Base = 8 cm
Height = 2.5 cm 
∴ Area of the parallelogram
=base \(\times\)height
=8\(\times\)2.5 = 20 cm'²

(ii) Here, Base (b) = 5 em and Height (h) = 4 em
\(\therefore\)Area of the triangle PQR=\(\frac{1}{2}\)\(\times\)b\(\times\)h
=​​​​​\(\frac{1}{2}\)\(\times\)5\(\times\)4 cm² = 10 cm²

20 cm, 94 cm.

We need to find the area of rectangular swimming pool to cover it with plastic sheer.

Given, length = 15 cm and breadth = 8 cm
\(\therefore\) Area of a rectangle = Length x Breadth = 15 x 8 = 120 cm²
Hence, area of rectangle is 120 cm².

Firstly, on counting the squares enclosed within the figures of the parallelogram, we find in each case these are 15 in numbers. So, the area of each parallelogram = 15 sq units, base = 5 units and height = 3 units in each case. Now, draw a perpendicular DM to the base AB or produced AB (if necessary) in each figure as shown below:

For figure (b), AD = \(\sqrt{MD^2+AM^2}\)
                         = \(\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}\) units
\(\therefore\) Perimeter = \(2(\sqrt{13}+5)\) units.
Hence, the required complete table is as follows:

Given, length of a rectangular piece of land, I = 500 m and breadth of a rectangular piece of land, b = 300 m
Cost of 1m² land = Rs 10000
\(\therefore\) Cost of 150000 m² land = Rs 10000 x 150000
                                       = Rs 1500000000
Hence, the cost of 150000 m² land at the rate of 10000 per m² is Rs 1500000000.

Area of a tile = 25 x 20 cm² = 500 cm²
Area of 2400 tiles = 2400 x 500 cm²
= 1200000 cm²
= \(\frac { 1200000 }{ 10000 } \)m²
 [\(\therefore\) 10000 cm² = 1 m²]
= 120 m²
Let the height of the wall be h metres then,
Area of the wall = 4h m²
Since 2400 tiles completely cover the wall.
\(\therefore\) Area of the wall = Area of 2400 tiles
\(\Rightarrow\) 4h = 120
\(\Rightarrow\) \(\frac { 4h }{ 4 } \)= \(\frac { 120 }{ 4 } \)
\(\Rightarrow\) h = 30
[Dividing both sides by 4]
Hence the height of the wall is 30 metre.

2346 m²

Given, ABCD is a square and AB = 15 cm
\(\therefore\) Diagonal of square ABCD = \(\sqrt{2}\) a
                                         = \(\sqrt{2}\) x 15 = 15\(\sqrt{2}\) cm
\(\therefore\) Area of the square BDFE = (Side)² = (15\(\sqrt{2}\))²
                                        = 15 x 15 x \(\sqrt{2}\) x \(\sqrt{2}\) = 225 x 2 = 450 cm².

A rectangular lawn of size = 72 m x 18 m
Length of rectangular lawn = 72 m
Breadth of rectangular lawn = 18 m
Perimeter of rectangle = 2 x (Length + Breadth)
\(\therefore\) Perimeter of rectangular lawn
            = 2(72 + 18) = 2(90) = 180 m
If 3 shrubs can be planted in a metre of hedge.
Then, number of shrubs = 3 x Perimeter of rectangular lawn
            = 3 x 180 = 540.

Length of rectangular metal plate = 7 cm
Breadth of rectangular metal plate = 5 cm
\(\therefore\) Area of a rectangle = Length x Breadth
                               = 7 x 5 = 35 cm²
\(\therefore\) Rate of the plate = Rs 75 per sq cm.
Then, total cost = 35 x 75
                       = Rs 2625.

Area of the squares MNOP= 144 cm² [given]
Since number of the triangles in square is 8.
So, area of each triangle will be \(\frac{144}{8}\) = 18 cm².

ABCD is the rhombus such that its diagonals AC and BD intersect at O. Here BD = 12 cm.

Suppose AC = x cm.
Since, the area of a rhombus =\(\frac{1}{2}\)\(\times\)[Product of 2 its diagonals]
\(\therefore\)Area of the rhombus ABCD =\(\frac{1}{2}\)\(\times\) 12 \(\times\) x cm²
2 But the area of the rhombus ABCD = 96 cm²
\(\Rightarrow\)\(\frac{1}{2}\)\(\times\)12 \(\times\) x = 96
\(\Rightarrow\)x=\(\frac{96\times2}{12}\)cm \(\Rightarrow\)x=8\(\times\)2cm
\(\Rightarrow\) x=16 cm
Since, the diagonals of a rhombus, bisect each other at right angles.
\(\therefore\)\(\angle \)COD = 90°, OC =\(\frac{1}{2}\)\(\times\) 16 cm=8 cm
and Od=\(\frac{1}{2}\)\(\times\) 12 cm =6 cm
Now, in right \(\Delta\)COD, we have
\(\Rightarrow\)OD² + OC² = CD² \(\Rightarrow\) 6² + 8² = CD²
\(\Rightarrow\)36 + 64 = CD² \(\Rightarrow\)100 = CD²
\(\Rightarrow\) 10² = CD² \(\Rightarrow\) CD = 10 cm.
Since, all the sides of rhombus are equal and perimeter of a rhombus = 4 x Side.
\(\therefore\) Perimeter ofthe rhombus = 4 x 10 em = 40 cm

Let ABCD be the rectangular park of sides 40 m and 25 m, and the shaded region represents the path 2.5 m wide.
Now, PQ = (40 + 2.5 + 2.5) m = 45 m
PS = (25 + 2.5 + 2.5) m = 30 m
\(\therefore\)Area of rectangle ABCD = I x b = 40 m x 25 m
= 1000 m2
Area of rectangle PQRS = 45 m x 30 m = 1350 m²
So, Area of the path = [Area of rectangle PQRS] - [Area of rectangle ABCD]
= 1350 m² - 1000 m² = 350 m²