1. Dr.Karmveer Bhaurao Patil
OUR INSPIRATION
Dr.Karmveer Bhaurao Patil

2. 9.MENSURATION Std.9th Sub- MATHEMATICS 9.Mensuration
Sadavarte M.D
9.Mensuration

3. MENSURATION Std-9th Sub -Geometry 9.Mensuration INTRODUCTION
9.1) AREA OF TRIANGLE
9.2) AREA & PERIMETER OF A QUADRILAERAL
9.3)AREA & PERIMETER OF THE CIRCLE & SEMICIRCLE
9.4) AREA OF REGULAR POLYGON
9.5) AREA OF n-SIDED REGULAR POLYGON
9.6) TO FIND THE AREA OF THE REGULAR POLYGON IN
TERMS OF RADIUS OF CIRCUMSCRIBED CIRCLE
9.7) PARTICULAR CASES
1. REGULAR HEXAGON
2. REGULAR OCTAGON
Std-9th
Sub -Geometry
9.Mensuration

4. Std-9th 9.Mensuration Sub -Geometry CONCEPT OF AREA.
Fig.1
Fig.2
Observe the
adjoining figure.
Triangular region
Triangle
Fig .1 represent a region which includes a triangle and its interior.
While fig.2 represents only a triangle.
The union of a triangle and its interior is called triangular region. It has definite area whereas triangle has no area.
Similarly rectangular, circular and polygonal region have area.
In this chapter we refer triangular region as area of triangle, circular area as area of circle etc.
Std-9th
9.Mensuration
Sub -Geometry

5. Std-9th Sub -Geometry 9.Mensuration AREA OF RECTANGULAR REGION.
UNITS OF AREA 1
Consider a square of side 1 cm . 1
Its area = 1 cm × 1 cm
4 cm
area = 1 sq. cm.
Thus , we get unit of area sq.cm
Area of square having side 1 cm is 1 sq.cm. area. B
6 cm C
Now we will find out the area of rectangle ABCD whose
length is 6 cm and breadth is 4 cm.
Rectangle ABCD can be divided into 24 squares with side 1 cm
Therefore area of rectangle ABCD is 24 sq.cm.
Now, length ×breadth=6 cm ×4cm
= 24.sq.cm.
= area of rectangle.
Here we get formula, area of rectangle = length× breadth
Like sq.cm. other units of area are sq.m. and sq.km.
Std-9th
Sub -Geometry
9.Mensuration

6. Revision – Some Formulae.
Sr.No.
Figure
Perimeter
Area a c
P = a+b+c h 1 b
Triangle a a
P = 3a 2 a
Equilateral
Triangle
Std-9th
Sub -Geometry
9.Mensuration

7. Revision – Some Formulae.
Sr.No.
Figure
Perimeter
Area a a a 3
P = 4a a
Square b 4 l
rectangle
Sub -Geometry
9.Mensuration
Std-9th

8. Revision – Some Formulae.
Sr.No.
Figure
Perimeter
Area a h 5
P =2(a+b) b
parallelogram a
P=a+b+c+d a d h 6 b
Trapezium
Sub -Geometry
9.Mensuration
Std-9th

9. Revision – Some Formulae.
Sr.No.
Figure
Perimeter
Area a 7 a
P =4a a a
Rhombus 8 r O
circle r O r 9
semicircle
Std-9th
Sub -Geometry
9.Mensuration

10. 9.1 Area of triangle Std-9th Sub -Geometry 9.Mensuration
Students must do activity given in the book and find the formula for area of triangle.
Observe the fig alongside
and write the name of
triangles. Find the area of
each triangle in terms of
base b and height h. what
do you find? H F E D A l h m B a C
When triangles lies between two parallel lines with common base or congruent bases, their areas are equal.
Find the parallelograms in the fig.Whether they have same area?
Find the relation between area of parallelogram and area of the
triangle with same or congruent base and lies between two parallel lines.
Area of triangle = ½ area of parallelogram.
(When their bases and heights are same.)
Std-9th
Sub -Geometry
9.Mensuration

11. Solved examples Std-9th Sub -Geometry 9.Mensuration
Ex-1 Area of right angled triangle is 240 sq.m. and its base is one
and half of its altitude. Find the base and altitude.
Solution-
Std-9th
Sub -Geometry
9.Mensuration

12. Ex-2) The polygonal field is as shown in figure. All
Ex-2) The polygonal field is as shown in figure. All measurements are given in meters. Find the area of the polygonal field. A B C D E F P Q R S 40 60 50 30 10 5 6 4 1 3 2
Sub -Geometry
9.Mensuration
Std-9th

13. Area of polygonal field ABCDEF
1) A(∆APF) = 1/2× × Formula
= 1/2× × A B C D E F P Q R S 40 60 50 30 10 =
2)A(□ABSP) = ×
= × … . =
3) A(∆ BSC) =1/2× . × =
4) A(∆ DRC) = 1/2× . . .×
5)A(□EQRD)= . .× ( )× (Trapezium) =
= ×( )× .
6) A(∆ FQE) =1/2× × = =
AREA OF POLOGONAL FIELD = =
Std-9th
Sub -Geometry
9.Mensuration

14. S P M Q R 9.2 AREA AND PERIMETER OF QUADRILATERAL Std-9th
EXAMPLES . EX-1
In rhombus PQRS, the perimeter is 20 cm and diagonal QS has length 8 cm. Find the length of the second diagonal.
Solution of example
Perimeter of rhombus = 4 x side
20 = 4 x side
side = 5 cm
Diagonals PR and QS bisect each other at right angles in point M. QS=8 cm ….. given
25 = PM
Area of rhombus
= ½ x product of diagonal
PM2 =
PM2 = 9
=1/2 x QS x PR
PM = 3 cm
=1/2 x 8 x 6
In right angled PMQ
= 24 cm2
PQ2= PM2 +QM2 (by Pythagoras thm)
diagonal PR =2x3
= 6 cm
Area of rhombus is 24 cm2 .
52 = PM
Std-9th
Sub -Geometry
9.Mensuration

15. Std-9th Sub -Geometry 9.Mensuration A B F G D C E H
EX-2) A rectangular lawn 75 m by 69 m wide , has two roads each 4 m wide running through the middle of the lawn, one parallel to the length and other parallel to breadth as shown in figure. Find the cost of gravelling the roads at the rate of Rs. 450 per sq.m.
Solution- A B F G D C E H
1)Area of rectangular Road ABCD= length x breadth
= 75mx 4m
= 300 sq.m.
2)Area of rectangular Road EFGH= length x breadth
= 60 m x 4 m
= 240 sq. m
3)Area of the common part of road i.e. square
∴Area of the road to be gravelled = 300 sq.m+ 240sq.m sq.m.
= 524 sq. m.
Cost of the road to be gravelled = rate x area.
= 4.50 x 524
Cost of the road to be gravelled = Rs.2358
Std-9th
Sub -Geometry
9.Mensuration

16. Std-9th Sub -Geometry 9.Mensuration Circumference –
9.3 AREA AND PERIMETER OF THE CIRCLE AND SEMICIRCLE
We know formulae ..
Circumference –
Consider a circular wire
of the copper as shown in the fig.
If we cut this circular wire at point A then
what will be the length of wire ? Observe
the activity carefully. ..
It will be equal to length of seg AB
which is equal to the circumference of circle and it is given by B A
Note- circumference of the circle is its perimeter too.
Std-9th
Sub -Geometry
9.Mensuration

17. Std-9th Sub -Geometry 9.Mensuration
Thus, circumference is the total length of the path of the circle which is given by the formula
But why is circumference is given by above formula. Why is the above formula?
For this ,see the proof carefully given in the book. In this proof, It is shown that the ratio of the circumference of the circle and its diameter is always constant. This constant is denoted by Greek letter pi ( )
The perimeter of a semicircle
= ½ x circumference + diameter
Std-9th
Sub -Geometry
9.Mensuration

18. 9.4 Area of a regular polygon
Polygon:: A simple closed figure formed by line segments is called polygon.
They are named according to the number of sides they have.
Triangle
Quadrilateral
Pentagon
Hexagon
Heptagon
Octagon
The line segments are called sides of the polygon .
Each common end point of two sides is called vertex of the polygon.
Regular polygon : A polygon in which all sides and all angles are congruent is called regular polygon.
Triangle
Quadrilateral
Pentagon
Hexagon
Heptagon
Octagon
Std-9th
Sub -Geometry
9.Mensuration

19. Circumscribed circle of polygon - E D
An important property of regular polygon is that we can draw a circle passing through its vertices called circumscribed-circle. o F C R
Incribed-circle r
A circle can drawn touching each side of polygon is called its incribed-circle . A P B
Central point of a polygon : The inscribed and circumscribed circles of a regular polygon have the same centre. This centre is called central point of a regular polygon.
In fig point o is the central point of regular polygon ABCDEG.
In-radius : The length of the perpendicular drawn from central point of a polygon to any of its side is called the radius of inscribed circle of the polygon it is denoted by r In fig OP = r
Circum-radius: The line segment joining central point of a polygon to any vertex is called the radius of the circumscribed circle. OA = R
Std-9th
Sub -Geometry
9.Mensuration

20. 9.5 Area of n-sided regular polygonB C P O a r R
In terms of in-radius
Let G,A,B,C be any four vertices of regular polygon of n-sides with central point o and side a
Let bisectors of ÐGAB and Ð ABC meet each other in point O.
Seg OP^ side AB
Let OP = r ( In-radius)
OG =OA =OB =OC = R (Circum-radius)
Now, A( ∆ AOB) = ½ x AB xOP
Area of regular polygon = n x A(DAOB)
= n x ½ x AB x OP
= n x ½ x a x r
Area of regular polygon(A)
= ½ x n a x r sq.units.
= ½ x perimeter x In -radius
∵Perimeter of n sided polygon = na
Perimeter = 2A/r
Std-9th
Sub -Geometry
9.Mensuration

21. G A B C P O Std-9th Sub -Geometry 9.Mensuration
9.6 Area of regular polygon in terms of circum-radius(R)
Here in-radius OP = r
Circum-radius OA = R G A B C P O a r R
DOAP is right angled triangle.
Std-9th
Sub -Geometry
9.Mensuration

22. 9.7 I )Area of Regular HexagonD B C O a a
Area of regular hexagon
= 6 x A(DOAB) a a a a a
DOAB is equilateral
Area of regular hexagon
sq.unit
Std-9th
Sub -Geometry
9.Mensuration

23. II) Area of regular octagon
ABCDEFDH is a regular octagon G D
Length of each side is ‘a ’ and its in-radius is ‘r
From fig. O a a
OP = r = OQ + PQ (PQ = MB why ?) M Q
r = OQ + MB why ? H C r
Where BC is diagonal of square MBNC a a P N A B
Area of regular polygon
Std-9th
Sub -Geometry
9.Mensuration

24. Std-9th Sub -Geometry 9.Mensuration
Ex.1) Find the area of a regular hexagon whose side is 5 cm. ( = )
Solution :
Side of regular hexagon =
=64.95
∴Area of regular hexagon is sq.cm.
Std-9th
Sub -Geometry
9.Mensuration

25. Std-9th Sub -Geometry 9.Mensuration
Ex.2) Find area of regular octagon each of its side measures 4 cm.
Solution :
Area of regular octagon
= 32 x 2.414 =
∴Area of regular octagon = sq.cm
Ex.3) Find area of regular pentagon whose each of measures
6 cm and the radius of incribed circle is 4 cm.
Solution :
Area of regular polygon = ½ x n x a x r
=1/2 x5 x6 x4
=60
Area of regular pentagon = 60 sq.cm.
Std-9th
Sub -Geometry
9.Mensuration

26. Std-9th Sub -Geometry 9.Mensuration
Ex.4) Find the area of regular polygon of 7 sides whose each side
measures 4 cm and the circum-radius is 3 cm( )
Here, a = 4 cm, R = 3 cm, n =7
Solution :
=14 x 2.236
=31.304
The area of regular polygon of 7 sides = sq.cm.
Std-9th
Sub -Geometry
9.Mensuration

27. PRESENTED BY – NEHRU VIDYALAYA HINGANGAON BK. MR. RASKAR .K.B
B.Sc.B.Ed.
NEHRU VIDYALAYA HINGANGAON BK.

28. THANK YOU
Std-9th
Sub -Geometry
9.Mensuration