\(\therefore\) Perimeter of a triangle = Sum of lengths of all three sides
                                    = 3 + 5 + 7 = 15 cm
So, perimeter of the triangle will be 15 cm.

Given, length of rectangle = 6 cm
and breadth of rectangle = 4 cm
\(\therefore\)  Area of a rectangle = Length X Breadth = 6 x 4 = 24 cm.
From the given figure, it is clear that rectangle is divided into 4 congruent polygons.
\(\therefore\)  Area of each polygon = \(\frac{1}{4}\times\) Area of a rectangle
                                    = \(\frac{1}{4}\times\) 24 = 6 cm².

27 m

We need to find the perimeter of a rectangular flower bed, which gives the required length of the wire required to fence a rectangular flower bed.

Here, the circumference = 88 em.
Let radius of the circle be 'r',
\(\therefore\)         2\(\pi\)r=88.
or  2\(\times\)\(\frac{22}{7}\)\(\times\)r=88
or  r=\(\frac{88\times7}{2\times22}\)=14 cm
Now,   area of the circle =\(\pi\)r²=\(\frac{22}{7}\)\(\times\)14\(\times\)14 cm²=22\(\times\)2\(\times\)14 cm²
=616 cm²​​
Thus, the area of the circle is 616 cm².

Given, radius of a circular sheet (outer circle), R = 4 cm and radius of removed circular sheet (inner circle), r = 3 cm.
We know that, Area of a circle = \(\pi\) x (Radius)2
\(\therefore\) Area of circular sheet of radius 4 cm = \(\pi\) x (4)²
                                                         = 3.14 x 16 = 50.24 cm²
and area of circular sheet of radius 3 cm = \(\pi\) x (3)²
                                                           = 3.14 x 9 = 28.26 cm²
\(\therefore\) Area of remaining sheet= Area of circular sheet of radius 4 cm - Area of circular sheet of radius 3 cm
                                      = 50.24 - 28.26 = 21.98 cm²
Hence, the area of the remaining sheet is 21.98 cm².

There are two examples as given below:
Example 1. Let the length of a rectangle = 5 cm and breadth of a rectangle = 3 cm
\(\therefore\) Area of a rectangle = Length X Breadth
                                = 5 X 3=15 cm²
and perimeter of a rectangle = 2 (Length + Breadth)
                                         = 2 ( 5 + 3) = 2 X 8 = 16 cm
Again, let the length of a rectangle = 6 cm and breadth of a rectangle = 5 cm
\(\therefore\) Area of a rectangle = Length X Breadth = 6 X 5 = 30 cm² 
and perimeter of a rectangle = 2 (Length + Breadth)
                                         = 2 (6 + 5) = 2 X 11 = 22 cm
Thus, for both rectangles, area increases as the perimeter Increases.
Example 2. Let the side of a square = 4 ern
\(\therefore\) Perimeter of square = 4 X Side = 4 x 4 = 16 cm
and area of square = (Side X Side)
                           = 4 X 4 = 16 sq ern
Again, let the side of a square = 5 cm
\(\therefore\) Perimeter of square = 4 X Side = 4 x 5 = 20 cm
and area of square = Side X Side
                           = 5 x 5 = 25 sq cm
Thus, for both squares, area increases as the perimeter Increases.

(a) In the parallelogram ABCD, AD = BC, AB = DC
Base = 300 m and height = 80 m
\(\therefore\)  Area of a parallelogram = Base x Height
so, area of parallelogram = 300 m x 80 m
                                    = 24000 m²
(b) The values are depicted here is that owner of the plot is a very helpful cooperative to the society.

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.

Perimeter = \(2\times 7+\frac { 22 }{ 7 } \times 7\)
= 14 + 22 = 36 cm

Side = \(\frac { 48 }{ 4 } \)= 12 cm
Area = 12 \(\times\)12 = 144 cm²

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