Class-VII Maths

PERIMETER AND AREA

PRESENTED BY

AMIT RANJAN

TGT(Maths),

JNV Cooch-Behar, W.B.

INTRODUCTION

AREA AND PERIMETER OF A SQUARE

Square is a quadrilateral, with four equal sides.

• Area = Side × Side = (Side)²
• Perimeter = 4 × Side

​

Example

Find the area and perimeter of a square-shaped cardboard whose length is 5 cm.

Solution

Area of square = (side)² = (5)² = 25 cm²

Perimeter of square = 4 × side = 4 × 5 = 20 cm

​

AREA AND PERIMETER OF A RECTANGLE

Example

What is the length of a rectangular field if its width is 20 ft and Area is 500 ft²?

​

Solution 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)

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 and equal sides.
• Also denoted as ∥ gm

​

Area of parallelogram = base × height Or b × h 

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 following figure:

​

Solution Base of ∥ gm = 8 cm

Height of ∥ gm = 6 cm

Area of ∥ gm = b × h

= 8 x 6 = 48 cm²

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 following figure :

​

Solution Area of triangle = 1/2 (base × height)

= 1/2 (12 × 5)

= 1/2 × 60

= 30 cm²

CIRCLE

• It is a round, closed shape.
• Radius: A straight line joining centre and other point on Circumference 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

CIRCUMFERENCE AND AREA �OF A CIRCLE

Example What is the circumference of a circle of diameter 12 cm? (Take π = 3.14)

​

Solution C = π d

C = 3.14 × 12 = 37.68 cm

Note: diameter (d) = twice the radius (r) i.e.  d = 2r

Example Find the area of a circle of radius 23 cm (use π = 3.14).

​

Solution Area of circle = πr² (r = 23 cm)

= 3.14 × 23²

= 1661.06 cm²

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

Conversion of Units

Example: Convert 3.5 ha in m²

Applications�

• 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.

Solution Area of path = Area of rectangle ABCD – Area of rectangle STUV

AB = 35 + 1.5 + 1.5 = 38 m AD = 20 + 1.5 + 1.5 = 23 m

Area of ABCD = 38 × 23 = 874 m² Area of STUV = 35 × 20 = 700 m²

Area of path = Area of rectangle ABCD – Area of rectangle STUV

= 874 – 700 = 174 m²

Thank you

Amit Ranjan

(TGT-Maths)

JNV, COOCH-BEHAR, W.B.

​