter = 2 (12 + 8) m= 40 m ... (i)
Now, size of lawn in front of Dolly's house
=15m\(\times\)5m
Perimeter = 2 (15+ 5) m = 40 m ... (ii)
From Eqs. (i) and (ii), we get = 40+ 40= 80 m
Hence, total length of bamboo fencing is 80 m.

Given, length of rectangular field (I) = 250 m and width is 150 m

Perimeter of this field = 2 (I + b) = 2 (250 + 150) m
= 2 \(\times\) 400 m = 800 m
Distance covered in one round = Perimeter = 800 m
Distance covered in three rounds
= 3\(\times\) 800= 2400 m
Now, number of rounds to cover 4 km, i.e. 400 m.
=\(\frac{4000}{800}=5\) [∵1 km= 1000 m]
Hence, she should run 5 times around the field to cover the distance of 4 km

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.