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MATHEMATICS
Mensuration
Presented By O.Raghava
Mensuration
• Area Of Rectangle And Square
• Area of Quadrilateral
Contents
Mensuration
•What Is Meant By Area
 A Figure Made up Of Straight Line Segment Called A Rectilinear Figure
The Surface Of the Boundary Is Called Area
AREA OF RECTANGLE
• Rectangle
Area=l x b
Perimeter = 2(l+b)
Mensuration
Where As l=Length
b=Breadth
N l O
b
M P
d
AREA OF SQUARE
• Area Of Square =S2
• Perimeter Of Square = 4 x S
Mensuration
Here S is the Side of the square
S
S
EXAMPLES
• Ex.1 Show That Area Of a Square =
1
2
x Diagnoal 2
.find the Area Of Square Whose Diagonol Is = 2.5
cm
• In Right Angle Triangle BCD
• Diagonal2 =DC2 + CB2 = S2+S2=2S2
• But The Area Of Square = S2
• Diagonal2=2 x Area
• Area = =
1
2
x Diagnoal 2
Mensuration
A S B
S
C
D
• If Diagonal = 2.5
• Area = =
1
2
x 2.5 2
• =
1
2
x 6.25
• Ans- 3.125
Mensuration
 If we take 2 to left side we get ,
 Area =
1
2
x Diagnoal 2
2nd Case
• Ex-2 the Length And Breadth Of a rectangular Field Is in the
ratio of 4:3 if the area is 3072 m2 find the cost of fencing
the field at the rate of 4 rupees Per meter
• Let the Length And Breadth Be 4x and 3x
• Area Of field = l x b
• 4𝑥 𝑋 3𝑥 = 12𝑥 2 = 3072
• Hence x = 16
Mensuration
• Length = 4x= 64m , breadth = 3x = 48m
• Length of fencing = Perimeter of the field
• 2(64+48)
• =224m
• Cost Of Fencing = 4(224)m
• = R896
Mensuration
Mensuration
AREA OF PARALLELOGRAM
A B
D E C
Consider Parallelogram ABCD.
Let AC Be the Diagonal
In ADC And CBA
AD=CB, CD=AB
AC is the common
 ADC = CBA
 Area Of Parallelogram ABCD
 =Area of ADC + Area Of ABC]
 = 2 x Area Of ADC
 =2 x (1/2 CD x AE) (Where AE|DC)
 = DC x AE
 i.e. Area Of Parallelogram = Base x Height
AREA OF RHOMUS
• Since rhombus is a parallelogram its area is given by
• Area Of rhombus = Base x Height
• Area Of Rhombus ABCD = Area Of ABD +Area of CBD
• => ½(BD x AO) + ½(BD x CO)
• => ½(BD(AO+CO))
• => ½ BD+AC
• i.e. Area Of Rhombus = ½ x diagonals = ½ (d1 x d2)
Mensuration
O
A
B
D C
AREA OF TRAPEZIUM
• Let ABCD be a Trapezium With AB || DC.
Draw AE and BF Perpendicular to DC
• Then AE=BF= height of trapezium =
h
• Area of trapezium ABCD = Area of ADE +
Area of rectangle ABFE + Area of BCF
Mensuration
A B
D E F
C
• ½ x DE x h + EF x h + ½ FE x h
• ½ h(DE+2EF + FC )
• ½ h(DE+EF+FC+EF)
• ½ h (DC+EF)
• ½ h(DC+AB) ( as EF = AB)
• i.e. Area of trapezium = ½ x sum of the parallel lines x Distance
Between Parallel Sides
Mensuration
A B
D E F
C
AREA OF
QUADRILATERAL
• Let ABCD be a Quadrilateral and AC be one of
its diagonal. Draw Perpendicular BE and DF
From B and D respectively to AC
• Area of Quadrilateral ABCD
Area Of ABC + Area of ADC
 ½ AC x BE + ½ AC x DF
 ½ AC(BE +DF )
If AC = d , BE = h1 , DF = h2 then
Area Of Quadrilateral = ½ d(h1+h2)
Mensuration
D
F
E
EXAMPLES PROBLEMS
Mensuration
SOLUTION
Mensuration
EXAMPLE NO 2
Mensuration
SOLUTION
Mensuration
EXAMPLE 3
Mensuration
SOLUTION
Mensuration
A SPECIAL THANKS TO MY
TEACHERS
Mensuration
Things I used in this ppt
1. An Olympiad book for questions
2. PPT Presentation
3. Reference books and my Brain for Matter
4. 3d Painter For Shapes

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MATHEMATICS MENSURATION

  • 2. Mensuration • Area Of Rectangle And Square • Area of Quadrilateral Contents
  • 3. Mensuration •What Is Meant By Area  A Figure Made up Of Straight Line Segment Called A Rectilinear Figure The Surface Of the Boundary Is Called Area
  • 4. AREA OF RECTANGLE • Rectangle Area=l x b Perimeter = 2(l+b) Mensuration Where As l=Length b=Breadth N l O b M P d
  • 5. AREA OF SQUARE • Area Of Square =S2 • Perimeter Of Square = 4 x S Mensuration Here S is the Side of the square S S
  • 6. EXAMPLES • Ex.1 Show That Area Of a Square = 1 2 x Diagnoal 2 .find the Area Of Square Whose Diagonol Is = 2.5 cm • In Right Angle Triangle BCD • Diagonal2 =DC2 + CB2 = S2+S2=2S2 • But The Area Of Square = S2 • Diagonal2=2 x Area • Area = = 1 2 x Diagnoal 2 Mensuration A S B S C D
  • 7. • If Diagonal = 2.5 • Area = = 1 2 x 2.5 2 • = 1 2 x 6.25 • Ans- 3.125 Mensuration  If we take 2 to left side we get ,  Area = 1 2 x Diagnoal 2 2nd Case
  • 8. • Ex-2 the Length And Breadth Of a rectangular Field Is in the ratio of 4:3 if the area is 3072 m2 find the cost of fencing the field at the rate of 4 rupees Per meter • Let the Length And Breadth Be 4x and 3x • Area Of field = l x b • 4𝑥 𝑋 3𝑥 = 12𝑥 2 = 3072 • Hence x = 16 Mensuration
  • 9. • Length = 4x= 64m , breadth = 3x = 48m • Length of fencing = Perimeter of the field • 2(64+48) • =224m • Cost Of Fencing = 4(224)m • = R896 Mensuration
  • 10. Mensuration AREA OF PARALLELOGRAM A B D E C Consider Parallelogram ABCD. Let AC Be the Diagonal In ADC And CBA AD=CB, CD=AB AC is the common  ADC = CBA  Area Of Parallelogram ABCD  =Area of ADC + Area Of ABC]  = 2 x Area Of ADC  =2 x (1/2 CD x AE) (Where AE|DC)  = DC x AE  i.e. Area Of Parallelogram = Base x Height
  • 11. AREA OF RHOMUS • Since rhombus is a parallelogram its area is given by • Area Of rhombus = Base x Height • Area Of Rhombus ABCD = Area Of ABD +Area of CBD • => ½(BD x AO) + ½(BD x CO) • => ½(BD(AO+CO)) • => ½ BD+AC • i.e. Area Of Rhombus = ½ x diagonals = ½ (d1 x d2) Mensuration O A B D C
  • 12. AREA OF TRAPEZIUM • Let ABCD be a Trapezium With AB || DC. Draw AE and BF Perpendicular to DC • Then AE=BF= height of trapezium = h • Area of trapezium ABCD = Area of ADE + Area of rectangle ABFE + Area of BCF Mensuration A B D E F C
  • 13. • ½ x DE x h + EF x h + ½ FE x h • ½ h(DE+2EF + FC ) • ½ h(DE+EF+FC+EF) • ½ h (DC+EF) • ½ h(DC+AB) ( as EF = AB) • i.e. Area of trapezium = ½ x sum of the parallel lines x Distance Between Parallel Sides Mensuration A B D E F C
  • 14. AREA OF QUADRILATERAL • Let ABCD be a Quadrilateral and AC be one of its diagonal. Draw Perpendicular BE and DF From B and D respectively to AC • Area of Quadrilateral ABCD Area Of ABC + Area of ADC  ½ AC x BE + ½ AC x DF  ½ AC(BE +DF ) If AC = d , BE = h1 , DF = h2 then Area Of Quadrilateral = ½ d(h1+h2) Mensuration D F E
  • 21. A SPECIAL THANKS TO MY TEACHERS Mensuration Things I used in this ppt 1. An Olympiad book for questions 2. PPT Presentation 3. Reference books and my Brain for Matter 4. 3d Painter For Shapes