Transcript

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö

Öß, ´Ö¬µÖ ×¾Ö³ÖÖÖ, �ÃÖÖŸÖÖ¸üÖ

Std-9

Subject -Geometry

OUR INSPIRATION

Dr.Karmveer Bhaurao Patil

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�´ÖÖ¬µÖ´Ö

‡µÖ¢ÖÖ 9 ¾Öß

×¾ÖÂÖµÖ -

Ö×ÖŸÖ � �

9.Mensuration

‡Ó �ÖÏ•Öß ‡µ

Ö¢ÖÖ ×¾ÖÂÖµÖ

9¾ÖßÖ×ÖŸÖ � �

(³Öæ×´ÖŸÖß )‘Ö™

üú-� 9.Mensur

ation

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

Std-9th Sub -Geometr

y

STD.9TH

Sub- MATHEMATICS

9.MENSURATION

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú´ÖÔ¾Öß¸ü � � �×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß, ´Ö¬µÖ

×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�MENSURATION

9.1) AREA OF TRIANGLE9.2) AREA & PERIMETER OF A QUADRILAERAL9.3)AREA & PERIMETER OF THE CIRCLE & SEMICIRCLE

9.7) PARTICULAR CASES 1. REGULAR HEXAGON 2. REGULAR OCTAGON

9.6) TO FIND THE AREA OF THE REGULAR POLYGON IN TERMS OF RADIUS OF CIRCUMSCRIBED CIRCLE

9.5) AREA OF n-SIDED REGULAR POLYGON9.4) AREA OF REGULAR POLYGON

INTRODUCTION

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�MENSURATION - INTRODUCTION

In our daily life we come across many situations like-1)Fencing a circular garden with barbed wire .

We have to consider the total length of wire required ,number of turns of wire and supporting poles etc and then we calculate the expenditure.

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�MENSURATION - INTRODUCTION

2) Painting a walls of drawing room .

we must know the area , colour required for unit area and the rate of painting. Then we can calculate the expenditure.

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�MENSURATION - INTRODUCTION

3)For purchasing land or plot

Std-9th Sub -Geometr

y

9.Mensuration

it is essential to know the area and rate

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�MENSURATION - INTRODUCTION

4)While tiling the floor

we must know the area of floor and also the area of each tile.

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�MENSURATION - INTRODUCTION

5)To decide the number of bricks to built a wall ,

we must know the volume of wall and a brick. Std-9th Sub -

Geometry

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�MENSURATION - INTRODUCTION

In above examples we need to measure area , perimeter and volumes. Such a measurements come under the topics mensuration.

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ� CONCEPT OF AREA.

Triangular region Triangle

Observe the adjoining figure.

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.

Fig.1 Fig.2

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�AREA OF RECTANGULAR REGION.

UNITS OF AREA1

1Consider a square of side 1 cm .

Its area = 1 cm × 1 cm area = 1 sq. cm.

Like sq.cm. other units of area are sq.m. and sq.km.

Area of square having side 1 cm is 1 sq.cm. area. 6 cm

4 c

m

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

Here we get formula, area of rectangle = length× breadth

Now, length ×breadth=6 cm ×4cm = 24.sq.cm. = area of rectangle.

Thus , we get unit of area sq.cm

A

B C

D

Therefore area of rectangle ABCD is 24 sq.cm.

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

hca

bTriangle

Perimeter Area

Revision – Some Formulae.Sr.No.

P = a+b+c

c)b)(sa)(ss(sA

hb2

1heightbase

2

1A

Figure

2

cbas where

a

a

a

EquilateralTriangle

P = 3a a243A

1

2

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

hca

bTriangle

Perimeter Area

Revision – Some Formulae.Sr.No.

P = a+b+c

c)b)(sa)(ss(sA

hb2

1heightbase

2

1A

Figure

2

cbas where

a

a

a

EquilateralTriangle

P = 3a a243A

1

2

Std-9th Sub -Geometr

y

9.Mensuration

v

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�Perimeter Area

Revision – Some Formulae.

Sr.No.

P = 4aα2Α

Figure

3

4

a

aa

a

Square2

(diagonal)2A

b

l

b)(lP 2 breadthlengthA

blA rectangle

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�Perimeter Area

Revision – Some Formulae.

Sr.No.

P =2(a+b)

Figure

5

6

parallelogram

heightsides) parallel

of lengths of (sum2

1A

heightbaseA

hbA

a

b

b

a

daP=a+b+c+d

hb)(a2

1A

h

h

Trapezium

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�a

dC

Perimeter Area

Revision – Some Formulae.

Sr.No.

P =4a

Figure

7

8

Rhombus

rC 2

diagonals) of

lengths of(product 2

1A

)dd (A 212

1

aa

Or

rr9 semicircle

circle

rπ rP 2dπ rP

rπ A 2

rπ A 22

1

a

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�9.1 Area of triangle

Area of triangles lies between two parallel lines with common base

F E D A H

B C

l

m

When triangles lies between two parallel lines with common base or congruent bases, their areas are equal.

h

a

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?

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 -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�Solved examples

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- xxx2

3

2

11 triangle of base be triangle of altitude theLet

height base ½ triangle of Area

xx 2

3

2

1 240

x2

4

3 240

320x2

58x

m. 58triangleAltitudeof

m. 58triangle of Base 2

3

m. 5triangle of Base 12

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�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

DE

FP

Q RS

4060

50 50

30 50 10 30601 2 3

456

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�Area of polygonal field ABCDEF

1) A(∆APF) = 1/2× × . . . . Formula = 1/2× . . . .×. . . . . . . . .

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 = . . + . . . +. . . . . .+ . . . . . .+ +

= . . . . . .

=. . . ..×( . .+ )× .

A B

C

DE

FP

Q R

S40

60

50 50

30 50 10 3060

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

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.

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.- 16 sq.m.= 524 sq. m.

16sq.m.4 2side2 square of Area

Cost of the road to be gravelled = rate x area.= 4.50 x 524

Cost of the road to be gravelled = Rs.2358

Solution-

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�9.3 AREA AND PERIMETER OF THE CIRCLE AND SEMICIRCLE

We know formulaer2 circle of Area

r2 π2

1semicircle of Area

.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.

A B

It will be equal to length of seg ABwhich is equal to the circumference of circle and it is given by

r π2circle of nceCircumfere d πcircle of nceCircumfere

Note- circumference of the circle is its perimeter too.

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ� Thus, circumference is the total length of the path of the circle which is given by the formula

rC 2

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 ( )

d

CThus,

d C r 2 C OR

The perimeter of a semicircle

d r 22

1

d r

= ½ x circumference + diameter

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

The line segments are called sides of the polygon .

9.4 Area of a regular polygonPolygon:: A simple closed figure formed by line segments is called polygon.

Regular polygon : A polygon in which all sides and all angles are congruent is called regular polygon.

Triangle

Triangle Quadrilateral

Quadrilateral Pentagon

Pentagon

Hexagon

Hexagon

Heptagon

Heptagon

Octagon

Octagon

They are named according to the number of sides they have.

Each common end point of two sides is called vertex of the polygon.

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

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.

An important property of regular polygon is that we can draw a circle passing through its vertices called circumscribed-circle.

A circle can drawn touching each side of polygon is called its incribed-circle .A B

C

DE

F Ro

Pr

Circumscribed circle of polygon -

Incribed-circle

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 -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�9.5 Area of n-sided regular polygon

G

A B

C

P

O

a a

a

r

RR

R RLet OP = r ( In-radius)OG =OA =OB =OC = R . . . . .(Circum-radius)

In terms of in-radiusLet G,A,B,C be any four vertices of regular polygon of n-sides with central point o and side aLet bisectors of ÐGAB and Ð ABC meet each other in point O.Seg OP^ side AB

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

(∵Perimeter of n sided polygon = na)Area of regular polygon(A) = ½ x n a x r sq.units.

= ½ x perimeter x In -radius

Perimeter = 2Area/r

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�9.6 Area of regular polygon in terms of circum-radius(R)

G

A B

C

P

O

a a

a

r

RR

R R

Circum-radius OA = R

22 APOAOP 2

2

2

a

Rr

narpolygonregularofArea 2

1

Here in-radius OP = rDOAP is right angled triangle.

2

2

22

1

a

RnapolygonregularofArea

22

42

1 aRna

4) (

2

1polygon of Area

22 side

radiuscircumperimeter

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

DOAB is equilateral

2

3ar

OPABOABA 2

1

ra2

1

Area of regular hexagon

= 6 x A(DOAB)ra2

16

ar3

aa 2

33

2

2

33a

Area of regular hexagon

2

2

33a

.

rA

F

E D

B

CO

aaaa

9.7 I )Area of Regular Hexagon

a

aa

a

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�ABCDEFDH is a regular octagon

Length of each side is ‘a ’ and its in-radius is ‘r From fig.

OP = r = OQ + PQ (PQ = MB . . . why ?)

r = OQ + MB

Where BC is diagonal of square MBNC

22

1 BCLMr

?..........2

whyBC

MB

II) Area of regular octagon

22

1 aar

22

2aa

r

2)21( a

r

Area of regular polygon )(8 OABA

OPAB21

8

ra2

18

2

)21(4

aa

unitssqa .)21(2 2

aP

a

a

a

a

a

a

A B

C

D

EF

G

H

L

MQ

O

N

r

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�Ex.1) Find the area of a regular hexagon whose side is 5 cm.

Side of regular hexagon = side)2

(2

33

5)2

(2

33

2

25732.13

=64.95

∴Area of regular hexagon is 64.95 sq.cm.

732.13 Solution :

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�Ex.2) Find area of regular octagon each of its side measures 4 cm. )414.12( Solution :

Area of regular octagon a2

)21(2

42

)414.11(2

= 32 x 2.414= 77.248

Ex.3) Find area of regular pentagon whose each of measures 6 cm and the radius of incribed circle is 4 cm.

Area of regular polygon = ½ x n x a x r Solution :

=1/2 x5 x6 x4=60

Area of regular pentagon = 60 sq.cm.

∴Area of regular octagon =77.248 sq.cm

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

2

2

42

1 aRnapolygonregularofArea

Ex.4) Find the area of regular polygon of 7 sides whose each side measures 4 cm and the circum-radius is 3 cm( )236.25

Solution :

22

4

4347

2

1

4927

514

=14 x 2.236

=31.304

The area of regular polygon of 7 sides =31.304 sq.cm.

Here, a = 4 cm, R = 3 cm, n =7

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

Do the activity to verify the formula-

Sum of angles of a polygon =180 x (n-2)

Std-9th Sub -Geometr

y

9.Mensuration

Q-2

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö

Öß, ´Ö¬µÖ ×¾Ö³ÖÖÖ, �ÃÖÖŸÖÖ¸üÖ

Std-9

Subject -Geometry

❀ TEST ❀ 20 marks

Q-1.Attempt any two 2 1) Find area of triangle whose height is 3 cm and base is 4 cm? 2) If the side square is 8 cm then find then find the length of its diagonal 3) If radius of circle is 7 cm then find ts area.Q-2 .Attempt any two 4 1) If the length of rectangular plot is twice of its breadth and its perimeter is 420 cm then find length and breadth 2) Find the perimeter of semicircle whose radius of is 14 cm

3) If the area of equilateral triangle is sq.cm then find its side.

336

9.Mensuration

Q-2

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö

Öß, ´Ö¬µÖ ×¾Ö³ÖÖÖ, �ÃÖÖŸÖÖ¸üÖ

Std-9

Subject -Geometry

Q-3.Attempt any two 6 1)Redii of two circles are 3m and 4m respectively .Find the radius of circle having area equal to the sum of area these two circles. 2) A regular hexagon is incribed in a circle of radius 6 cm. Find the area of shaded portion.

3) The length of rectangular hall is 5 m more than its breadth. The area of hall is 750 m2. Find the perimeter of hall.

A

B

C D

E

F

)14.3,73.13(

9.Mensuration

Q-2

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö

Öß, ´Ö¬µÖ ×¾Ö³ÖÖÖ, �ÃÖÖŸÖÖ¸üÖ

Std-9

Subject -Geometry

Q-4 . Attempt any two 8 1) Find the length of side of an equilateral triangle △ABC incribed in a circle of radius 4 cm. 2) Two parallel chords AB and CD are drawn on the same side of centre O of a circle. Radius of a circle is 65 m , length of chord are 112 m and 126 m. Find the area of quadrilateral ABCD.

3) The length of sides of a triangle are in the ratio 3:4:5 and its perimeter is 144 cm. Find the area of triangle.

)73.13(

A

B C

A B

CD

O

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

THANK YOUStd-9th Sub -

Geometry

9.Mensuration

prepared by-

MAHARAJA SAYAJIRAO VIDYALAYA,SATARA

MR. SADAWARTE M.D.B.Sc.B.Ed

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�

THANK YOU

Std-9th Sub -Geometr

y

9.Mensuration

¸üµÖŸÖ ×¿ÖÖÖ ÃÖÓÃ£ÖÖ, ú� � �´ÖÔ¾Öß¸ü ×¾ÖªÖ¯ÖÏ²ÖÖê×¬Ö Öß,

´Ö¬µÖ ×¾Ö³ÖÖÖ, ÃÖÖŸÖÖ¸üÖ�´ÖÖ¬µÖ´Ö

‡µÖ¢ÖÖ 9 ¾Öß

×¾ÖÂÖµÖ -

Ö×ÖŸÖ � �

‡Ó �ÖÏ•Öß ‡µ

Ö¢ÖÖ ×¾ÖÂÖµÖ

9¾ÖßÖ×� �

ÖŸÖ

9.Mensuration

Geometry

Top Related