\(\because\) Side of the square = 40 cm.
\(\therefore\)Area of the square = (Side)²= (40 cm)²
= 40 cm x 40 cm = 1600 cm²
Since [Area of the rectangle] = [Area of the square]
\(\therefore\)Area of the rectangle = 1600 cm²
or  I x b = 1600 \(\Rightarrow\) l x 20 = 1600
or   l=\(\frac{1600}{20}\)=80 cm
\(\therefore\)Length (l) of the rectangle = 80 cm
Now, perimeter of the rectangle = 2(l + b) = 2(80 + 20) cm
= 2 x 100 cm = 200 cm.

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.

Area of one tile (parallelogram shape)
= base x height
= 3 x 4.5
= 13.5 cm²
Area of such 2760 tiles = 2760 x 13.5
= 37,260 cm²
= 3.726 m²
Cost of polishing = 3.726 x 20
= Rs. 74.52.

Area enclosed the copper wire
In square shape = (side)²
\(\therefore\)(side)² = 121 cm²
\(\Rightarrow\) side = \(\sqrt { 121 } \) = 11 cm.
Hence length of wire = 11 x 4
= 44 cm
Now this length = Circumference of the circle
\(\Rightarrow\) 2\(\pi\)r = 44
\(\Rightarrow\) 2 x \(\frac { 22 }{ 7 } \) x r = 44
\(\Rightarrow\) r = \(\frac { 44 }{ 2\times 22 } \times 7\)
Thus, r = 7 cm
Hence, area enclosed by the wire when it is bent in circular shape
= \(\pi\)r²
= \(\frac { 22 }{ 7 } \)x (7)²
= \(\frac { 22 }{ 7 } \)x 7 x 7
= 154 m²

(a) Given, length of the side of the square platform
                                       = 95.40 m
\(\therefore\) Area of square platform = Length x Length
                                      = 95.40 x 95.40 =  9101.16 sq m
Hence, the area of square platform is 9101.16 sq m.
(b) Given, floor area of the main building'
                                      = 3214 m²
Now, area of the part of the platform that is not covered by the main building
                                     = Area of square platform - Floor area of the main building
                                     = 9101.16 - 3214 = 5887.16 sq.

\(\therefore\)  Area of a triangle = \(\frac{1}{2}\) x Base x Height
\(\therefore\)  Area of \(\Delta\)MNO = \(\frac{1}{2}\) x NO x MO
                            = \(\frac{1}{2}\) x 4 x 5 = 10 cm²
 Now, Area of \(\Delta\)MPO = \(\frac{1}{2}\) x OP x MO
                                = \(\frac{1}{2}\) x 2 x 5 = 5 cm²
and Area of \(\Delta\)MPQ = \(\frac{1}{2}\) x PQ x MO = \(\frac{1}{2}\) x 6 x 5
                             = 15 cm²
\(\therefore\)  Ratio of areas of the triangles = \(\Delta\)MNO:  \(\Delta\)MPO:  \(\Delta\)MPQ
                                                = 10 cm²: 5 cm²: 15 cm²
                                                = 2: 1: 3.

A wall of a room is of dimensions 5 m x 4 m.
Length of the room = 5 m
Breadth of the room = 4 m
\(\therefore\) Area of the room = 5 x 4= 20 m²
Length of the window = 1.5 m
Breadth of the window = 1 m
\(\therefore\) Area of the window = 1.5 x 1 = 1.5 m²
Length of the door = 2.25 m
Breadth of the door = 1 m
\(\therefore\) Area of the door = 2.25 x 1 = 2.25 m²
The area of the wall to be painted = Area of the room - Area of the window - Area of the door
                                                 = 20 - 1.5 - 2.25
                                                 = 20 - 3.75 = 16.25 m².

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 paper = 324 cm
Breadth of paper = 172 cm
Total, area of paper for envelopes = 324x 172 cm²
If each envelope requires a piece of paper size 18 cm by 12 cm.
So, length of each envelope = 18 cm and breadth of each envelope = 12 cm
Area of paper required for each envelope
= 18 x 12 cm².
\(\therefore\) Number of envelopes = \(\frac{Total\ areas\ of\ paper}{Area\ of\ paper\ required\ for\ one\ envelope}\)
= \(\frac{324\times 172}{18\times 12}\) = 258.

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.