Let width of rectangle (b)= x cm
Then, length of rectangle (I) = 3x cm
∴ Perimeter = 2 (l + b)
⇒ 40 =2 (3x + x)
⇒ 8x =40
⇒ x=\(\frac{40}{8}\)=5cm
Hence, length is 15 cm and width is 5 cm.

Given, figure is covered by 5 full squares.
∴ Area = (5 \(\times\) 1)sq units = 5 sq units

Pentagon

The perimeter of the given figure is 60 cm

Perimeter of a squared park = 4 \(\times\) Length of a side = 4 \(\times\) 60 = 240 m
Distance covered in one round = 240 m
Distance covered in three rounds = 3 \(\times\) 240 m = 720 m

Given. sides of three squares are 4 cm, 10 cm and 3 cm, respectively.
Total perimeter of given squares
= Sum of all outer sides of the figure
=4+ 4+ 4+ 6+10+ 1+ 3+ 3+ 3+10
=54cm

Area of room = 1.4 \(\times\) 9.5 m² = 1.4\(\times\) 9.5\(\times\) 10000 cm²
Area of one tile = 20\(\times\) 10= 200 cm²

Number of tiles\(=\frac { Area\quad of\quad the\quad room }{ Area\quad of\quad tile } =\frac { 74\times 95\times 100 }{ 200 } \)
\(=\frac { 74\times 95 }{ 2 } =95\times 37=3515\)

24 sq m

Let ABC be the given triangle and its sides AB = 12 cm and BC=14cm.
Also, perimeter of triangle = 36 cm

We know that,
Perimeter of a triangle = Sum of all its sides
⇒ AB + BC + CA = 36cm
⇒ 12 cm + 14 cm + CA = 36 cm
⇒ (12+14)cm+CA=36cm
⇒ 26 cm+ CA = 36 cm
∴ CA = (36 - 26) cm = 10 cm
Hence, the third side of the triangle is 10 cm.

Here, AB and CD are equal and DA and BC are equal.
On measuring, we get
AB = CD = 60 cm and DA = BC = 30 cm
Now, the sum of the lengths of the four sides
= AB + BC + CD + DA
=60 cm + 30 cm+ 60 cm + 30 cm
= (60 + 30 + 60 + 30) cm = 180 cm
∴ Perimeter of the table =Sum of the lengths of the four sides
= 180 cm
Hence, the perimeter of the study table is 180 cm.

16 cm²

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

(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.

Area = 3 x -\(\frac{1}{2}\)+1x2+3x\(\frac{1}{2}\)=5

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

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