Mathematics AA SL IB Diploma Program - Topic 3: Geometry and Trigonometry

topics 3 :




Geometry
and
trigonometry

, Wednesday ,
13
January


-
circles raeecsranrd sectors ,


⑨ For
a circle with radius r :




⑤
the circumference G- 2T r '

radius
⑤
the area A- = IT r2




⑨
An arc is a part of a circle which joins any 2
different points
beg
.




r
⑤
O subtended by
o
It be measured the the points at the center
can
using .
. O

⑤ O
Arc length
-
-


x stir
360




⑨
and the
Agysector
is the
region between 2 radii of a circle arc between them .




Perimeter of
3%
RF
a sector -
-
2rt X 21T r
r
⑨ O O
Area of a sector = X Thr
2
.


360




⑨
Worked example for : the given figure find to ,
3 significant figures :


0
⑤ the × 2K r
length of the arc : 360
48
×
360 2 IT (3) =
2.51cm



0
⑤ X 21T r
the perimeter of the sector : 2 r t 360
ago

213) t 34680 ×
217131=8.51 cm
3cm

O
⑨ Z
the area of the sector : 360 X Mr


48
2
360 X IT (3) =
3.77cm




Exercise 6A : SL H ease core 2019 .




1. Find the
length of :




A. the blue arc .
b. the red arc .




O
Arc length 36% x 21T r Arc length = 360 X 2Thr

60 142
360 X 21T (5) 360 X 21T (5)


=
5.24cm =
12.4cm

,2. Find the perimeter of :


o o o
A. f- 2Mt b. PoS=2rt360X2Tr c. Perimeter of a sector --2rt360X2Tr d. Perimeter of a sector --2rt360X2TLr
180 125 32
18.2
( 21T 213.41+360×21713.4) =2( 8) +360×21718 )
e-
2141+360×21714)
'
=
2 -




C- 21719.1 ) 217.5cm -33.5cm 10.2cm
'
'
-




G- 57.2mm


e. Perimeter
of a sector --2rt3%X2Tr
260=216.21+360×21716.2)
-
-
40.5M


3. Find the radius of the circle .
4. Find the perimeter of the sector .




O
0=360 0=1270 Perimeter 21Mt
length length
* *
Arc --
360 X2tr -_ arc

36
* Arc length --
26cm 26=360×2 Thr - Arc length -36mm =
2116.2411-36
26 O
O.li/2TL=r Arc
length --
360 X2TLr e-
68.5mm
127

41.38 er 36=360×2 Thr
36
127 =p
360×212
16.24mm er



5. Find the area of :


TLRZ O O
TLr2
A. A=TLr2 b. A= 2 C. A= 4 d. Area of a sector -_ 360 Xtra e. Area of a sector -_ 360 XTLRZ
2
Z TLC 8.212 TLC 3. 5) 67 215
A- IT 123 A= 2 A- =
4 =
360×1711012 =
360 XTC ( 85


A IT (6.512 9.6mi
'
58.5cm
' '
-_ A -_ 705.6mm A -
-
= =
120.1M

'
A' 132.7cm
-




o

f. Area of a sector -_ 360 XTLRZ
260
0=360 -
100 = 360×1719.212

0=2600 492cm
?




6. Find the radius : 7. Find the perimeter .




O O
*
0=670 Area of a sector 360 XTLRZ * 0=1360 Perimeter of a sector --2rt 360 XZTLR

67 136
*
16.2=360 XTLRZ *
? ?
Area -_
16.2cm A- 28.8cm =2rt 360 x2TLr
16.2 O 136

×t=rZ Area of sector 360 XTLRZ =2( 4.931+360×2544.93)
36670 a --




136
28.8=360 XTLRZ
'
16.2 '
-
21.5cm
67 =r 28.8
XTL
360 136 =p
2
XTC
360
'
5.262 r 28.8
136 =r
×t
360



4.93cm er

8. Running track

Total perimeter of track
.




- ygoom .
€ 500
① d

, b. Calculate Jason 's !
Determine the diameter of the semicircular ends
average speed in
'


a. . MS

distance
TP -1600M d=2r Speed
-
-
time Time --4m25sec
1600M
1 circle --
600M D= 95.5×2 f- 2655 -265 Sec
'




2TLr=6OO d- 791M f- 6.04ms "



600
✓ = 21T


re 95.5 Thursday ,14 January

thecieecle
⑤ Radian measure of angles :




⑨
The circle is divided into 360 sectors and the central each sector is 19
equal angle of
⑨
This unit of does not relate to the circle itself ; it
measure is
only dividing it into 360 equal parts .




⑨
There is another unit of measure for angles and it is derived from the relationship between the arc
length and the radius of the

circle This unit of
. measure is called the radian .




⑨
Radian when the : arc of a circle is the same length as the radius then the, angle subtended bythearcis 't radian .




⑤ The radian is the SI unit of
angle .




⑨
Itis the subtended circular the radius
measure of a central angle by a arc which has the same length as .




⑨ When the arc
length is equal to half the circumference of a circle 11809 the , corrispondihg angle ist radians .




⑨ the circle 13609 the radians
If arc
length is equal to the whole circumference of a , corrisponding angle is 21T .




⑤ Conversion between
degrees and radians :




⑤ 1800 IB notation :
IT radians =




⑨ 360°
*
Radians -_ rad
21T radians =




⑨
I radian = 790=57.30
*
Angle given in terms oft .




Example 1 : An
angle is given as 40 ? Example 2. An angle is given as
4Th radians .




¥ TL TL
radians ( 3 significant
=
*
Find its value in .


figures) Find its value in
degrees .
A
3180
°
X TL
7¥
4 IT
=

1rad
*
= 400 -1800 IT radians -- 1800 *
A = IT


401T 180
1800×34
3

Xrad =
400 *
X =
180 4Th radians -
*
A 1350

*
X = GET =135°
*
Xx 0.698rad

Exercise 1 Exercise 2 Exercise 3 Exercised

What is 1200in radians ? What equation represents the what is 0.15rad in degrees ? Convert 75 to radians .




R IT 180 180 R IT
= 0.15 TL
=
=

degrees ? 1rad 1rad
* *
1200 180 conversion of 1.2rad to -_ IT A 180 = IT 73 180

1201T 180 1.2 IT 0.15×180 731T
=
R= 180 1rad =
IT
*
A 180 0.15rad -_
A * A- =
IT Rrad=73° *
R= 180
1. 2×180

12=-23 'T
7.2rad -
A
*
A = IT * A- 28.590
*
12=1.270