Perimeter and Area Class 7 Notes: Chapter 11

Perimeter

• Perimeter is the total length or total distance covered along the boundary of a closed shape

The perimeter of a Quadrilateral 

Area

• The area is the total amount of surface enclosed by a closed figure.

Areas of a closed figure

The perimeter of Square and Rectangle

• Perimeter of a square = a + a + a + a = 4a, where a is the length of each side.

• Perimeter of a rectangle = l + l + b + b = 2(l + b), where l and b are length and breadth, respectively.

Rectangle with length ‘l’ units and breadth ‘b’ units

• Area of each triangle = $\frac{1}{2}$ (Area of the rectangle).
  =  $\frac{1}{2}(length\times breadth)$
  =  $\frac{1}{2}(10cm\times 5cm)$
  =  $25c{m}^2$

Area of a Triangle

• Consider a parallelogram ABCD.
• Draw a diagonal BD to divide the parallelogram into two congruent triangles.

• Area of triangle ABD = 1/2 (Area of parallelogram ABCD)

Area of triangle ABD  = 1/2 $(b\times h)$

Conversion of Units

• Kilometres, metres, centimetres, millimetres are units of length.
• 10 millimetres = 1 centimetre
• 100 centimetres = 1 metre
• 1000 metres = 1 kilometre

Life of Pi

Terms Related to Circle

• A circle is a simple closed curve which is not a polygon.
• A circle is a collection of points which are equidistant from a fixed point.

• The fixed point in the middle is called the centre.
• The fixed distance is known as radius.
• The perimeter of a circle is also called as the circumference of the circle.

Introduction and Value of Pi

• Pi $\left(\pi \right)$  is the constant which is defined as the ratio of a circle’s circumference $\left(2\pi r\right)$ to its diameter(2r).

$\pi =\; Circumference\; (2\pi r)/Diameter\; (2r)$

• The value of pi is approximately equal to 3.14159 or 22/7.

Problem Solving

Cost of Framing, Fencing

• Cost of framing or fencing a land is calculated by finding its perimeter.
• Example: A square-shaped land has length of its side 10m.
  Perimeter of the land = 4 $\times$ 10 = 40m
  Cost of fencing 1m = Rs 10
  Cost of fencing the land = 40 m $\times$ Rs 10 = Rs 400

Cost of Painting, Laminating

• Cost of painting a surface depends on the area of the surface.
• Example: A wall has dimensions $5m\times 4m$.
  Area of the wall = $5m\times 4m=20m^2$
  Cost of painting $1m^2$ of area is $Rs20$.
  Cost of painting the wall $=20m^2\times Rs20=Rs400$

Area of Mixed Shapes

• Find the area of  the shaded portion using the given information.

Solution: Diameter of the semicircle = 10cm
Radius of semicircle = 5cm
Area of the shaded portion = Area of rectangle ABCD – Area of semicircle
Area of the shaded portion  = ($l\times \mathrm{b)}-\; (\pi r2/2)$
= $\mathrm{30\times 10}-\; (\pi \times 52/2)$
= $300- (\pi \times 25/2)=\; (600\; –\; 25\pi )/2=\; (600\; –\; 78.5)/2$
= $260.7c{m}^2$