This document provides an overview of mensuration topics including the formulas to calculate the area of rectangles, squares, parallelograms, rhombuses, trapezoids, and quadrilaterals. It includes the definitions of these shapes and derivations of the area formulas. Examples of application problems are shown with step-by-step solutions. The key areas and formulas covered are the area of a rectangle, square, parallelogram as base times height and the area of a rhombus, trapezoid and quadrilateral involving the sums of lengths and distances.

1. MATHEMATICS
Mensuration
Presented By O.Raghava

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

15. EXAMPLES PROBLEMS
Mensuration

16. SOLUTION
Mensuration

17. EXAMPLE NO 2
Mensuration

18. SOLUTION
Mensuration

19. EXAMPLE 3
Mensuration

20. SOLUTION
Mensuration

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