Perimeter and Area Class 7 Notes Maths Chapter 9

It refers to the length of the outline of the enclosed figure.

Area

It refers to the surface of the enclosed figure.

Area and Perimeter of Square

Square is a quadrilateral, with four equal sides.
Area = Side × Side
Perimeter = 4 × Side
Example: Find the area and perimeter of a square-shaped cardboard whose length is 5 cm.

Sol: Area of square = (side)²
= (5)²
= 25 cm²
Perimeter of square = 4 × side
= 4 × 5
= 20 cm

Area and Perimeter of Rectangle

The rectangle is a quadrilateral, with equal opposite sides.
Area = Length × Breadth
Perimeter = 2(Length + Breadth)

Example: What is the length of a rectangular field if its width is 20 ft and Area is 500 ft²?
Sol: Area of rectangular field = length × width
500 = l × 20
l = 500/20
l = 25 ft

Triangles as Parts of Rectangles

If we draw a diagonal of a rectangle then we get two equal sizes of triangles. So the area of these triangles will be half of the area of a rectangle.
The area of each triangle = 1/2 (Area of the rectangle)
Likewise, if we draw two diagonals of a square then we get four equal sizes of triangles .so the area of each triangle will be one-fourth of the area of the square.
The area of each triangle = 1/4 (Area of the square)
Example: What will be the area of each triangle if we draw two diagonals of a square with side 7 cm?
Sol: Area of square = 7 × 7
= 49 cm²
The area of each triangle = 1/4 (Area of the square)
= 1/4 × 49
= 12.25 cm²

Congruent Parts of Rectangles

Two parts of a rectangle are congruent to each other if the area of the first part is equal to the area of the second part.
Example
The area of each congruent part = 1/2 (Area of the rectangle)
= 1/2 (l × b) cm²
=1/2 (4 × 3) cm²
= 1/2 (12) cm²
= 6 cm²

Parallelogram

It is a simple quadrilateral with two pairs of parallel sides. Also denoted as ∥ gm
Area of parallelogram = base × height
 Or b × h (bh)
We can take any of the sides as the base of the parallelogram. And the perpendicular drawn on that side from the opposite vertex is the height of the parallelogram.
Example: Find the area of the figure given below:

Sol: Base of ∥ gm = 8 cm
Height of ∥ gm = 6 cm
Area of ∥ gm = b × h
= 8 × 6
= 48 cm

Area of Triangle

Triangle is a three-sided closed polygon. If we join two congruent triangles together then we get a parallelogram. So the area of the triangle will be half of the area of the parallelogram.
Area of Triangle = 1/2 (Area of ∥ gm)
= 1/2 (base × height)
Example: Find the area of the figure given below:

Sol: Area of triangle = 1/2 (base × height)
= 1/2 (12 × 5)
= 1/2 × 60
= 30 cm²

Circles

It is a round, closed shape.

The circumference of a Circle
The circumference of a circle refers to the distance around the circle.

• Radius: A straight line from the Circumference till the centre of the circle.
• Diameter: It refers to the line from one point of the Circumference to the other point of the Circumference.
• π (pi): It refers to the ratio of a circle's circumference to its diameter.

Circumference(c) = π × diameter
C = πd
= π × 2r
d = 2r

Example: What is the circumference of a circle of diameter 12 cm (Take π = 3.14)?
Sol: C = πd
C = 3.14 × 12
= 37.68 cm

Area of Circle

Area of the circle = (Half of the circumference) × radius
= πr²

Example: Find the area of a circle of radius 23 cm (use π = 3.14).
Sol: R = 23 cm
π = 3.14
Area of circle = 3.14 × 23²
= 1,661 cm²

Conversion of Units

Sometimes we need to convert the unit of the given measurements to make it similar to the other given units.

Conversion of Units

Example 1: Convert 70 cm² in mm²
Sol:
1 cm = 10 mm
1 cm² = 10 × 10
1 cm² = 100 mm²
70 cm² = 700 mm²

Example: 2 Convert 3.5 ha in m²
Sol: 1 ha = 10000 m²
3.5 ha = 10000 × 3.5
ha = 35000 m²

Applications

We can use these concepts of area and perimeter of plane figures in our day to day life.

• If we have a rectangular field and want to calculate that how long will be the length of the fence required to cover that field, then we will use the perimeter.
• If a child has to decorate a circular card with the lace then he can calculate the length of the lace required by calculating the circumference of the card, etc.

Example: A rectangular park is 35 m long and 20 m wide. A path 1.5 m wide is constructed outside the park. Find the area of the path.
Sol: Area of rectangle ABCD – Area of rectangle STUV
AB = 35 + 2.5 + 2.5
= 40 m
AD = 20 + 2.5 + 2.5
= 25 m
Area of ABCD = 40 × 25
= 1000 m²
Area of STUV = 35 × 20
= 700 m²
Area of path = Area of rectangle ABCD – Area of rectangle STUV
= 1000 – 700
= 300 m²