Perimeter and Area Class 7 Notes: Chapter 11

Perimeter

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  • Perimeter is the total length or total distance covered along the boundary of a closed shape.
    Perimeter of a Quadrilateral    
    The perimeter of a Quadrilateral    

Area

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

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The perimeter of Square and Rectangle

  • Perimeter of a square = a + a + a + a = 4a, where a is the length of each side.
The perimeter of Square and Rectangle
Square with side length ‘a’ units
  • 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
Rectangle with length ‘l’ units and breadth ‘b’ units

Area of Square & Rectangle

Area of square = 4a2

Here a is the length of each side

Square with the length of each side 'a' units
Square with the length of each side ‘a’ units

 

Area of rectangle = Length(l) √ó Breadth(b) = l√ób

 

Area of rectangle = Length(l) √ó Breadth(b) = l√ób
Rectangle with length ‘a’ units and breadth ‘b’ units

Area of a Parallelogram

Area of a Parallelogram

  • Area of parallelogram ABCD =¬†(base√óheight)

Area of parallelogram ABCD  = (b×h)

Triangle as Part of Rectangle

  • The rectangle can be considered¬†as a combination¬†of two congruent triangles.
  • Consider a rectangle ABCD, it is divided into 2 triangles ACD¬†and ABD.
    Triangle as Part of Rectangle
    Triangles as parts of Rectangle
  • Area of each triangle = 12 (Area of the rectangle).
    =  12(length×breadth)
    =  12(10cm×5cm)
    =  25cm2

Area of a Triangle

  • Consider a parallelogram ABCD.
  • Draw a diagonal BD to divide the parallelogram into two congruent triangles.
Area of a Triangle
Area of Triangle = 1/2 (base×height)
  • Area of triangle ABD = 1/2¬†(Area of parallelogram ABCD)

Area of triangle ABD  = 1/2 (b×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.

Circle

  • 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.

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Circumference of a Circle

  • The circumference of a circle ( C )¬† is the total path or total distance covered by the circle. It is also called a perimeter of the circle.

Circumference¬†of a circle = 2√óŌÄ√ór,

where r is the radius of the circle.

Visualizing Area of a Circle

Area of Circle

  • Area of a¬†circle is the total region enclosed by the circle.

Area of a circle = ŌÄ√ór2, where r is the radius of the circle.

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Introduction and Value of Pi

  • Pi (ŌÄ)¬†¬†is the constant which is defined as the ratio of a circle‚Äôs circumference¬†(2ŌÄr)¬†to its¬†diameter(2r).

ŌÄ= Circumference (2ŌÄr)/Diameter (2r)

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

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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 √ó 10 = 40m
    Cost of fencing 1m = Rs 10
    Cost of fencing the land = 40 m × 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√ó4m.
    Area of the wall = 5m√ó4m=20m2
    Cost of painting 1m2 of area is Rs 20.
    Cost of painting the wall =20m2×Rs 20=Rs 400

Area of Mixed Shapes

  • Find the area of¬† the shaded portion using the given information.
Area of Mixed Shapes
Area of the shaded portion

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√ób)¬†‚ąí (ŌÄr2/2)
=¬†30√ó10¬†‚ąí (ŌÄ√ó52/2)
= 300¬†‚ąí¬†(ŌÄ√ó25/2)
= (600 – 25ŌÄ)/2
= (600 – 78.5)/2

= 260.7 cm2

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