B.Ed. course

1. Relation between distance between
the chord and the centre of the circle,
length of the chord and radius

2. d l/2
OB= radius = r
OC = distance between the centre and
chord= d
AB = length of the chord =l
BC = AB/2 =l/2
O
C B
l

3. OBC Right angled triangle
we can Apply Pythagoras
theorem
O
d
l/2
O
d
C
l/2
C B
If we do not know
By Pythagoras theorem
Where r is the hypotenuse of OBC
B
r
풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ
풉풚풑풐풕풆풏풐풔풆 = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ
OB= 푩푪ퟐ + 푶푪ퟐ
= dퟐ + l/2 ퟐ r

4. O
d
l/2
O
d
C
l/2
C B
If we do not know d
By Pythagoras theorem
Where d is the altitude of OBC
풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ
풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ
B
풂풍풕풊풕풖풅풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풃풂풔풆ퟐ
풂풍풕풊풕풖풅풆 = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풃풂풔풆ퟐ
푶푪 = 푶푩ퟐ − 푩푪ퟐ
= rퟐ − l/2 ퟐ d

5. O
d
l/2
O
d
C
l/2
C B
If we do not know
l/2
By Pythagoras theorem
Where l/2 is the base of OBC
풉풚풑풐풕풆풏풐풔풆ퟐ = 풂풍풕풊풕풖풅풆ퟐ + 풃풂풔풆ퟐ
B
풃풂풔풆ퟐ = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풂풍풕풊풕풖풅풆ퟐ
풃풂풔풆 = 풉풚풑풐풕풆풏풐풔풆ퟐ − 풂풍풕풊풕풖풅풆ퟐ
푩푪 = 푶푩ퟐ − 푶푪ퟐ
= rퟐ − dퟐ
l/2

6. rퟐ − l/2 ퟐ =
= rퟐ − dퟐ
O
d
C
l/2
B

7. REVIEW AND APPLICATION
8CM
P Q
A B
16CM
AB is the diameter
PQ is a chord
AB PQ
Area of ABPQ ?

8. 8CM
16CM
P
Trapezium
a
h
b
Area=h/2(a+b)
a=16cm
b=8cm
h = ?

9. P 8CM
Q
o
A B
16CM
To find ‘h’
P
4 cm T
O
Where TO is the
height of the
trapezium also
the distance
from the centre
to chord.
= rퟐ − l/2 ퟐ d
풅 = rퟐ − l/2 ퟐ
= 8ퟐ − (8/2) ퟐ
= ퟔퟒ − ퟏퟔퟐ
= ퟒퟖ
= ퟐ ퟑ cm

10. 8CM
16CM
P
Trapezium
a=8 cm
h=2 ퟑ
b = 16cm
Area=h/2(a+b)
= ퟐ ퟑ /2(8+16)
= ퟑ (ퟐퟒ)
=24 ퟑ cmퟐ a=16cm
b=8cm
h = ퟐ ퟑ

11. Thank you
Submitted by ,
priya